Masayoshi NagataLocal Rings (Tracts in Pure & Applied Mathematics)John Wiley & Sons Inc (1962).pdf
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LOCAL RINGS MASAYOSHI NAGATA Kyoto University, Kyoto, Japan
INTERSCIENCE PUBLISHERS a division of John Wiley & Sons, New York· London
IN'n:II~C
:mNc:E 'l'UAc :'I'~ IN III lin: ANn (\IIIILII from Minto N is said to be surjective if cf> ( M ) = N. Adapting the definition of zero divisors and nilpotent elements, we define: An element a of a ring R is called a zero divisor with respeet to a given Rmodule 111 if there is a nonzero element m of M such that am = 0; a is said to be nilpotent with mopect to M if a'M = 0 for some natural number r; and a is said to be an annihilator of ill[ if aM = o. The intersection of submodules is again a submodule. We say that the intersection NI n ... n N, is irredundant if nir'j Ni 7'" ni Ni for any j = 1, 2 , ... , r. In the following, let R be a ring, M an Rmodule and let N, N).. be submodules of M. The sum of Embmodules N).. is the snbmodule generated by the N, , and is denoted by L N)... The sum of NI , ... , N, may be denoted by NI N,. Note that NI N, is settheoroti(:ally nr ni ENd· (nl A subset {u,} of N which generates N is called a basis for N. A
+ ... + + ... +
+ ... +
I
llJ
·)
GENERAL COMMUTATIVE RINGS
basis eonsisting of a finite number of elements is called a finite basis. X i;;: called a finite module if N has a finite basis. A basis for N is said to be a minimal basis for N if any proper subset of the basis is not a ba;;:is for N. Note that the submodule of a module generated by the empty set consists of zero. If a is an ideal of R, we define the product aN to be the sub module of S generated by {an I a E a, n E N}, this last set itself may not be a module. Xote that if x is an element of R, then xN = {xn In E N} is an Rmodule. 'lYe call the following wellknown theorem the isomorphism theorem: 1·_Y1 + N 2 )/N1 is naturally isomorphic to N 2 /(N 1 N 2 ). Xext we observe some facts about ideals. In the following, German letters denote ideals of R, and A, B, HI>, C, D, E, denote subsets of R. [a:B]c denotes the set of elements c of C such that cB C a; this set is obyiouslyan ideal if C is an ideal. [a:B]R is denoted simply by a:B. The following equalities can be verified easily:
n
. 1.1 )
nC
[a:B]c
=
(a:B)
[[a:B]b:CJE
=
En [a:BClb:E
[(na}.) :BJD = n[(a,:B)]D
[a: (L B,R)]D
=
n[a:B']D
·We generalize tht above to the case of modules, using the following priueiple, which we call the principle of idealization: When an Rmodule lVf is given, let R* be the direct sum R EEl M as modules. In R*, we introduce multiplication defined by (r + m)(r' + m') = rr' + rm' + r'm, and R* becomes a ring containing Rand M, in which 111 is an ideal and M2 = O. Furthermore, submodules of Mare nothing but ideals of R* contained in 111; the structure of 1J.1. as an Rmodule is substantially the same as that of 111 as an R*module beea.use R* / 111 = Rand 1112 = O. Xow, for submodules N, N' of 111, we defined N :N' to be {x I x E R, xX' eN}; if A is a subset of R, then N:A, which may better be denoted by [N:A]M, is defined to be {m I m E 111, am eN}. Then, in R* = R EEl 111 as above, our N:N' is nothing but [N:N'JR and our X:A. is nothing but [N :A]M' Therefore (1.1) can be generalized to the ..ase of modules in the following forms:
3
CHAPTER I
(l.2)
(N:A):B
=
N:AB,
(N:A) :N'
(nN,.):A
=
n(N,,:A),
(nN,,):N
(L: A,,)
=
n(N:A,,),
N:
N:
=
(L: N,,) a+ 0 =
N:AN'
=
n(N,,:N) =
n(N:N,,)
a + e = R. Then We go back to ideals: Assume that R = R2 = (a + 0) (a + c) C a + oe C R, hence we have a + oe = R. On the other hand, since (a + 0) (a no) C ao Can 0 in general, we have, by our assumption that a + 0 = R, that ao = an o. Furthermore, the isomorphism theorem shows that R/ao = (a + 0) / (a n 0) = (a/(a no» + (o/(a no» = ((a + 0)/0) + ((a + o)/a) = R/o + R/a; this sum is a direct sum because, in R/ao, (a/(a no» n (o/(a n h» = O. Thus we have: (l.3) If a + 0 = R, a + e = R, then a + oe = R (hence a + 02 = R), ao = an 0, R/ao = R/a EEl R/o. From this, we deduce: 0.4) If, for given ideals al , . . . , an; it holds that (ni",j ai) + aj = R for every j = ], ... , n, then, for any power a: of ai, we have (a) (n i ",] a:) + a; = R for any j, (b) II a: = na', and (c) R/(na:) = R/a~ EEl ... EEl R/a~ . . " Proof: If n = 2, then a repeated application of a + 02 = R in (l.3) gives (a); and (b) and (c) follow from (a) and (l.3); the general case can be proved easily by induction on n. N ext we give an application of the isomorphism theorem: (l.5) Let M be a module over a ring R. Let N be a submodule of M and let m be an element of M. Then (N + mR)/N is isomorphic to R/(N:mR). Proof: (N + mR)/N = mR/(mR n N) by the isomorphism theorem. Let cf> be the homomorphism from R/(N:mR) onto mR/ (mR N) such that cf>(x modulo N:mR) = (mx modulo mR N). cf>(x modulo N:mR) = 0 implies that mx E mR N, whence mx E N and x E N:mR, and therefore (x modulo N:mR) = O. Thus cf> is an isomorphism, and the assertion is proved. Similarly we have: (l.6) With M, N, and R as above, if a E R, then (N + aM)/N is isomorphic to M/(N:aR). We add here some definitions and remarks concerning tensor products. Let R be a ring and let R' be a ring which is an Rmodule. Then for an Rmodule M, M ® R R', which is usually denoted simply
n
n
n
GENERAL COMMUTATIVE RINGS
by M ® R ' , is naturally an R'module, though it is still an Rmodule. ,Ye treat M ® R' as an R' module, unless the contrary is explicitly stated. Let cf> be a homomorphism from an Rmodule M into an Rmodule N. Then there is a uniquely determined homomorphism cf>* from M ® R' into N ® R' such that cf>*(m ® r') = cf>(m) ® r'. This cf>* is denoted by cf> ® R'. It is obvious that: (1. 7) If cf> is surjective, then so is cf> ® R'. We say that ® R ' , or more exactly ® R R ' , is exact if for any identity map cf> from a finite Rmodule iVI into a finite Rmodule N such that JI eN, the tensor product cf> ® R' is an isomorphism. (This definition can be adapted to the case where R' is just an Rmodule. ) In closing this section, we define the length of a module. If an Rmodule M has a composition series M = Mo:J M1:J ... :J lJ;In = 0, then the length n is independent of the choice of the composition series by the well known JordanHoelderSchreier theorem. The length n is called the length of the Rmodule M and is denoted by length R M or simply by length M. E:S:ERCISE: Let R be a ring. A sequence of Rmodules Ni accompanied by homomorphiBms d i , which is denoted by No ~ N, ~ Nt ~ ... ... ~ N,_! ~ N r , is called exact if the kernel of di! is the image of di for i = 2, .,. , r. Prove that, in such a case, if ®R' is' exact, then No ® R'~N! ®R'~ '" ~ N,®R'isexact.
"e
2. Prime ideals and primary ideals
observe first that the following conditions for an ideal \J of a ring R (lJ ~ R) are equivalent to each other: (1) \J is a prime ideal of R. ':2) If ab E \J(a, b E R), then either a E \J or b E \J. (:3) If a and b are ideals of R such that ab C \J, then either a C \J or b C \J. (4) If 1) C a and \J C b, then ab %\J. Though the ring R itself satisfies the above conditions, we exclude it from th6 set of prime ideals. Let S be a multiplicatively closed subset of a ring R. An ideal \J of R is called a maximal ideal with respect to S if \J does not meet S and if eyery ideal of R properly containing \J meets S. :2.1 i THEOREM: Let S be a multiplicatively closed subset of a ring R. If an ideal a of R does not meet S, then there is a maximal ideal \J with "t,[yd tl} S sllch that \J contains a. Such a \J is necessarily prime. P1 \ p') .
Proof: Let x be an element of R. x is nilpotent modulo q ~ xn E q for some n ~ cf>(xf E q' for some n p cf>(x) E ~'~ x E p. Thus we see that ~ is the radical of q. If ab E q, a Et ~,then cf>(a)cf> (b) E q', cf>(a) Et ~', hence cf>(b) E q', and therefore b E q. Thus we see that q is primary to p. (2.13) If q is a primary ideal of a ring R with prime divisor ~ and if a is an element of R which is not in q, then q: aR is primary to p. Proof: Since a EE q, we have q C (q: a) C ~, and p is the radical of (q:a). Now be E (q:a), bEEp imply ca E q because bca E q.
Hence cEq: a, proving our assertion.
3. Noetherian rings We say that a ring R is Noetherian if it satisfies the maximum condition for ideals, namely, any nonempty set of ideals of R has maximal members. We say that a module JJ1 over a ring R is Noetherian if it satisfies the maximum condition for (R )submodules. Observe that a ring R is a Noetherian Rmodule if and only if R is a Noetherian ring. (3.1) THEOREM: Let JJ1 be a module over a ring R. Then the following three conditions are equivalent to each other: (1) JJ1 is Noetherian. (2) If NI , ... , N n , ••• are submodules of JJ1 such that Ni C NiH for any i = 1, 2, ... , then there is an n such that N j = N n for any
j
2:. n.
(3) Every submodule of JJ1 is a finite module. Proof: Equivalence between (1) and (2): If there are N i as in (2), but no such n, then the set of the N j has no maximal member. This shows that (1) implies (2). Assume, conversely, that (2) is true and let F be any nonempty set of submodules of JJ1. Let NI be any arbitrary member of F. When Ni is defined, we define N i+1 to be such that: (i) NiH = Ni if Ni is maximal in F, (ii) NiH is a member of F such that Ni C NiH if Ni js not maximal. Then, by the validity of (2) we see that some N n must be maximal in F. Thus (2) implies (1). Equivalence between (2) and (3): If an Rsubmodule N of M is not finite, then we see easily that (2) is not true. Thus (2) implies (3).
8
GENERAL COMMUTATIVE RINGS
Conversely, assume that (3) is true and let Ni be as in (2). The union N of the Ni is a submodule of M, hence N is generated by a finite number of elements, say Xl, ... , x m • Then they are in some Ni , say N n , whence N n = N, which implies that N j = N n for any j 2:: n. Thus the proof is complete. (3.2) COROLLARY: If an Rmodule M is Noetherian, then any (R ) homomorphic image of M and any submodules of Mare Noetherian. (3.3) Let a be an ideal of a ring R and let b be an element of R. If both a + bR and a: bR have finite bases, then a has a finite basis. Proof: Let ai and cj(i = 1, ... , r; j = I, ... , s) be such that a + bR = L: aiR + bR and a:bR = L: cjR. We may assume that the ai are in a. Let a' be the ideal generated by the ai and the bCj . Then we have a' C a. Let a be an arbitrary element of a. Then, since a E a' + Rb, a == rb (modulo a') with an r E R. Since a E a, we have rb E a, hence r E a:bR = cjR. Therefore rb E a'. Thus a E a', and we have a = a' and the proof is complete.
>.:
(3.4) TH} r, let a be an element of anN N'. Since a E an, there is anf E Sn such that f( al , .. , ,as) = a. Sincef E S, f = L figi with gi E R[x]. Comparing the degrees, we may assume that gi is a homogeneous form of degree n  d i . Thus we have a = f(al, ... , as) = Lfi(al, ... ,as) X gi(al, ., . , as) E L andi(adiN N') c anT(aTN N'). Thus we see that anN N' C anT(aTN N'). Since the converse inclusion is obvious, we have completed the proof. (3.8) Let a be an ideal of a Noetherian ring R and let M be a finite anM. Then aN = N. Rmodule. Set N = Proof: By (3.7) (the lemma of ArtinRces), we have anM N = anT(arM N) for n > r. Hence we have N = anTN, which proves our assertion. (3.9) Let aij and tj(i, j = 1, ... , s) be elements of a ring R. If Lj aijt j = 0 for i = 1, ... , s, then, denoting by d the determinant I aij I , we have dtj = 0 for any j. Proof: Let dij be the cofactor of aij in the determinant I aij I . Then Li dijaij = d, Li dijaik = 0 (j ~ k). Therefore 8 )
n
n
n
n
n
n
nn
n
n
o=
Li dij(Lk aiktk) = dt j
•
(3.10) Let M be a finite module over a ring R such that aM = M for an ideal a of R. If no element a of R such that a  I E a is a zero divisor with respect to M, then M = O. Proof: Let tl , . . . , ts be elements of M such that M = Rti . Then the relation aM = M shows that there are elements aij E a such that ti = L: aijtj for every i = 1, ... , 8. Let d be the determinant I Oij  aij I (Oij being the Kronecker 0). Then d == 1 (mod a), hence d is not a zero divisor with respect to M. On the other hand, dti = 0 by (3.9), which implies that ti = 0 by our assumption. Thus M = O.
L
(3.11) THEOREM: Let M be a finite module over a Noetherian ring R and let a be an ideal of R. Then anM = 0 if and only if no element a of R such that a  I E a is a zero divisor with respect to M. (INTER
nn
SECTION THEOREM OF KRULL)
Proof: Assume at first that such an element a is not a zero divisor with respect to M. Set N = anM. Then aN = N by (3.8). Since
nn
11
CHAPTER I
111 is X oetherian, N has a finite basis. Therefore the relation aN = N shows that N = 0 by (3.10). Conversely, assume that there is a zero divisor a with respect to jJ{ .such that a  1 E a. Let m be a nonzero element of M such that am = O. Then we have (1  a)m = m, hence (1  a)nm = m for any natural number n. Therefore m E anM for any n, and anjJ{ :;t' O. (3.12) THEOREM: Let 111 be a finite module over a Noetherian ring R, let a be an ideal of R and let x be an element of R. Then there is an integer r such that anM:xR C [O:XR]M + anrM for n > r. Proof: x(anM:xR) = anM xM = anr(a™ xM) C xanTjJ{ by (3.7) (the lemma of ArtinRees). Therefore, if y E anM:xM, then there is an element y' E anTM such that xy = xy', whence y  y' E [0: XR]M , which proves the assertion.
nn
n
n
(3.13) COROLLARY: With the same notation as above, if N is a submodule of M, then there is an integer r such that (N + anJl;I): xR C [N::l~R]M + anrM for n > r. Proof: Applying (:3.12) to 111' = JJIIN, we prove the assertion. (3.14) COROLLARY: Let 111, N, a, and R be the same as above and let x be an element of M. Then there is an integer r such that (N + anM) :xR C (N:xR) + an r for n > r. Proof: The proof of (3.12) can be applied to the case where N = 0 if we replace xM with xR, whence we prove the assertion in the same way as in (3.13). (8.15) If an ideal a £s generated by a finite number of nilpotent elements, then a is nilpotent. Consequently, if a' is the rad£cal of an ideal a in aN oetherian ring, then a' is nilpotent modulo a. Proof: Let aj , ... , am be a basis for a such that each of the ai is nilpotent. Then there is a natural number n such that a7 = 0 for every i. If r is greater than m( n  1), then any monomial of degree r in the ai has a formal factor a7 for at least one i, which shows that such a monomial must be zero, and aT = O. The last assertion follows from (3.1) and what we have proved. We add here the following theorem and an application:
ni bl
(3.16)
bi
=
THEO REM: If a ring R has ideals Ih, . . . , bn such that 0, and such that each Rlb i is Noetherian, then R is Noetherian.
Proof: Using induction on n, we may assume that n = 2. The sum + b2 is a direct sum in that case. There are a finite number of ele
1:2
GENERAL COMMUTATIVE RINGS
L
ments bi of OJ such that OJ + 02 = biR + 112 . From the properties of direct sums, we see that the b; generate l1 J . Thns l1J has a finite basis; similarly 02 has a finite basis. Let now p be an arbitrary prime ideal of R. Then p contains one Oi . Since lJ i and P/Oi have finite bases, we see that P has a finite basis, which proves the assertion by virtue of (3.4) (the theorem of Cohen). (3.17) COROLLARY: If M is a Noetherian module over a ring R, then R/(O:M) is a Noetherian ring and M is a finite R/(O:M)module. Proof: The finiteness of M is obvious by (3.1). Let UJ, ... ,Un be a basis for M. Then RUi is Noetherian and is isomorphic to R/ (O:RUi). Hence R/(O:Rui) is Noetherian for each i. l\'ow, O:M = ni (O:RUi) by (1.2), and we have the proof by (3.16). We note that we have made use of the following in the above proof: (3.18) The structure of the Rmodule 111 is the same as the structure of the R/(O:M)module M. EXERCISES: 1. Prove (3.7), (3.8), and (3.11) without assuming that the ring R is Noetherian but assuming that M is a Noetherian module. 2. Let a and bI , . . . , bn be ideals of a Noetherian ring R. Prove that there exists a natural number 1" such that n i nUbi = nnr( n i nrb i ) for n > r. (Hint: Consider the direct sum of rn copies of R as an Rmodule.) 3. Let a and b be ideals of a Noetherian integral domain R. Prove that there exists a natural number r such that uu:b = anr(ur:b) for n > /".
4. Jacobson radicals The intersection m of all maximal ideals of a ring R is called the ,Jacobson radical of R. It is obvious, by virtue of (2.2), that if a i :: m, then a is a unit in R. ( 4:.1) THEOREM: Let m be the ,Jacobson radical of a ring R and let JI be a finite Rmodule. If N is a submodule of M such that M = mJI N, then M = N. (LEMMA OF KRULLAzUMAYA) Proof: Set M' = M/N. Then M' is a finite module and M' = m1l1'. Hence, by (3.10), we have 111' = 0, namely, 111 = N.
+
(4:.2) THEOREM: If m is the ,Jacobson radical of a Noetherian ring R, and 1f 111 is a finite Rmodule, then nn mnM = 0; in particular,
nn mn =
O.
This is an immediate consequence of (3.11). 4:.3
J
THEOREM:
Let a be an element of the ,Jacobson radical of a
13
CHAPTEH I
Noetherian ring R and assume that a and 0 are ideals of R such that b C a and a CaR 0. If a:aR = a, then a = b. Proof: a modulo 0 is in the Jacobson radical of Rio and therefore we may assume that b = O. Then a eRa, which implies that a = a(a:a) = aa. Therefore we have a = 0 by (3.10), which proves our assertion.
+
5. The definition oj local rings Local rings and semilocal rings, which are defined below, are naturally topological rings. But the topology will be introduced in Chapter II, §16. We give here only ringtheoretic definitions of them. A ring R is called a quasisemilocal ring if it has only a finite number of maximal ideals; it is called a quasilocal ring if it has only one maximal ideal. A Noetherian quasisemiIocal ring is called a semilocal ring; a Noetherian quasilocal ring is called a local ring. When we say that (R, lh, '" , prj is a quasisemiIocal ring (or semilocal ring, or quasilocal ring, or local ring), we mean that R is a quasisemiIocal ring (or semilocal ring, etc.) and that the maximal ideals of R are the Pi . It is sometimes convenient to have a name for quasilocal rings (R, m) such that m = O. Note that, by virtue of (4.2), local rings are included in the family of such quasilocal rings. Therefore we call such quasilocal rings local rings which may not be Noetherian. Similarly, a quasisemiIocall'ing R with the Jacobson radical m is called a semilocal ring which may not be Noetherian if mn = O. We give some applications of the lemma of KrullAzumaya (4.1): (5.1) Let (R, m) be a quasilocal ring and let M be a finite Rmodule. Then a subset {u}J of M is a basis for M if and only if the set of residue classes {u~l of {u}.} is a basis for MlmM over Rim. Consequently, the u" form a minimal basis for M if and only if the u~ form a linearly independent basis for MlmM over the field Rim. Proof: Setting N = 1: RUi, the assertion follows immediately from the lemma of KrullAzumaya. We note that the above proof shows that: (5.2) The first half of (5.1) is true for any ring R if m is the Jacobson radical of R. The following is an immediate corollary to (5.1) and shows a nice property of quasilocal rings:
nn n
nn
14
GE)[ERAL COMMUTATIVE RINGS
(5.3) If Mis a finite module over a quasilocal ring R, then any basis for NI contains a minimal basis for JJ;I as a subset; if UI , . . . , U m and 1'1 , .•• , Vn are minimal bases of M, then m = n and there is an invertible linear transformation T ouer R such that (Uj, ... , Un) T = (1\ , ... , Vn). Let R be a subring of a ring R'. We say that an ideal a' of R' lies Ol'CF an ideal a of R if a = a' R. We say that a ring R is dominated by another ring R' if: (I) R C R', (2) every ideal of R which is different from R generates an ideal in R' which is different from R', and (3) every maximal ideal of R' lies ove'.' a maximal ideal of R. In that case, we write R ::; R'; R < R' means that R ::; R' and R ~ R'. Therefore, a quasilocal ring (R', m') dominates a quasilocal ring (R, m) if and only if R C R' and m = m' R.
n
n
6. Rings oj quotients Let R be a ring and let S be a multiplicatively closed subset of R which does not contain zero. Let U be the set of nonzero divisors of R. Let n be the set of elements a of R such that there is an element s of S with as = O. Since S is mUltiplicatively closed, n becomes an ideal of R. Let cf> be the natural homomorphism from R onto R/n. Then we have the following lemma: (6.1) If a is an element of the multiplicatively closed set generated by e and S, then cf>(a) is not a zero divisor in R/n. Proof: Since U and S are multiplicatively closed, a = us with II E U, s E S. Assume that cf>(a)cf> (b) = 0 (b E R). Then ab = usb E n, hence there is an element s' of S such that usbs' = O. Since u is not a zero divisor, we have ss'b = O. Since ss' E S, we have bEn, hence cf>(b) = 0, which proves that cf>( a) is not a zero divisor in R/n. Now we define the notion of rings of quotients; we must treat three eases. In the set P = {( a, u) [ a E R, u E Ul, we introduce an equivalence relation such that Ca, u) is equivalent to (b, v) if and only if av = bu. We denote the equivalence class of (a, u) by a/u. The set Q of the equivalence classes becomes a ring under the operations such that the 8lill and the product of a/u and b/v are (av + bu)/uv and ab/uv, respectively. Q is called the total quotient ring of R. Elements a of R ran be identified with a/I of Q. Thus Q contains Rand Q is generated by R and the inverses l/u of the elements u of U.
15
CHAPTER I
Now we consider the general case: By (6.1), ef>(S) consists only of nonzerodivisors, hence the subring of the total quotient ring of ef>(R) generated by ef>(R) and the inverses of the elements of ef>(S) is well defined. This subring is called the ring of quotients of R with respect to S. Note that Rs = lef>(a)!cf>(s) I a E R, s E S} and thatef>(a)!cf>(s) = ef>(a) ·ef>(S)I. The third case occurs when S is the complement of a prime ideal ~. Though this is a special case of the above, we use a different notation; Rs is called the ring of quotients of R with respect to ~, and is denoted by R~. A ring R* is called a ring of quotients of R if there is a multiplicatively closed subset S of R such that (0 ~ Sand) R* = Rs. We use the following notation: (a) If a is a subset of R, then the ideal of Rs generated by ef>( a) is denoted byaRs. (b) If a' is an ideal of Rs , we denote by a' R the ideal ef>l( a') == ef>l(a' ef>(R)). The following is a characteristic property of Rs : (6.2) If there are a ring R' and a homomorphism (f from R into R' such that for any s E S, (f( s) has an inverse in R', then there is a homomorphism r from Rs into R' such that (f = ref>. Proof: Let a be the kernel of (f. Since all elements of S are mapped to units, we see that (f(n) = 0; i.e., n C a, which means that there is a homomorphism r' from ef>(R) into R' such that (f = r'ef>. We define a mapping r by the relation r(ef>(a)!cf>(s)) = r'ef>(a)[r'ef>(s)r 1 and we see easily that this r is the required homomorphism. As a consequence, we have: (6.3) Let a be an ideal of R which does not meet S and let (f be the natural homomorphism from R onto Ria. Then (f(R)u(s) = RslaRs.
n
n
n
(6.4) THEOREM: If a' is an ideal of R s , then (a' R)R8 = a'. Proof: We have (a' R)Rs = ef>(ef>l(a' ef>(R) )Rs = (a' ef>(R))R s ~ a'. If ef>( a) Ief>( s) E a', then ef>( a) is in a' ef>(R), and therefore ef>(a) !cf>(s) is in (a' R)Rs (because l!cf>(s) is in Rs), which proves that (a' R)Rs :::> a', and the assertion is proved.
n
(6.5)
n n
COROLLARY:
n
n
n
I; R is Noetherian, then Rs is Noetherian.
16
GENERAL COMMUTATIVE RINGS
'13,61 THEOREM: Assume that q is a primary ideal of R belonging to a primE ideal ~. Then: I'a) If p meets S, then ~Rs = qRs = Rs. Ib ,J If ~ does not meet S, then q contains n, ~Rs is a prime ideal, qRs is primary to ~Rs , ~Rs n R = ~,and qRs n R = q. Proof: If ~ meets S, then q meets S because any power of an elemen t of S is in S. Therefore we have (a). Assume tha t ~ does not meet S. If a is in n, then there is an s in S such that as = 0, hence as :: q. Since s EE ~, it follows that a E q, hence n C q. Let b be an elementofqRsnR. Then Cb) = (q)!(R') is integral over cf>(R) and (2)
30
GENERAL COMMUTATIVE RINGS
for any mUltiplicatively closed subset S of R which does not contain zero, R~ is integral over Rs . Now we have the following important result. THEOREM: Assume that a ring R' is integral over its subring be a prime ideal of R and let S be the complement of P in R. Then a prime ideal p' of R' lies over p if and only if p' is maximal with respect to S. Proof: Assume at first that a prime ideal p' of R' lies over p. Then p' does not meet S, therefore p'R~ is different from R~. Hence pRs C p'R~ Rs c Rs ,which proves that p'R~ Rs = pRs . Hence R~/p'R~ is integral over the field Rs/pR s , which shows that R~/p'Rs is a field, hence p'Rs is maximal, which means that p' is maximal with respect to S. Conversely, assume that p' is maximal with respect to S. Then p'R~ is a maximal ideal and the field R~/p'R.g is integral over RS/(plR~ Rs). (10.5) implies that this last integral domain is a field, which means that p'R~ Rs = pR s , which implies that p' R = p.
(10.7)
R. Let
~
n
n
n
n
n
(10.8) COROLLARY: With the same R, R ' , and p as above, there are prime ideals of R' which lie over p; there is no inclusion relation between any of these prime ideals of R'. (LYINGOVER THEOREM) (10.9) COROLLARY: With the same R, R' as above, if Po c '" c Pc is an ascending chain of prime ideals in R and if a prime ideal p~ of R' lying over Po is given, then there is an ascending chain of prime ideals lchich begins with p~ such thal p~ R = Pi for each i. If, in this case, there is no prime ideal between Pi and Pi+! , then there is no prime ideal between p; and Pi+l . (GOINGUP THEOREM). Proof: The existence is easy by induction on n, while the last assertion follows from (10.8) (the lyingover theorem).
n
p;
(10.10) COROLLARY: If a ring R' is integral over a ring R, then altitude R' = altitude R. This follows from (10.9). _~nother immediate consequence of (10.8) i,,:
'10.11) i.~
COROLLARY:
If a ring R' is integral over a ring R and if a
an ideal of R different from R, then aR' j1 1.
Let R be an integral domain. A ring R' containing R is called an ,"'d'gral extension of R if R' is an integral domain and if R' is integral ,~,l,er R. If the field of quotients of R' is finite over that of R, we say :kt R' is almost finite over R.
31
CHAPTER I
An integrally closed integral domain is called a normal ring. The integral closure of an integral domain R in its field of quotients is called the derived normal ring of R. When R is a normal ring, an integral extension R' of R is called a Galois extension of R if R' is the integral closure of R in a Galois extension (not necessarily separable) of the field of quotients of R, in the sense of Galois theory: the Galois group of the field extension is called the Galois group of the integral extension. ( 10 .12) THEOREM: Let R be a normal ring and let R' be a Galois extension of R. Then for any prime ideal p of R, the prime ideals of R' which lie over p are conjugate to each other; that is, if prime ideals p~ and p~ lie over p, then there exists an automorphism 0" of R' over R such that p~u = p~. Proof: We consider at first the case where R' is almost finite over
R. Assuming the contrary, let p~, ... , p~ be all of the prime ideals of R' which are conjugate to p~ . Since there is no inclusion relation among the p; by the lyingover theorem «(10.8)), there is an element a of p~ which is not in any of the p~, ... , p~. Then no conjugate of a is in any of the p~, ... , p~ , whence the norm a* of a with respect to R is not in p = p~ R. This is a contradiction because a E p~ implies that a* E p~ R = p. Thus we have proved this case. Let us turn to the general case ..Consider the set F of pairs (S, 0") of Galois extensions S of R contained in R' and an automorphism 0" of S over R such that (p~ Sr = p~ s, for all possible Sand 0". Then we introduce an order in F as follows: (S, 0") ::; (S', 0"') if and only if S C S' and the restriction of 0"' to S coincides with 0". With this order, we see that F is an inductive set, hence there is a maximal member, say (S*, 0"*) of F. It is sufficient to show that S* = R'. Let S" be a Galois extension of R such that S* C S" C R' and such that S" is almost finite over S*. There is an automorphism of S" over R such that its restriction to S* is 0"*; we denote such an automorphism by the same letter 0"*. Then (p~ S"r* and (p~ S") lie S*), hence there is an automorphism over the same prime ideal (p~ 0"" of S" over S* such that (p~ S" = p~ S". By the maximality of S*, we have S" = S*, which implies that S* = R', and the proof is complete.
n
n
n
n
n
n
n r*u"
n
n
(10.13) THEOREM: Let R be a normal ring and let R' be a ring such that (1) R C R', (2) R' is integral over R and (3) no nonzero ele
32
GENERAL COMMUTATIVE RINGS
ment of R is a zero divisor in R'. If a prime ideal ~~ of R' and a descending chain PI :J P2 :J .. , :J Pr of prime ideals in R are given such that PI = P; n R, then there is a descending chain of prime ideals P; in R' which begins with P; and such that p~ R = Pi; here if pr = 0 alld zj a prime ideal P' of R' which lies over 0 and such that p' C P; is preassigned, then there is such a chain with p~ = p'. (GOINGDOW~
n
THEOREM)
Proof: Since
P;
contains zero,
p;
contains a minimal prime divisor
q' of zero by (2.5). Since nonzero elements of R are not zero divisors
in R', q' lies over zero by (7.1). Thus it is sufficient to prove the last assertion. R'/p' is an integral extension of R; if we see the existence of such a chain in R'/p', then, considering the inverse image of the chain, we prove the assertion. Thus we may assume that R' is an integral extension of R. Let R" be a Galois extension of R containing R' and let p~ be a prime ideal of R" which lies over P; . Let q~ C ... C a; be a chain of prime ideals in R" such that q~ R = Pi by the goingup theorem (( 10.9)). Since p~ and q~ lie over the same prime ideal PI of R, there is an automorphism 0" of R" over R such that q~" = p~ . Then, obviously the chain of (q~U R:) is the required one.
n
n
dO.14) THEOREM: With the same Rand R' as in (10.13), let a' an ideal of R' and set a = a' R. Then height a = height a'. Proof : We first consider the case where a' is a prime ideal. For a chain a' = p~ :J p; :J ... :J p; of prime ideals in R', we have a chain II = lJ~ R :J p; R :J '" :J p; R by the lyingover theorem 10.8)), whence height a' ::; height a; similarly the converse int"'iuality follows from (10.13), and height a' = height a in this ":i"e. ~ ow we consider the general case. Let p' be a prime divisor of J' ",ueh that height a' = height p'. Since P' R contains a, we have height a ::; height p' n R = height P' = height a'. Conversely, let P ["E' a prime divisor of a such that height a = height p. Let p' be a prillle ideal of R' such that a' C p' and such that p' n R = p; the exi",tenee follows from the lyingover theorem applied to R'I a' and R II with the prime ideal pia. Then we have height a' ::; height p' = height b' n R = height a. Thus~eight a' = height a.
n
i)r,
n
n
n
n
10.15 I THEOREM: Let R be a normal ring and let f(x) be a monic .:'i from R* into L such that 1>i(a) = ai. Kow let j b be an arbitrary element of R*. Then b = uja (Ui E K, n = degree of f( x ) ). Then 1>,( b) = U ja~ and we regard that these equalities as linear equations in the unknown Uj . The determinant D of the coefficients is IIii(b) and ai are integral over R, the dUi are integral over R. Since dUi are in K and since R is a normal ring, we see that dUi are in R, hence db is in R', whence dR* is contained in R'.
L
2:;1
(10.16) COROLL,\.RY: If R' is an almost finite separable integral extension of a Noetherian normal ring R, then R' is a finite Rmodule. Proof: Since R' is separable, there is an element a of R' such that R' and R[a] have same field of quotients. Let f be the irreducible monic: polynomial over R which has a as a root and let d be the discriminant of J(x). Because of the separability, d ;;t' O. (10.15) implies that dR' C R[a], whenee R' is a submodule of the finite Rmodnle L:;I (aijd)R (n = deg f(x)). Since R is Noetherian, it follows that R' is a finite Rmodnle. ( 10 .17) A ssume that f (x) is a monic polynomial over an integral domain R. Let the roots of f( x) be Ui (i = 1, ... , r) and let l' (x) be the derivative of f(x). Then the discriminant d of f(x) coincides with (_1)(1+2+'" +r1) IId'(ui). Proof: f(x) = (x  Ul) .,. (x  u r ). Therefore, setting gi(X) = f(x)j(x  Ui), we have1'(x) = Lgi(X), hencef'(ui) = 9i(Ui) for each i. Therefore 1I1'(ui) = IIgi(Ui) = (_I)(I+2+···+r1) IIi bR and that aR + bR = max (aR, bR). Thus we see that {aRI is linearly ordered and that a + b E max (aR, bR). Therefore we see that v* can be modified to be a valuation v (by the opposite order and changing multiplication to addition), and the valuation ring of v is R. v is unique up to equivalence. Thus we have (1l.8) There is a oneone correspondence between valuation rings R
37
CHAPTER I
of a given field K and equivalence classes of additive valuations v of K in such a way that R corresponds to the class of v if and only if R is the valuation ring of v. We prove next the following existence theorem of valuation rings: (11.9) Let R be a subring of a field K and let 0 C \h C \.12 C .. , C \ls be an ascending chain of prime ideals in R. Then there exists a valuation ring V of K· such that V has prime ideals nl, . .. , n. which lie over \l1, . . . , P8 respectively. Proof : We prove the assertion by induction on s. When s = 1: Considering Rp with \l = \ll, we may assume that \ll is the unique maximal ideal of R. Let F be the set of sub rings S of K such that \lIS ;;z:: S and such that R c S. F is an inductive set, and therefore there is a maximal member S* of F. Then S* is quasilocal, for, otherwise, if m is a maximal ideal of S* containing \lIS*, then S: E F, which contradicts to the maximality of S*. Let x be an element of K which is not in S*. x ~ S* implies that \lIS*[X] contains 1, namely, PIX there are elements po, ... ,pn of \lIS* such that 1 Po Pnxn = O. Since S* is quasilocal, 1 + Po is a unit in S*, which means that XI is integral over S*. Hence lhS*[XI] does not contain 1 by (10.11). The maximality of S* implies that X  I E S*. Thus S* is a valuation ring, and is obviously the required one, because we assumed that \ll is the unique maximal ideal of R. Now, we assume that such a V, say V' with prime ideals nl , ... , nS1 , exists for the chain \l1 C \l2 C ... C \l'I. Considering V~'_l , we may assume that n,_1 is maximal. R* = R/\lsI is a subring of the field V' /nH and has a prime ideal \ls/psI . Hence, by the case where s = 1, there is a valuation ring V* of V' /nsI such that V* has a prime ideal which lies over \l,/\l81 . Then, as is easily seen, the composite of V' with V* is the required valuation ring. (11.10) Let RI , . . , , Rn be valuation rings of the same field K. For a given element a of K, there exists a natural number s such that both a/(l a asI) and 1/(1 a asI) are in the intersection D of the rings Ri . Proof: Let \li be the maximal ideal of Ri . We consider an arbitrarily given i (::;; n). If a ~ R i , then by a valuation Vi with the valuation ring R i , 0 = Vi(l) > vi(a) ;:::: vi(a 1 ) = vi(1 a asI) for any s ;:::: 2, whence is;:::: 2, these elements are in the R, . Consider the case where a E Ri . Since 1 a asI = (1 a')/(l  a), when a modulo \li is a primitive (ei)th root of 1 with
+ +
+ + ... +
+ ... +
+ + ... + S

+ + ... +
+ + .,. +
38
GENERAL COMMUTATIVE RINGS
ei ~ 2, then for s which are prime to ei , these two elements are in the Hi . When a  1 E ~i, for s which are not multiple of the characteristic of R/~i , these two elements are in the Ri . In the other case, these two elements are in the Ri for any s. By the finiteness of the munber n of the Ri , there is surely an s which satisfies the above requirement for any i, and the assertion is proved.
(11.11) ThEOREM: Let RI , . . . , Rn be valuation rings of the same field K and assume that Ri R j for any (i ~ j). Let ~i be the maximal ideal of Ri , let D be the intersection of the Ri and set qi = ~i D. Then we have (1) an ideal m of D is maximal if and only if m = qi for some i and (2) Ri = D iq • (THEOREM OF INDEPEXDEXCE OF VALUATIOXS) Proof: We prove (2) at first. Let a be an arbitrary element of Ri . Let s be such that both 1/(1 a asI) and a/(l a as!) are in D, by virtue of (11.10). Since a E R i , we have 1 a asI E R i , whence it is a unit in R i . Therefore 1/(1 a asI) Et qi and therefore a is in D qi • Thus Ri ~ D qi • Since the converse inclusion is obvious, we have (2). (2) implies that q, qj for any (i ~ j). Therefore there is an element ei of nj,ei qj which is not in qi for each i. Let a be an ideal of D which is not contained in any of the qi . Then a contains an element ai which is not in qi , hence a contains aiei. aiei is not in any qi . It follows that aiei is a unit in any Ri , and therefore it is a unit in D, too. Therefore a = D. This implies that any maximal ideal m of D is one of the qi . Since qi Q; qj for any (i ~ j), we see the converse, and the proof is complete. 'iVe add here some remarks on normal rings.
%
n
+ + ... +
.. . +
+ + ...
+ + .. , +
+ + ... +
%
L
L
L
(11.12) THEOREM: An integral domain R is a normal ring if and ollly if R is an intersection of valuation rings of the field of quotients K of R. Proof: If R is the intersection of valuation rings VA , then, since each VA is normal, R is normal. Conversely, assume that R is normal and let b be an element of K which is not in R. Set R' = R[l/b]. If there is a relation (l/b)(L ai(l/b)i) = 1 (ai E R), then we 5ee that b is integral over R, which is not the case. Therefore l/b is not a unit in R', whence there is a valuation ring V of K such that 1b is a nonunit in V and such that R c R' c V by (11.9). Thus we 5ee that b is not in V, whence b is not in the intersection of valuation
CHAPTER I
39
rings of K which contain R. Therefore we see that R is the intersection of all valuation rings V of K such that R C V. (11.13) Assume that R is a normal ring and that b is an element of the field of quotients K of R. Then the kernel n of the Rhomomorphism cf>, from R[x] (x being a transcendental element over R) onto R[b] such that cf>(x) = b, is generated by polynomials ex  d such that b = die (c, d E R). i Proof: Assume H:at L:~ a;x E n (ai E R) and let V be an arbii trary valuation ring of K such that R C V. Since L aib = 0, we n 1 have anbnV = (an_1b  + ... + ao) V C bn1V, which implies that anbV C V, i.e., anb E V. Since V is arbitrary, we have anb E R by (11.12), whence (1;nX  dEn with d = a"b. Therefore we complete the proof by induction on n. EXERCISES: 1.
Willi the same notation as in (11.11), prove that if
0.1 , '"
... , an are ideals of R1 , . " , Rn respectively such that there is no inclusion
relations among the minimal prime divisors of them, then D / n (ai) is the direct sum of the Ri/ai . 2. Prove that a quasilocal integral domain with the field of quotients K is a valuation ring of K if and only if any ring R' such that R c R' k: K contains the inverse of some nonunit of R. 3. Assume that a ring R' dominates a valuation ring R of a field K. Prove that R' n K = R. 4. Let R be a valuation ring of a field K and let K' be an algebraic extension of K. Let R' be the integral closure of R in K'. Prove that valuation rings of K' which dominate R are just rings of quotients of R' with respect to maximal ideals, in the following way: (1) When K' is a finite Galois extension of K; let V be a valuation ring of K' dominating R and let D be the intersection of all the conjugates of V. Prove that D = R'. Then apply (11.11). (2) When K' is finite over K; apply the result in (1) to the smallest Galois extension of K containing K'. (3) Prove the general case, applying (2) to finite subextensions. 5. Let R be a valuation ring of a field K and let K' be an overfield of K. Prove that there is a valuation ring R' of K' such that R < R' and such that the residue class field of R' is algebraic over that of R, in the following way: Let {u~l be a transcendance base of K' over K and set K" = K({u~l). Prove the existence of such a valuation ring of K". Then apply 4 above.
12. Noetherian normal rings Let us begin with the following remark: (12.1) A local ring (R, m) which is not a field is a valuation ring if and only if height m ;:::: 1 and m is principal. In this case, height m = 1. Proof: The only if part is obvious. We prove the if part. height
40
GENERAL COMMUTATIVE RINGS
m = 1 by the altitude theorem of Krull. Let q be a minimal prime divisor of zero and let p be a basis for m. Since p ~ q, we have q:p = q. On the other hand, q cpR, hence q = 0 by (4.3). Thus R is an integral domain. Let a ~ 0 be an ideal of R, and let n be such that a C pnR, a Q; pn+lR. Then a = (a:pn)pn. If a:pn ~ R, then a:pn C pR and a C pn+lR which is not the case. Thus a = pnR. Therefore we see that every nonzero ideal of R is a power of pR, hence R is a ,~aluation ring. Next we have the following lemma: (12.2) Let a, b, c, d be elements of a ring R such that ad = bc. If a is not a zero divisor, then aR: cR C bR: dR. Consequently if both a and b are nonzero divisors, then we have aR:cR = bR:dR. Proof: x E aR: cR implies cx = ay with y E R, hence ayb = bcx = adx. Therefore by = dx, and x E bR: dR. (12.3) Let ~ be a prime ideal of a Noetherian ring R. Assume that Rp is not a valuation ring. Then for any nonzero divisor a E p and for any element b of aR: p, bI a is integral over R and the conductor of R in R[bla] contains ~. Proof: Let p be any arbitrary element of ~. Then bp = ar with an r E R. If r ~ p, then p C aR:bR C pR:rR C ~ (by virtue of (12.2)), which implies that ~ = pR:rR and we see that pRp is generated by p, hence Rp is a valuation ring by (12.1), which is a contradiction. Thus (bla)p C ~, and the assertion is proved by (10.4). ~~s an immediate consequence of (12.3), we have (12.4)
THEOREM:
e12.5)
THEOREM:
A normal local ring of altitude 1 is a valuation ring. Let p be a prime ideal of a Noetherian ring Rand assume that p contains an element a which is not a zero divisor. Then ~ is a prime diviso: of aR if and only if either height ~ = 1 and Rp is a 1'01 nation ring or there exist a, b E R such that bI a is integral over R and such that the conductor of R in R[bla] coincides with p. Proof: Assume at first that ~ is a prime divisor of aR and that Rp i" not a valuation ring. Let b be an element of R such that ~ = aR: bR Iby (8.8)). Then b E aR:p, hence, by (12.3), bla is integral over R and the conductor c of R in R[bla] contains p. If c E c, then cbla E R, i.e., cb EaR, hence c E aR: b = p. Thus the conductor c coincides ,yith p. Conversely assume at first that height p = 1. Then obviously :J is a minimal prime divisor of aR. Assume next that ~ is the conductor of R in R[bla]. Then pRp is the conductor of Rp in Rp[bla].
CHAPTER I
41
Since (bjd)'p C R, we have bp CaR, and 'p C aR:bR, and therefore 'pRp CaRp: bRp ~ R p , which implies that 'pRp = aRp: bRp ; hence 'pR p is a maximal prime divisor of aRp , and the assertion is proved. (12.6) THEOREM: If R is a Noetherian ring and if a prime ideal 'p is a prime divisor of aR with a nonzerodivisor a of R, then for any non.zerodivisor b contained in 'p, 'p is a prime divisor of bR. Proof: By (8.8), there is an element c of R such that aR: cR = 'po Then, since b E 'p, there is an element d E R such that bc = ad. It follows from (12.2) that aR:cR = bR:dR, hence 'p = bR:dR, which implies that 'p is a prime divisor of bR by virtue of (8.8). (12.7) COROLLARY: Let R be a Noetherian ring and let a, b be elements of R such that a is not a zero divisor, b/a ~ R and such that b/a is integral over R. Then either there is a minimal prime divisor p of aR such that Rp is not a normal ring or there exists an imbedded prime divisor of aR. Proof: Assume that the prime divisors 'pi of aR are all minimal and that Rp are normal for any 'p = 'pi. Let qi be the primary component of aR belonging to 'pi . Since b/a is integral over R, it is integral over R Pi whieh is a normal ring, hence b/a E R Pi , and b E aRpi R = qi. Thus bEn qi = aR, and b/a E R. Thus we have proved the assertioll. (12.8) If R is a normal ring, then any ring of quotients of R is also a normal ring. Proof: If a is integral over R s , then there is an element s E S such that as is integral over R, which proves the assertion.
n
(12.9) THEOREM: A Noetherian integral domain R is a normal ring if and only if the following two conditions are satisfied: (1) If 'p is a prime ideal of height 1 in R, then Rp is a normal ring and (2) if 'p is a prime divisor of a principal ideal ~ 0, then height p = l. Proof: Assume that R is normal. Then (1) holds good by (12.8), while the validity of (2) follows from (12.5). The converse is an immediate eonsequence of (12.7). A Koetherian normal ring of altitude 1 is called a Dedekind domain. (12.4) and (1.4) imply that: (12.10) Any ideal a ~ of a Dedekind domain R is the product of maximal ideals of R, and such an expression is uniquely determined
°
bya. EXERCISES:
1. Let \l be a prime ideal of aN oetherian ring R and assume that
42
GENERAL COMMUTATIVE RINGS
height p;::: 1. Prove that Rp is a valuation ring if and only if there is no primary ideal q such that p(2) C q C p. 2. Prove that a Noetherian ring R is integrally closed if and only if R is the direct sum of finite number of Noetherian rings Ri such that either Ri is a normal ring or every maximal ideal of Ri is a prime divisor of zero (namely, every nonunit of Ri is a zerO divisor). 3. Let R be a Noetherian normal ring. Prove that every principal ideal aR of R is an intersection of symbolic powers of prime divisors of aR. 4. Let R be a Noetherian integral domain. Assume that every ideal of R is a product of prime ideals. Prove that R is either a Dedekind domain or a field. 5. Let R be a Dedekind domain with field of quotients K. Prove that the set of nonzero finite Rsubmodules of K forms a multiplicative group by the natural multiplication. 6. Assume that M is a finite module over a Noetherian ring R, that a E R is not a zero divisor with respect to M and that p is an associated prime ideal of aM (in M). Prove that if b E p is not a zero divisor with respect to M, then p is an associated prime ideal of bM.
13. Unique factorization rings ~~n element a of a ring R is called irreducible (in R) if a is not a product of any two nonunits of R. We say that an integral domain R is a unique factorization ring if (1) every element a ~ 0 of R is the product of a finite number of irreducible elements and (2) if a = al . . . am = bl . . . bn with irreducible elements ai and b j , then Tn = n and there is a permutation 7r of the i such that a"(i)R = b,R. _~n element a ~ 0 of an integral domain R is called a prime element ii aR is a prime ideal of R. It is obvious that a prime element is an irreducible element, but not conversely. :\"ote that the condition (1) above is satisfied by any Noetherian integral domain.
(1:3.1) THEOREM: A Noetherian integral domain R is a unique factorization ring if and only if every prime ideal ~ of height 1 in R is principal. Proof: Assume that R is a unique factorization ring. For any prime ideal P of height 1, let a be an irreducible element of R contained in ~. Let b be an element of aR: p which is not in aR. Assume for a moment that aR ~ ~. Then let c E p such that c ~ aR. Then be EaR, and Thpreiore there is an element d E R such that be = ad. But a is not an irreducible factor of b or c, which is a contradiction. Thus ~ = aR and hE' '1I!1y if part is proved. Assume conversely that every prime ideal 01 height 1 is principal and let a1 ••• am = bl • . . bn be factoriza
CHAPTER I
43
tions of an element c as products of irreducible elements ai and bj • We prove the uniqueness by induction on n. \Vhen n = 1, c is irreducible and the assertion is obvious. By assumption, irreducible clements wh;ch are nonunits are prime elements. Since al ... am E blR, which is prime, some ai E blR; we may assume that a l E blR. Since aiR is prime, we have aIR = blR, and therefore there is a unit u such that a2 ... am = (u~) ba • . • bn , whence, by induction, we have the proof of the uniqueness. (13.2) If R is a unique factorization ring, then for any multiplicatively closed set S which does not contain 0, Rs is a unique factorization ring. Proof; A prime element of R is either a unit or a prime element of Rs . Furthermore, ideals of Rs are generated by ideals of R, hence the assertion follows from the definition. (13.3) A unique factorization ring R is a normal ring. Proof: Assume that alb (a, b E R) is integral over R: (alb)n + cI(alb)n1 + ... + Cn = (Ci E R). We may assume that a and b have no common prime factor. Assume that p is a prime factor of b. Then, since an + ClanIb + ... + cnb n = 0, we have an E pR and a E pR, which is a contradiction, and alb E R.
°
(1:3.4) COROLLARY; The polynomial ring in a finite n'tlmber of algebraically independent elements over a field is a normal ring. EXERCISES: 1 . Let R be aN oetherian normal ring and let S be a multiplicatively closed subset of R which does not contain O. Assume that (1) there is a natural number e such that q(') is principal, for any prime ideal q of height 1 in R such that q meets S and that (2) there is a natural number f such that p'(f) is principal for any prime ideal p' of height 1 in Rs . Prove that for any prime ideal p of height 1 in R, p(,j) is principal. 2. Assume that p, , ... , pr are prime elements of an integral domain R. Prove that an ideal a of R has prime divisors only among the piR if and only if a is a principal ideal generated by the product of some powers of Pi . 3. Prove that an integral domain R is a unique factorization ring if and only if (1) R satisfies the maximum condition for principal ideals and (2) every prime ideal of R different from zero contains a principal prime ideal different from zero. . 4. Give an example of a normal ring R, which is not a unique factorization ring, in which every prime ideal of height 1 is principal.
14. A normalization theorem When I is a subring of an integral domain R and when K and L are the fields of quotients of I and R respectively, then we know
GENERAL COMMUTATIVE RINGS
the transcendence degree of Lover K, which is defined to be t,he transcendence degree of R over I. (14.1) Let K be a field and let Xl , ... , Xn be algebraically independent elements over K. If YI is an element of K[x] = K[XI, ... , X"] Ichich is not in K, then there are elements Y2 , ... , Yn of K[x] such that (1) Yi = Xi x7 i for some natural numbers mi (i = 2, ... , n) and (2) K[x] is integral over K,[YI , ... , Yn] (and therefore YI , ... , Yn are algebraz'cally independent over K). Furthermore, (3) if a natural number s > 1 is given, all the mi can be chosen to be powers of s. Proof: We write YI as ai111i, where ai E K, ai ~ 0 and the kI; are monomials in the Xi . "\111 e define weights ml = 1, m2 , ... , mn of XI, X2 , ••• , Xn such that one Mi , say MI , has greater weight than
+
1:
the others; let t be a power of s such that t is greater than the degree i d of YI and set mi = t  l for each i. Then, since weight x~ < weight Xi+l for any i = 1, 2, ... , n  1, we see that the mi satisfies the requirement on the weights (considering a lexicographical order of the monomials Mi)' Set Yi = Xi x7 i for i = 2, .. , , n. Then YI w l ran be written ±alxr + flx + fw where w = weight MI and thefi are polynomials in 1)2, ... ,Yn with coefficients in K. Therefore XI is integral over K[YI , ... , Yn], whence the Xi , which are in K[.Ih , ... , y" , xd, arc integral over K[YI , ... , Yn]. Thus the Yi are the required elements.
+ + ...
(14.2) ThEOREM: Let K be a field and let XI , . . . , Xn be algebraically independent elements over K. If a is an ideal of height r in the polynomial ring K[x] = K[xI, ... , Xn}, then there are elements YI , ... , Yn of K[.r] such that (1) K[x] is integral over K[y] = K[YI, ... , Yn], (2) L1 K[y] is generated by YI , ... , Yr and (3) Yr+j = Xr+j fj with polynomials fj in XI , . . . , Xr with coefficients in the prime integral domain 7r of K for each j = 1, 2, ... , n  r. If K is of characteristic p ~ 0, then the fj can be chos6n such that fj E 7r [xl, ... , x;'J. (1\ OR
n
+
~L\..LIZATlON THEOREM FOR POLYNOMIAL RINGS)
Proof: We prove the assertion by induction on r. If r = 0, the asis obvious. Assume that I' 2:: 1 and let a' be an ideal of K[x] :ou('h that height a' = r  1 and such that a' c a. Then, by induction, chere are elements YI, ... , YrI, Y~ , ... , y~ of K[x] which satisfy the o::onditions in our assertion with a' instead of a. Since height a = r, we han' by (10.14) that height (a K[YI, ... , YrI , y;,J) = ~. (In the other hand, we have Yi E a' k a(1 ~ i ~ r  1) by con~:n(·tion. Therefore there is an element Yr of a n K[y;. , ... , y~] ~ertion
n
y;., ... ,
45
CHAPTER I
'which is not zero. Then, applying (14.1) to Yr and K[y~ , . " ,Y~], we see the existence of Yr+l, ... , Yn of K[y~, ... , y~] such that (a) Yr+j = y;.+j + y~mi (if the characteristic p ~ 0, mj can be powers of p) and such that (b) K[y~, ... , y~] is integral over K[Yr, ... ,Yn]. The first condition (a) implies the validity of (3) (and the statement in parentheses implies the validity of the last statement in the theorem), while the condition (b) implies the integral dependence of K[x] over K[y]. Since YI , ... , v,. are in a, (t K[y] contains them. Since Lr yiK[y] is a prime ideal of height r and since height a n K[y] = r, the inclusion implies the equality. Thus the theorem is proved.
n
(14.3) COROLLARY: Let I be an integral domain and let XI, '" , Xn be algebraically independent elements over I. Let K be the field of quotients of I. If a is an ideal of I[x] = I[xI, .,. , Xn] such that a n I = 0, then there are elements YI, ... , Yn of I[x] and an element a( ~O) of I such that (1) I[aI][x] is integral over I[aI][y], (2) aI[aI][x] I[aI][y] is generated by YI, ... ,Yr with r = height aK[x] and (3) Yr+j = xr+j + fj with polynomials Ii in XI, ... , x,. with coefficients in the prime integral domain for each i = 1, '" , n  r. Proof: Set a' = aK[x], and let YI, ... , Yn be as in (14.2) applied to a' and K[x]. Then (3) is valid for them. Since YI, .,. , Yr are in a', there is an element ai ~ of I such that aiY E a for each i. Since K is a field, aIYI, ... , ary,. are as good as YI, ... , Yr . Therefore we may assume that YI, '" , V,. are in a. Since Xi is integral over K[y], there is an element CiC ~o) of I such that CiXi is integral over I[y] for each i. Let a be the product of all the Ci . Then we see easily that this a and the above Yi are the required elements.
n
°
(14.4) THEOREM: Assume that a ring R is generated by elements JI , . . , , bn over an integral domain I. Assume furthermore that no elenent a( ~o) of I is a zerO divisor in R. Then there are elements ZI , .. , • . • • , Z t of 1r[b l , ... , bn] (where 1r is the prime integral domain of 1) which are algebraically independent over I and an element a( ~o) of I such that R[a I] is integral over I[a\ ZI, ... , Zt]. (NORMALIZATION THEOREM FOR FINITELY GENERATED RINGS)
Proof: Let XI, ... , Xn be indeterminates. Then there is a uniquely determined homomorphism c/> from I[x] = I[xI, '" , Xn] onto I[b] such that c/>(Xi) = bi for each i. Let a be the kernel of C/>, and let YI, '" , Yn , and a be elements as in (14.3) applied to a and I[x]. Set Zi = c/>(Yr+i). Since Yi E a for i ~ r, I[aI][b] is integral over
46
GENERAL COMMUTATIVE RINGS
n
I[a I][zl. Since a I[Yr+l, ... , Ynl = 0, the Zi are algebraically independent over I. Since Yr+j E 1r[XI' ... , xn], we have Zi E 1r[b l , '" , bn ], and the assertion is proved. (14.5) THEOREM: Let R be an integral domain which is generated by a finite number of elements bl , ' " , bn over a field K. If po, '" , Pt are prime ideals such that Po = 0, PI is maximal, Po C PI C .. , C Pt and such that there is no prime ideal q such that PiI C q C Pi for any i, then t must be the transcendence degree of Rover K. Proof : We prove the assertion by induction on t. There are algebraically independent elements ZI, ... , Zu over K (Zi E R) such that R is integral over K[z], by (14.4). If t = 0, then K[z], is a field by (10.5) and the assertion is true. Assume that t > 0. Then apwhence u = plying (14.1) to PI K[z] and K[z], we may assume that ZI E PI. Then PI K[z] = zIK[z] by (10.14). Then, applying the induction assumption to R/rl1 , which is integral over K[zl/zIK[z], and to the prime ideals pi/PI, we see that t  1 = u  1, whence t = u, which proves the assertion.
°
n
n
(14.6) COROLLARY: If an integral domain R is finitely generated over a field K, then for any prime ideal P of R, height p depth p is equal to the transcendence degree of Rover K and depth P is equal to the t,.anscendence degree of R/p over K.
+
(14.7) THEOREM: If an integral domain R is generated by n elements a field K, and if m is a maximal ideal of R, then m is generated by n elements and R/m is algebraic over K. Proof: That R/m is algebraic over K follows from (14.6). Let bl , ' " , bn be such that R = K[b l , • . • , br.l. Let Ci be (b i modulo m) for each i and let f;(xi) be the irreducible monic polynomial over K[ci , ... , cid which has Ci as a root. Let fi be the monic polynomial in b; with coefficients in K[b l , ' " , bil] which is obtained from f: replacing CI, '" , CiI , Xi with bl , . . ' , bi I , bi , respectively. Then we see that R/ ( L fiR) = K[CI' '" , cn ], and therefore m = L fiR, which proves the assertion. Ol'e!'
(14.8)
COROLLARY:
Let
Xl, ... ,
Xn be algebraically independent ele
mJ?llts over an integral domain I. Let p be a prime ideal of I[x] I~Il'
=
'" , xnl. If IcpnI) is a regular local ring, then I[x]P is a regular local
,:,illg. Proof: We may assume that I = IcpnI) . Let h,
...
,fs be a regular
47
CHAPTER I
n
system of parameters of I. We set q = l1 I, K = I/q, and Zi = Xi modulo qI[xl. Then, obviously the Zi are algebraically independent over K and I[xl/qI[xl = K[zl. We set ~' = ~/qI[xl, and apply (14.2) to ~' and K[zl; let Yl , ... , Yn be as in (14.2) applied to our case. Then K[zl = K[ZI' ... , Zr , YrH, ... , Ynl. Let L be the field of quotients of K[Yr+l, ... , Ynl. Then ~'L[ZI' ... , zrl is a maximal ideal of L[ZI, ... , zrl, and therefore it is generated by r elements, whence ~'K[z]P, is generated by r elements, which implies that pI[x]P is generated by r + s elements. Since height p' = r, we see that height ~ ~ r + s by (6.15), whence I[xlp is a regular local ring. (14.9) THEOREM: If a ring R is finitely generated over a field K, then the Jacobson radical of R is the radical of R. (HILBERT ZEROPOINTS THEOREM)
Proof: We first assume that R is an integral domain. Let f be an element of R which is not zero and consider R[l/fl. (14.6) implies that maximal ideals of R[l/fllie over maximal ideals of R, which implies that there are maximal ideals which do not contain f. Since f is arbitrary, it follows that the Jacobson radical of R is zero. Now we consider the general case. The above result shows that if ~ is a prime ideal of R, then the intersection of maximal ideals of R which contain ~ coincides with ~, and therefore we get the result. (14.10) Let Xl, ... , Xn be algebraically independent elements over an integral domain I. Then there is a maximal ideal m of I[xl such that m I = 0 if and only if there is an element a( ~O) of I such that I[ a11 is the field of q11/Jtients of I. Proof: If there is such an element a, then a maximal ideal m of I[x], such that aXl  1 E m, lies over zero of I. Assume conversely that there is a maximal ideal m of I[xl which lies over zero of I. Then, applying (14.4) to I[xl/m, which is a field, we see that the t in (14.4) must be zero in this case, and I[a 11 is a field by (10.5). Thus the proof is complete.
n
EXERCISE: With the same K, Xi and 11 as in (14.2), aswme that K contains infinitely many elements. Prove that there are elements Yl , ... , Yn of K[xj satisfying (1) and (2) in (14.2) and such that the Yr+i (j = 1,2, ... , n  r) are linear combinations of the Xi with coefficients in K.
CHAPTER II
Completions 15. Formal power series ring Let R be a ring and let Xl, '" , Xr be indeterminates. Let Fd be Lilli module of homogeneous forms of degree d in the Xi with coeffiI',inll('s in R for every d = 0, 1, ... ,n, .. , . The set F of infinite sums L: ai with ai E Fi forms a ring by the obvious operations ( L ai) + (Lb i) = (ai + bi ), (Lai)(Lb i ) = Ln (Li+j~naibj). This /1' is called the formal power series ring or merely the power series ring ill 1,ho Xi with coefficients R, and is denoted by R[[XI, ... , xrll or I'illlply by R[[xll. Elements of R[[x]] are called (formal) power series ill Lhexiwith coefficients in R.For an element at of R[[x]](ai E F i ), Llw Ilumber n, such that an ~ 0 and such that ai = 0 for i < n, is (lnllcd the leading degree of the element, and the an is called the leadin(J form of the element. The leading form of 0 is defined to be O. ao il' (:alled the constant term of the element. Note that R[x] is a sub ring of R[[x]] by the obvious identification "hut a finite sum L~ ai is identified with infinite sum L~ ai with !/'i = 0 for i > r. The above notation will be fixed throughout this section.
L
L
(15.1) THEOREM: There is a oneone correspondence between all ma.Timal ideals m of R and all maximal ideals m* of R[[xll in such a way that m* corresponds to m if and only if m* is generated by m and the Xi . Proof : We see obviously that, when m is a maximal ideal of R, (,he ideal mR[[xll + L xiR[[xll is a maximal ideal of R[[x]]. Let m* be a maximal ideal of R[[xll and let 111 be tho set of constant terms of elements of 111*. Then m is obviously an ideal of R. If we know that lit ~ R, then 111* is contained in mR + xiR[[x]], and we seo that m* = mR[[xll + xiR[[xll and that 111 is maximal. Therefore it is I'ufficient to show that 111 ~ R. Assume the contrary. Then there is ILIl eloment ai of 111* with ao = 1. Set b =  (L~ ai). Then we
L
L
L
[49]
J
~'I;'
()'
L (.,
10 ",IiI'I'I' ('11.1'11 ('" iH til' ~::' h' 1'111' Hii iii . II, 1It'III'(' 1'01' allY . eJ' ,., '~I/,{' I ('\",,(1 I (,,11 'i/l, 2:. n. ,'IIH'n (~III)( ~II I) ( ))( L.,.II I) I , wn K(~(: Lhal, /:1) O~ 1+ (~.: (LI:I'IIIK 01' dl:gn:
ing as n is zero or not. Thus we see that the element L a,. it:: unit, which contradicts to the assumption that [: ai E m*, and the proof is completed.
(15.2) COROLLARY: An element f E R[[x]] is a unit if and only if the constant term of f is a unit in R. (15.3) THEOREM: If R is Noetherian, then R[[x]] is also Noetherian· Proof: Let a be an arbitrary ideal of R[[x]]. It is sufficient to show that a has a finite basis. Let a* be the set of leading forms of elements of a. Then a* generates an ideal of the polynomial ring R[x], which has a finite basis (over R[x]) , say ii, ... ,ii, consisting of elements of a*. Let fi be an element of a which has f.i as its leading form and let a' be the ideal of R[[xll generated by the f1, '" ,ft. Let g be an arbitrary element of a. By the choice of a', there is an element [!ihi,o of a' which has the same leading form as g, and the same is applied to g  L fih i .O and so on. Thus we see that there is a sequonce of elements Lfih.,.,n(h.i,n E R[[x]]; n = 0, 1,2, ... ) such that (1) Lfihi,n has leading degree greater than that of Lfihi,nl and (2) Lfihi,n and g  Lj1< ( M>I< (:.l) 11/* iH I'.olllplnl.!', (:l) III iN II, tlnllHn HI I hHPH("l'. of 111*, HIHI Cl) if ,r* nlld HI'(, lillliLH of (:nlll'.hy H('1'1'01111' I''''H.\'
oj'
"'lid
('/11.
Till' oLlII'" "'H,'iI'i·iioll.'I 1.111' d!~LaiIN.
ill 0111'
LI 1l'1I !'! 'II I
Ilav!' III1IY
\1'1' ollli!.
'l'III':OIII'IM: ,,1 88 If, II 1.1: thal (1":8 11.1/, '£deaf OJ' H and that the metric III /8 /1£1 1('11 b!f thl' nruJ£c topology (by ( I G.1)). If a has a finite basis, Ihl'lI Ihl' tnl!11fnu.IJ I~r tlw (;nml!felion M* is the aadic topology. l'l'oo\': II. i:; HufficiPllt, by virtue of (17.:3), to show that the closure N~ oj' n"M ill M* is a"M*. Let x* be an arbitrary element of N!. 'I'bnll :1:* = lim :r;i with (Xi) such that Xl E a"M and such that Xi n .1"; II ( (1'11 i'iJYI = a ( aiM). Let ai, ... , at be a basis for an. Then ,I'j . :1:j'H = Lj ajb ji with b ji E aiM. For each j, the series Li bji iH I:()II vergent (i.e., { L~~l bj;} is a Cauchy sequence) and express an l'lnnwnt of M*. Then we see that x* = L ajb;, which is in a"M*. 'l'huH N! C anM*. Since the converse inclusion is obvious, the proof
(,,'.,1)
I~r
b;
iH (\omplete. (17.5) THEOREM: Assume that the aadic topology oj' R is To and that a has a finite basis ai, ... , aT . Let Xl, .. . , Xr be indeterminates and consider the formal power series ring R[[x]]. Then the completion U* of R is (isomorphic to) the ring R[[xll/n*, where 11* is the closure oj' tlu! ideal n = L (x,;  ai)R[[x]] in R[[xll with the (n aR[[x]])adic topology (= the (aR[[x]] LXi R[[x]])adic topology). Proof: We consider the map cf> from R[[x]] into R* such that if J = Lfi(x) E R[[x]] (fieX) being homogeneous form of degree i) I.Illm cf>(f) = Lfi(a). Then we see that cf> is a homomorphism from UI[x]l onto R* and that the kernel n' of cf> contains x  a, hence n. f = L!i(X) is in 11' if and only if Lf;(a) = O. Let f,,(x) be the Imding form of f. Then Lf;(a) = 0 implies that fuCa) E a"H, and Lh ere is a homogeneous form hUH (x) of degree u 1 such that .fAa) = hUH(a). f  f,,(x) huH(x) has leading degree greater Lhan u and f == f  fu(x) hU+JCx) modulo 11. Thus, repeating the Hame, we see that if fEn', then fEn (L xiR[[x]]) " for any n. Conversely, if f E n( n L xiR[[x]]) "), then we see that L fie a) E an = O. Thus we have n' = nn (n + (L xiR[[x]])n) = nn (n + n"R[[x]]) = n*.
+
+
+
n"
+(
+
+
+
(17.6) COROLLARY: Assume that R is a semilocal ring with J acobNon radical m and assume that elements ai, ... , aT generate an ideal which has 111 as its radical. Then the completion R* of R is a semilocal ring and is (isomorphic to) the ring
1111~11'1.lli'I'III~n\
NII.I'I, .. , , .1',11/( ~
(.1',
where the Xi are indetc'I'm'inllJi:8. '/'/11' IO/lo{o(1!I ol /tl. 118 1/1.1' t:IIllIllll'lio/l. IJ' /?, coincides with that of /t* as a semifo('(d 1'/11(1. I'roof: Set a = L aiR. Then since tlw rail il'al or n iK Ill, Lhe nadie Lopology is equivalent to the madic topology. TIII,rdo\'(, WI) eall apply (17.4) and (17 ..5). Since R[[x]] is a semilocal ring by (L5.4), every idnal of R[[x]] is closed, whence we have R* = R[[x]l/(L (Xi lI,i)R[[xlJ) by (17.5). The coincidence of topology follows from (17.4).
Another important fact concerning the completions of semilocal dngs is: (17.7) THEOREM: Let (R, ~1, ••• , ~r) be a semilocal ring. Then the completion R* of R is the direct sum of completions Ri of local rings Ri = Rpi. Proof: Let m be the Jacobson radical of R, i.e., m = i ~i • Then, for each natural number n, Rim" is the direct sum of R/~~ and there n are ei.n E R such that 1  Li ei,n E m , ei,n E ~j if i ~ j, hence 1  ei,n E ~~ . Then for each i, ei,n  ei,n+! E ~~ ll~ = mn and therefore the sequence {ei,,,l has a limit ei in R*. Since
n
n .,. n
.n,
we have Lei = 1. Since ei,nej,n E mn (i ~ we have eiej = 0 if i ~ j. Hence we have e~ = ei. Thus R* is the direct sum of ideals R*ei, which are rings with identities ei. We shall prove now that R*ei is (isomorphic to) the completion of Ri . Let {c n} be a Cauchy sequence in Ri such that Cn  Cn+l E ll~ Ri . Since R/~~ = Ri/~~ Ri for any n, there is a sequence {c~} of elements c~ of R such that (c n modulo ~~Ri) = (c~ modulo ~n for every n. Then the sequence ' } 'IS a C auc h y sequence In . R and l'1m Cnei,n I l'1m Cnei,n I 2 {cnei,n = = ei(lim C~ei,n)' Hence the limit of {c~ei,nl is in R*ei . It is easy to see that the map lim Cn '> lim C~ei,n gives a oneone correspondence between the completion Ri and R*ei. Then we see easily that is isomorphic to R*ei . Now we go back to the completions of modules:
R.;
( 17.8) THEOREM: A ssume that R is a semilocal ring and that M is a finite Rmodule. Let R* be the completion of R. Then M ® R R* is the completion of M, whose topology as a finite R*module coincides with its topology as the completion of M.
('II;\I"I"'iI(, II l'I'IIl,I': 1,1,1.1111,1' UII'.JHI·(tI,~iI)lll'lI,dil'lI.llIr N, '1'111'11 IIIN'" is UII!.l:I.(~lIh fl I'ntiil'allll' HI' I,y (17,(j). 1"IIl'LiI('I'IIIIlI'n N*/III"N* ~~ N/llt l{ Now, 11 (III (,"'! N*)/IIII/(M C9 U*) = (M /1II J111) ® (/l*/m"R*) = M/mnM. Hill(:I~ III" III = 0, WI) s(~() that M ® R* contains M (by the identifica!.illil 'In 'fn ® I for any m EM), and furthermore that mn(M ® Hilil
n
N*) n M = mnM. This shows that M is a subspace of,M ® R*. An 1I.1'i>iLra,I'y clement c* of M ® R* can be expressed as L mi ® wiLh mi E M and r; = limn n,n with ri,n E R. Then c* is the limit of I L miri,n} as is easily seen. Thus M is dense in M ® R*. Let {c!} be II. mgular sequence in M ® R*. Let Ul, . , . , Ut be a basis for M. Since (M ® R*)/mn(M ® R*) = M/mnM, there is a sequence {c n} in M Hlldl that c!  Cn E mn(M ® R*). cn+!  Cn is in mnM, hence is expnlHHed as L Uimin(min E mn). For each i, L min has limit r; in U*. Furthermore, we see that ~ Ui ® is the limit of the sequence I (:';}. Thus M ® R* is complete.
r;
r;
(17.9)
COROLLARY:
Let a be an ideal of a semilocal ring R. Let R*
III: the completion of R. Then the completion of a is aR* and aR* is iso
n
'/lwrphic to a ® R*. Furthermore aR* R = a, and R* / aR* is the completion of R/a. Proof: The first assertion follows from (16.5), hence the second HHHertion follows from (17.8), Since a is a closed set by (16.7) and :;ince aR* is its closure in R*, we see that aR* R = a. The completion of Ria is (Ria) ® R*, which is obviously R*laR*. Similarly, we have for modules:
n
(17.10) COROLLARY: Let N be a submodule of a finite module M IIvcr a semilocal ring R. Then the closure N* of N in the completion M* (~r M is the completion of N and we have N* M = N. Furthermore, M* I N* is the completion of MIN. If we identify M* with M ® R*, then N* is surely identified with N ® R* in M ® R* as is easily seen. That N* is the completion of N implies that the usual tensor product N ® R* is isomorphic to the Hubmodule N ® R* in M ® R*. Hence we have
n
(17.11) COROLLARY: If R is a semilocal ring, then ®n R* is exact. (17.12) Let (R, lh, ... , prj be a semilocal ring and let R* be the completions of R. Then we have altitude R* = altit.ude R. Proof: altitude R = max {altitude R~J. On the other hand, R* is Lhe dircet sum of completions of R~i . Therefore it is sufficient to prove the assertion in the case where R is a local ring with maximal
.'I'MI'I,I'I'I'I.INII l ill. '11,1 III ).11. Wn PI'O\'I' Ih .. IIHlll'l'Lillll II,\' illdlH'lioll Oil II.ll,il.lldn It '. l II' nll,il,ll be the nat'uTal !J,o'/l/'(l'llIo'l'jlh£sln Jl'lrill NI:/lto R* (cf>(r) = '1'·1 with 1 E R*). Let a be an ideal I({ U and lel 8 and 8* be multiplicatively closed subsets of Rand R* respectively 8uch that cf>(S) C S* and such that S a = S* aR* = empty. Set R' = U"/llU 8 and R" = R~* I aR~* . Then ® R' R" is exact. Proof: If M is an Riamodule, then M is an Rmodule and
n
M
®Rln
(R*laR*)
n
=
M
®R
R*.
Therefore we see easily that ® Rln R* I aR* is exact. Hence, we may assume that a = O. Since ®R* R" is exact by (6.18), we see that ®R R" is exact. If M is an R'module, then M is an Rmodule and M ® R' R" = M ® R R", and therefore the assert.ion is proved easily. (18.11) THEOREM: Let Rand R* be Noetherian rings such that R* is an Rmodule and such that ® R R* is exact. Let a be an ideal of Rand let p* be a prime ideal of R*. Then p* is a p'l'ime diviso'l' of aR* if and only if there is a p'l'ime divisor p of a such that p* i8 a p'l'ime divisor oj pR*. p* is a minimal prime divisor of aR* if and only if there i8 a minimal prime divisor p of a such that p* is a minimal prime divisor of pR*. Proof: Assume that. p* is a prime divisor of 'pR* wit.h a prime divisor p of a. Then t.here are element.s b E Rand c* E R* such t.hat. p = a:bR, p* = pR*:c*R* by (8.8). Since pR* = aR*:bR* by (18.1), we have p* = (aR*:bR*) :c*R* = aR*:bc*R*, and p* is a prime divisor of aR*. Conversely, assume t.hat. p* is a prime divisor of aR*.
Let. ql n ... n qr be a short.est primary decomposition of a. Then aR* = nqiR*, whence p* is a prime divisor of some qiR*, say, qIR*. Let p be the prime divisor of q = ql. We shall prove that p* is a prime divisor of pR* by induction on length Rpl qRp . We may a88ume that q = 0 by (18.10). If the length is 1, then the aS8ertion is obvioutl beeause p = q. Let a be an element. of p such t.hat p = 0: aR, and let q' = aR)) n R. Then there is an element d of R which is not in p 8ueh that dq' CaR. Let qi n ... n q; be a shorte8t primary decomposition of zero in R* such that the prime divisor pi of q; i8 a prime divisor of pR* if and only if i ::::; t. Since p = 0: aR, pR* = n (q;: aR*) by (18.1), whence by our assumption on t.he prime divisors pR* qi : aR*) n ... n (qi: aR*). Let b* be an arbitrary element of
1'11.\1"1'1'111 "
qt ,
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tlil'l'('
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n '" n q;) :IIN*
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0, WIIl'III'n b>l>d . O,III'III:n b* ( O:dN*' (O:dfnu+ 0, alld s .. I, whinb PI'OVI'H the HKKm(,ioll, Now we l'OIlHidl'l' t.lH~ InHt IIHYl'rtioli. AHHllrne thaL ~* il'l a lI\illimall)l'iml~ divil'lo]'o!' IlN*. 'l'III'1l p* il'l a pl'imc diviHot' of ~/(,* with a prime tiivitlo]' ).I of 11. 'I'lw llIillillIItiity or p* impliel'l Chat p* i.'l a miJlimal prime divisor of ).IU* alld that ~l iH /I. minimal prime diviso], of a. Conversely, assume that ~ iH H, minilll:d primo divisor of a and that ).1* is a minimal prime divisor of ).1n*, L.. t. ).1** be a minimal prime divisor of aRt such that p** C ).1*. Let, q) II11 Lbo natural homomorphism fromR into R* (cf>(a) = a·l with 1 ( U*), and let p' be cf>l(p**). Then ).1' is a prime ideal by (2,12) and p' iN ('.011,' I.nined in p because p = cf>l(p*) (for, nonzero elements of H/~l :IXI' 1I0!; zero divisors with respect to R*/pR* by (18,1), By the millillIality of ).1, we have ).1' = p, whence p** is a prime divisor of ).IN*, whieh implies that p** = p* by the minimality of p* and by (,haL p** C pt. Thus p* is a minimal prime divisor of aRt. We say that a module M over a ring R is torsionfree if every 11011;!,prodivisor in R is not a zero divisor with respect to M, '1'111'1'101'01'1' Ilt*
(18.12)
THEOREM:
Let Rand R' be Noetherian rings such that H'
is an Rmodule and such that ® R R' is exact, If M is a torsionfree Hmodule and if every prime divisor of zero in R is of height at most I I then M ® R R' is a torsionfree R' module. Proof: Let R* and R'* be the rings R EB M and R' EB (M ® R') I'nspectively in the principle of idealization. If N is an R*module , Llten N is an Rmodule, and N ® R R' is naturally identified with N 0 R* (R' ®R R*) = N ®R' R'*. Therefore we see that ®R* R'* is exact. In order to prove (18.12), we may assume that M is a finite module. Assume that a E R' is a zero divisor with respect to M ® R'. Then a is a zero divisor in R'*, and there is a prime divisor p'* of ;!'ero in R'* such that a E p'*. (18.11) implies that there is a prime divisor p* of zero in R* such that p'* is a prime divisor of p*R'*, rlince M2 = 0, we have M C ).1*, and therefore p* = p EB M with a prime ideal p of R. Similarly, ).1'* = p' EB (M ® R') with a prime ideal ~' of R'. Since M is torsionfree, we see that ).1 consists merely of zero
til tlil'iHIII'H, 111'111'1' i:1 II. prillll' tlil'illlll' IIi' '/,111'11 II,\' 0111' 1I:1rllllllpLiIIIl oil
N,
fJ" HIHI :lilll'I' U+ III ,~, 11', Lill' 1'1101'1. 1.1111,1. ~I'''' iH :I. pl'illll' dil'i,'1ol' 01' \l+N'+ ililplil'r1 l,illl,L 11' iel II, WillII' tliviHol' or tIlt,', Hillel' ¢Il", N' iH I'xni'!, HIHI Hilll'(' P iN H pl'illll' diviHlIl' oi' iWI'o ill N, WI~ H('(', Lila!. ).1' iN a WillI(' diviHOl' 01' :t,l'I'o ill H' Ily (IK,II), WIIl'III'!' I/,
Hilll'l' ,,",1,/( II'
(il
,,"
I '"
1
iH a l',('I'O diviNOl' ill N', wllil'll 1',OIlIPld(~H (.1)(' pi'ool'. 1';XI,JIWlf;JC::: I. AHHll1nc l.lmL U, U*, U** are riJl~H KJleh thaI. ®u U* and ® u* fl*" are exact" Prove that ® II H.** iH exact, 2, Let x, , .,. , x" be algebntically independent elements over a ring R. PI'ove that ® R R[x] and ® R R (x) are exact. (Hint: Prove first the exactness of ®R R[x] using the fact that R[x] has a free base over R.) 3. Let R* be a ring which is a module over a Noetherian ring R. Prove that ®R R* is exact if (a:bR)R* = aR*:bR* for any ideal a and any element b of R, 4. Let (R, a) be a Zariski ring and let M be a finite Rmodule. Let S be the set of nonzerodivisors with respect to M and let N* be the completion of a submodule N of M. Prove that (M ® Rs) n N* = N, where, M ® Rs and N* are naturally imbedded in M ® R~ (R* being the completion of R). 5. Let (R, a) be a Zariski ring and let R* be the completion of R. J~et b be an ideal of R. Prove that if bR* is principal then b is principal. (Hint: b/ba = b/ba ® Ria = b/ba ® R* /aR* = bR* IbaR*.) 6. Prove that if the completion R* of a Zariski ring (R, a) is a unique factorization ring, then R is also a unique factorization ring. (Hint: Let lJ be an arbitrary prime ideal of height 1 in R lind let S be the complement of lJ in R. Then, show that lJR~ is principal and that every maximal prime divisor of \JR* does not meet S, Then conclude that \JR* has only prime divisors of height] and then that \JR* is principal.) 7. Let R* be a ring containing a ring R such that (a) ®R R* is exact and (b) aR* n R = 11 holds for any ideal a of R. Let M be an Rmodule. Prove that if M ® R* is a finite R*module then M is a finite Rmodule.
19. The theorem oj transition We say that the theorem of transition holds for rings Rand R' if R is dominated by il' and (2) if q is a primary ideal in R such that the prime divisor of q is a maximal ideal, say m, then ~ 1)
lengthR' R' / qR' is finite and it holds that lengthR' R' / qR' (19.1)
THEOREM:
=
(lengthR' R' / mR') (length R R / q) .
Let Rand R' be Noetherian rings such that
R :s; R'. Assume that, for any maximal ideal m of R, the length of R' / mR' is finite (i.e., the prime divisors of mR' are all maximal). Then
(II L\ 1"1'11111
II
(\:)
,~j' l/i" follol/lill(/i{ i8 " II"",'IIS"I'!/ 1/1/(1 iill.lli,.;,.1I1 ,.o/It/ifio// Jor III,. I'Irlitlilli ,(/' Iii,' Ih"o/'t'1II qj' 1/'IIIINil;oIIIo/' l/i,' /'i//flN It allrl N': (a) II q /S 1/ /lrilll((I'!1 itll'lll hdo//I/illf/ 10 (('1/11/,,/,/11111'/ ir/I'II./ 111 in It I(.I/.({ 1/ q: hh~· 111 Iol' an d,:II/,'II'/' h ,~r H, the//. qU': hl{" = III /{,'. ( I») M I,: N' /:8 (',!>ad. ((~) 11'111' IJ://'!I '/II.(u'£'ma.1 ideal 111' (if Il', the theorem of transition holds JOI' Hlm'n,,: and N,~" . (d) If Il and 0 arc idcals of R, then (a:o)R' = aR':oR', ('(/I'll
1'1'001': We show first that the validity of the theorem of transition illiplins (a). Set q' = q + bR. Then length R/q = length R/q' + 1, !Il'III:n length R'/qR' = length R'/q'R' + length R'/mR', which impline; that length R'/mR' = length q'R'/qR' = length bR'/(bR' q/(,') = length R'/(qR':bR'), which implies that mR' = qR':bR'. Thie; proof being reversible by induction on length R/q, we see the eqllivalence with (a). (a) implies (b) by (lS.7). (b) implies (d) by (IR.l), while (d) implies (a) obviously. Thus (a), (b) and Cd) HI'P equivalent. It is obvious that (c) implies the validity of the theorem of transition; the converse is also obvious except for the i'lwt that R* = R(lIlnR) C R** = Rill" which is proved as follows: Q
O. 1,('1. ('
d, II' d "II
0, 1.11('11 I,hll I,hll IIOl'lJinil'll1. or
xd • Thon I(x)  I(:I: 1+ (1.1'1'1111' or lowI'!' dllJ,!;I·('('H). Since f(x)  I(:c  1) it:: aiHo II\UXWl'i(lal, ('0 c'c (:(d!) iN :1.11 illl.l'g(,I·.
Therefore f(
x)  (x ! d) is a numeriwl polYliomial or dngn'(' 1(,1'1' Co
than d, and we prove the assertion hy our indtwtion.
21. APolynomials Let a he an ideal of a ring R and let M he an Rmodule. Set F n an/an+! and Gn = anM/an+!M, for n = 0,1,2, ., .. When a E F m , b E F n , we define ab as follows: let a' and b' be elements of am and an respectively such that a = (a' modulo am+1 ), b = (b'modulo an+!) and then we define ab = (a'b' modulo am+ n+1 ) ( E Fn+m). This multiplication defines a ring structure in the direct sum F of all the F n , and F becomes a graded ring. Furthermore, since Fn = an/a n +\ it follows that Fn = (F1)n, which implies that F is a homogeneous ring. This homogeneous ring F is called the form ring of R with respect to a. When a E an and a ~ an+!, n is called the degree of a with respect to a and (a mod an+!) is called the aform of a. The direct sum G of all the Gi becomes similarly a graded module over F. This G is called the form module of M with respect to a. It is obvious that length M/a n +1M = length ai M/ai+ 1l\1 1:~ K(G; i). Therefore we see by virtue of (20.5) that (21.1) A ssume that Rand Mare Noetherian and that depth a = O. Then there is a numerical polynomialf(x) such that length M/an+1M = f( n) for all n that are greater than a fixed integer; f( x) is nothing but the O"polynomial of the form module G. Now, applying (20.6) and (20.7) to the above result, we see that:
1:;
(21.2) THEOREM: Let M be ajinite module over a Noetherian ring R and le,t a be an ideal of R such that depth a = O. Furthermore let R' be a Noetherian ring such that (1) R is an R' module, (2) with a'
=
(a I a E R', aR
c al,
R' / a' is dominated by R / a and (3) [R / mi: R' / m: 1 are finite, where mi (i = 1, ... , s) are the prime divisors of a and m; are the maximal ideals of R' such that m;/a' = (mi/a) (R'/a'). Then: (a) there is a numerical polynomial f(x) such that fen) = lengthR' M/anN\.for
n
71
('11111"1'1')11, III
if (/ iN thl! fOI'/11 I//lIdlll,' I~j' IlIu'illil'l'.Y/u'('I/IIIl,I/t"II./'(n I I) 1111'1",((,';//) • ./'(1/ I I) /(1/) h/,"III'( 0; 1/.) JII/' iiI/JII('':I'/lII!! 11/('(/1' //IIIIII'//.( //11I1I.{/('/,8 n I/l/.tI (('.) J( 1/) LIN/III i: N' /111; I· (1'lIlIlh J,'i • III (>0 10' Njn" AI C>() II Ni , I/Ihl '/'1 , Hi Nil, ( , Wn dnlloCn LlIl'/(.r) J.!:iVI~II.iIlld, ahovn hy IIIIJI NII:f!ir'/I'ltllllirll'{/I' //1111/./'111 /llllI/hr'I' II. (I,)
AIo:'(O; Ai; .r), whidl i:; ('alled a A]!II(.l/II,(lfIl:ia( or M. A/,:(n; AI; .r) Illay "I~ dn1ioj,I'd Hilllply by A(11; M; .r) HIHj 111 may 1)(: omiCCI~d ir 111 = H. By our defillition, we Nne iIIlln t u v, then at least one of l i ) is not in any of the N j for j > v. By the choice of g*, we see that the element is not in any of the N j for j = 1, ... , t u. Thus the existence of g is proved. Now letfbe an element of R such thatfmodulo5tR = g. Sincef ~ \Ji for any i by our choice, 0 :fR coincides with 0 up to primary components whose prime divisors contain Rl ; namely, there is an ideal b which contains a power R7 of RI such that b n (O:fR) = 0, whence (O:fR) n Rn C (O:fR) n b = 0 for n ~ m, and f is superficial. When a is an ideal of a Noetherian ring R, an element a of a is called a superficial element of a if there exists a natural number c such that (an: aR) n aC = a,,l for any natural number n > c. (22.2 ) With the notation as above, a E a is a superficial element of a 2 if a modulo a is a superficial element of the form ring F = L F" of R with respect to a. Proof: Set a' = a modulo a2 • Let a natural number c be tlllchLhat (O:a'F) n Fn = 0 for n ~ c. Let n be an arbitrary natural number such that n > c. It is obvious that a n  1 C (an:aR) n aC • Let b be an arbitrary element of (an:aR) n aC • Then ab E an. Let m be such that m l b E a  and that b ~ am. Set b' = b modulo am. Then b' E F ml .
+
+
+
+
N;
N;
+
+
7:1
.'11,11"1'1111(, III
n
Hilll'.' II I 1\", \1'1\ 1111,1'.' III . 1', ",lil'lll'(\ (.0:0'/,') I,'", I O. II' III • II, Ihl'll oh ( n" illl"lil'~j Lhnl. 1//1/ 0, whidl iN n I·OIlI.J'wlidioli. 'l'hnl'l'foJ'l'/II, . 1/., HIIII h ( nil I. ,I'hll:l I.hn pmol' iH I~OII1J1ltd,n. (:.l:.l.:\) '1'''Wllt,I')M: 1"ll (\ Iw ((,/1. 'I:t!(/(/,l I~r Illljill!. z(/ro 1:'1/. a Nor;lIW1"£an '/''://(/ N, and lel I./u: 7!1'irrw d£/r£,w)'/'S I!/, (1 he 1111, ... , 1\l". Let 8 he lhe inler.wlci'ion I~r til.!: co'fnpl(mwnls I~r 11\.[. I,et 01, ... , 0 t be ideals oj R wlf.ch lho,l aa" is nol contained in any oiRs . If R/mi contains infinitely IWIny elements Jor every i, then there is a superficial element a of a which .,:.~ not in any oj the OJ . I)roof: Let F = F n be the form ring of R with respect to a. Set 2 2 2 Iii = FI ((OJ )/a ). If nj = F I , then a C OJ a , hence a = 2 (Uj a) a , whence aRs = (Uj a)R8 by the lemma of KrullJ\,mmaya, and it is a contradiction. Thus nj ~ FI for any j. Therefore there is a superficial element a' of F which is not in any of the IIj by (22.1). An element a of a such that a' = a modulo a is a superfieial element of a by (22.2). It is ohvious that a ~ OJ, and the [I$sor
n n +
L +(
+
n
!,ion is proved. (22.4) With the same R and a as ahove, if (t = aR, then a is a .mper.ficial element of a. Proof: It itl sufficient to show that, in the form ring F = Fn of
S
Il with respect to a, a' = a modulo a2 is a superficial element of F. fi' is generated by a' over Fo = R/aR. Therefore, if \.1 is a prime divisor 0[' zero in F which does not contain F I , then a' Et \.1, hence 0: a' F (:oincides with 0 up to primary components whose prime divisors nontain Fl' Therefore, a' is a superficial element of F, and a IS a Kuperficial element of a. (22.5') With the same notation as in (21.2), iJ b E R, then length w R/(a n
+ bR)
= length w
R/a n  length1/' R/(an:bR).
This follows from (1.5). (22.6) THEOREM: With the same notation as above, assume that a E a. Set a' = a/aR. Then A1/'(a'; n) ~ Aw(a; n)  A1/'(a; n  1) for large n. If a is a superficial element of a, then, with the intersection S of the complements of the prime divisors of a, kngthR' (O:aRs) is finite and AR,(a'; x) = Aw(a; x)  Aw(a; x  1) length1/' (O:aRs). Proof: The firtlt assertion is immediate from (22.5). Let c be a
+
sufficiently large natural number which is fixed and let n be,greater Lhan c. Then (an:aR) n a = anI. By (22.5), we have C
7,1
MIII!I'll'ld('I'I'II'ill
'\"" (n;
/I )
'\10" (n;
I)
1/
ill Hill", I.hi:; length i:; a eOlli:iLan(" flay C:, for large n. We have only to prove that C = length w c (O:aRs). Since Rla = RslacRs we may assume that R = Rs. Then an:aR is contained in (O:aR) + a by (3.12) for large n, whence C = length w (O:aR)/(O:aR) a Since this C is independent of c C (when c is large), (O:aR) a must be zero, and the assertion is proved.
Situ: O. Let m be an element of M such that 0: mRp = pRp for one p, say q. Then there is an element c of It which is not in q such that O:cmR = q. Set N = cmR. Then N is isomorphic to R/q, and }tll'(aj N) = }tR'(a; R/q) = }tw((a + q)/q). This equality and our induction applied to M/N proves the assertion by virtue of (2:3.:3). The following lemma is immediate from the definition: (2:1.()) If 0 is an ideal of R such that 0 c a and if a and 0 have the same radical (or, moregeneralty, if altitude 0 = altitude a), then}t",(a; M) :::; }tR'(o; M).
L.
(2:1.7) THEOREM: With 0 as above, assume that J.!(a) }tR'(aj M) = }tR'(o; M). Proof: By the pin (23.5), we have }tea)
=
Lp
}tea
=
}teo). Then
+ p/p) . length R p , }teo) =
Lp }t(o + p/p) . length Rp.
SinceJ.!(o + p/p);:::: }tea + p/p) by (23.6),}t(a) = }teo) implies that J.!(o + p/p) = }tea p/p) for every p. By (23.1) we see, in the same way, that J.!/?'(o + p/p) = i!w(a + p/p). Therefore, again by (23.5), we h:we the equality J.!R'(aj "~f) = }tR'(o; M).
+
77
1'11,11"1'1111[ III
WI' ndd 111'1'1' Llil' !'oll:;wili/o!: l'I'IIIIU'I,: If '" ~ /I'" 1I11t/ (: 111'1' lit ( fU1'11I '1'111(/ I~r N /1111/1/11 flll'lll //Ilit/llit' 4 II! ,1'1'81/1'l'Iil'I'/,I/, '11)illt '1'1'8/11'1'1 III n, I/II'//. 1l1",(Il; AI) Il""( /1'1"'; (il, op. all, III op. all, (/, 1I,lId ,ill. 'l)((,l'ti(IIi./U/I', aiLiLIIII t, and "w(a/aR; n) ;::: "R'(a; n)  AR' (a; n  t) for sufficiently large n by (22.6), and we prove (1). (2) and (3) follow from the last assertion in (22.6). We note, by the way, that if altitude alaR = altitude a, then J.!w (a/ aR) :::; J.!R' (a) by the definition of multiplicities (or by (23.3». Now we prove (24.1). We prove the assertion by induction on r. If r = 0, then the empty set is the required system of parameters. Assume that r ~ 1. Let al be a superficial element of a; the existence follows from (22.3). Assume at first that r = 1. Applying (2) ill (24.2), we have lengthR'R/aIR = J.!w(a/aIR) = J.!w(a) + lengthw(O: aIR). Since al is a superficial element of aIR by (22.4), we have by (22.6) that lengthR'R/aIR = J.!w(aIR) + lengthR'(O:aIR), and we have J.!w(a) = J.!w(a;R). Assume now thatr > 1. 'rhenJ.!/t,(a/aIR) = J.!w(a) hy (24.2). By induction, there is a system of parameters I '" ' a2, ' .. , arI of a/aiR such that J.!R,(a/aIR) = MR'( "L.. ai(R/adl». Let ai be elements of a such that a; = ai modulo aIR. We are to prove that al , ... , ar are the required elements. Since a; E a, we have MR'( LaiR) ;::: J.!w(a). On the other hand, by (1) in (24.2) we have J.!1l'( (L aiR)/aIR) ;::: J.!R'( L aiR), whence J.!w(a) = J.!R,(a/aIR) = J.!w(La~(R/aIR» = J.!wC(LaiR)/alR) ;::: J.!R,(LaiR), which implies that J.!w( L a;R) = J.!w(a), and the proof is complete. (24.3) We use the same notation as in (21.2). If a E at and if altitude a/aR = altitude aI, then Mw(a/aR; M/aM) ;::: t·,uu,(a; M). If al , ... , ar form a system of parameters of a and if ai E ani, then lengthwM/( L aiM) ;::: nl ... n r • J.!R,(a; M). Proof: (22.5) can be generalized to the case of modules by virtue of (1.6), and we have lengthwM/(an + aR)M = lengthR'M/anM lengthwM/(anM:aR). Since a E at, we have length M/(anM:aR) :::; length M/antM, and Aw(a/aR; M/aM; n) 2:: AR,(a; M; n) Aw(a; M; n  t), and we have the first result. By a repeated application of the first result, we see that lengthR'M/( aiM) = Mw(O; M/C L aiM» ~ nl ... nr·Mw(a; M), which proves the last assertion.
L
71l
( 'II" I "1'11111, III
(:JI.,I) lI'ill! Ih" N!I,~/I'III
of
Nil 1111'
UN O,bOl'(', ,1/ n 'iN Ihl'!/' J,!",,(Il; AI)
N, U' , 11, (/,nd M
/)(//,(/,1111'/1"1'1/ 01, '"
/I:
,
(/'",
I/I'/I.("/'(/'[I'd
bl/ (/'
lilllll';I1(",:) .,"
a;'iil! )/(111'" '/I.,,). (LIOMMA 01" LW:II) 1'!"Ool': 'l'hnC 1011I1:CII/,.,/I1 a;'iM ;:::: 'ILl'" nr'J,!W(!l; M) follow/,; 1'1'01\1 (:J,I.:n, alld wn hav~ lim inl" (lengthwM/L aiM)/nl ... nr ;:::: It"" (11; M ). Therefore, we have only to show that (IClII/l:UI1,,,M
/L
lim Nllp (IengthwM/( L afM) )/nl ... nr ::; J,!w(a; M).
L
IJd /I' = /1'" be the form ring of R with respect to a and let G = 0" be the form module of M with respect to a. Set Xi = 2 ((,l modulo a • Then the form module of M/( L a;iM) is a homolIIorphic: image of G/(L xtiG), whence longthwG/(L x;iG) ;:::: 1(~lIgt.hIl'M/( L a;iM), whence it is sufficient to show that
L
lim sup (lengthwG/( L x;iG) )/nl ... nr ::; Itw(F I ·/?; G), I)(~(:ause Itw(a; M) = ItR,(FI·F; G) by (23.8). Thus we may HNsume that F = R, M = G, and ai = Xi. R is then a homomorphic image of the polynomial ring in r indeterminates Xl , . " , XI' over Fo. We prove the ease where M = mR (m E Go) by induction on Inngth Fo. M is isomorphic to R/(O:M). If r = 0, then the assertion il'l obvious, and we assume that r > 0. Assume that length Fo = 1, 'i.e., a is maximal. Since Fo is a field in this case, the Xi must be algebraically independent over Fo because altitude a = r. If O:M = 0, Chen obviously M is isomorphic to the polynomial ring, Itw(a; M) length w Fo , and
longth w M/( =
L
x;iM)
=
lengthRI R/( L xfiR)
length w (Fo[xd/x?lFo[x])
(9 Po ••• (9 Po
=
(Fo[xrl/xrnrFo[xrD nl ... n r • length R, F o ,
und we have the required equality in thiN case. Assume that 0: M ~ 0, and let .r be a homogeneous form of degree, say 8, which is ill O:M (f T" 0). Thell M is a homomorphic image of R/JR. Then Icngthll' M / (L x;'iM) ::; length w R/(Lxi'iR)  lengthw(.fR + L xtiR)/( L Xl'iR). lcngth w R/ ( L Xfi R) = n l . . . nr length R, Fo as was shown above, and length w (fR
+L
x;'iR)/(L x;iR)
=
length w R/((
L xi'iR) :JR)
by (1.5). Since Fo is a field, (L xtiR) :!R C L X;'i R and therefore length R, R/ ( ( LXi" R) :!R) is not smaller than 8
HO
MIII,'I'll'ldlll'I'lllili
N) • ,. (II, TItIIH
Iilll
HlIll
(1(~llgLh M / (
L
a. W(~ lIav(~ 11'llgth /(/11 = /I({',/') C~ p.(l',/'; II) p.(I',/'; F) by (:~:~.:)). Obviow;Iy, P.(J'IJ>; F) iH 1'1111:1,1 to M( .r/P), whidl iH eqlml to 101(0.) by (2:3.8). Therefore It'IIJ,!;1.11 N/n 101(11) if and ollly if p.(J\J>; n) = O. Assume that n 7"" o tl.lld Id J be a homogeneous element of n. Multiplying a suitable llil'lnclI!; or Po if necessary, we may assume that fm = 0 with a IIIlI,ximaJ ideal l1l of Po. Then O:fP = mP, and fP is isomorphic to 1'/1111>, whenceM(PIP; 11) 2: M(PIP; fP) = 101 (PIP; (Po/m) [XI, ... , X,.I) = 1. Therefore, M(PIP; n) = 0 if and only if n = 0, and the tl.H,~I~l'tion is proved. We have by (25.1) and (25.2) the following: illti!"j,I'I·llIillal.l'K .\', , .,. ,
L
+
7=
(:25.3) COROLLARY: If a semilocal ring R has a distinct system of 7111:mmeters al , ... , aT of the Jacobson radical of R, then R is unmixed.
(21).4) THEOREM: A system of parameters al , ... , aT of the J acoblion r'adical of a semilocal ring R is distinct if and only if ai is not a
~I'I'O divisor modulo L~I ajR for each i = 1, 2, ... , r. Proof: Set 0. = L~ aiR. We prove the assertion by induction on r.
II' r = 0, then the assertion is obvious because the empty set is a diHLinct system of parameters of a ring of altitude zero. Assume that 'I' ::::: 1. Assume at first that the ai form a distinct system of parameters. Thon, length R/a 2: M(a/aIR) 2: p.(a) by (24.3), whence the equaliCieH hold, and the ai modulo aIR (i = 2, ... , r) form a distinct H'yHtem of parameters of R/aIR. Therefore, by induction, ai is not a v,nl'O divisor modulo L~I ajR for each i = 2, ... , r. Thus it is suffiI'ient to show that al is not a zero divisor, which is obvious by virtue of (25.2). Conve~sely, assume that each ai is not a zero divisor modulo I ajR. Then al is not a zero divisor and, by induction, the ai modulo aIR (i 2: 2) form a distinct system of parameters of R/aIR. L(~t F = L F n be the form ring of R with respect to 0., let XI, ... ,Xr 1)(\ indeterminates and lete/> bethe homomorphism from P = (R/a) [Xl 2 Ollto F such that e/>(X i ) = ai modulo 0. • Since the ai modulo aIR U 2: 2) form a distinct system of parameters of R/ aIR, the form ring /1'* of R/aIR with respect to a/aIR may be identified with P/XIP,
Li
HI
~1111
11'11'1 ,\1 '1'1'1 WI
wllil'lI HiIlIWH 1,1111.1. 1.11(' 111'1'11(,1 II III' (I' in 1'!llIin.illl,d ill .\' 1/' /1,11(1 Llia!. /1"1' idl'lliili('d wilit /I'/viollK thuJ, (I) implinK (:2). AKHlllllP (,haL (~)
holds good. H iN i'Hlff1(:inllt (,0 show that R is a Mal:au'lay ring. Ln(, rat, ... , ar \ be a distinet system of parameters and set It = a,;R. Theil !l(a) = length R/a = L length Rn/aR n , where 11 rum, through all prime divisors of a. On the other hand, by (23.1), )L( a) = L !l( aR n,), where 11' runs through all prime divisors of a such that height 11' = r. Rince length Rn./aR n, 2: !l(aR n,) by (24.3), we see that height 11 = r for any 11 and that the ai form a distinct system of parameters of R" . Hence (2) implies that height 111 = height 111' and that Rm , R"" are Macaulay local rings for any two maximal ideals m and m', whence R is a Macaulay ring. Thus the proof is completed. The following remark is immediate from the definition: (25.8) A semilocal ring R is a Macaulay ring or a locally Macaulay ring if and only if so is the completion of R. On the other hand: (25.9) If R is a locally Macaulay ring, then every ring of quotients of R is also a locally Macaulay ring. Proof: By the definition, it is sufficient to prove that Rp is a Macaulay ring for any prime ideal p of R. Let r be the height of ).l and let al , ... , a,. be elements of ).l such that Lf ajR is of height s for any s :::;; r (by (9.5». Since the unmixedness theorem holds in R, ai is not a zero divisor modulo L~l ajR for each i, and therefore the ai form a distinct system of parameters of Rp by virtue of (25.4), which proves the assertion. The following is a generalization of the classical unmixedness theorem, because a field is obviously a Macaulay ring:
L
(25.10) THEOREM: Let Xl, '" , Xn be algebraically independent elements over a Noetherian ring R. If R is a locally Macaulay ring then so is the polynomial ring R[x). Proof: The general case follows easily from the case where n = ], and we assume that n = 1. Xl is denoted by x. It is sufficient to prove that if m is a maximal ideal of R[x), then R[xl m is a Macaulay loeal ring. Set \l = m R. Then Rp is a Macaulay loeal ring. In order to
n
'/J,7
I'll ;\1"1' 1 m, there is a relation II.j = Lf=1 aijVj , whence the relation module N of the Vi contains ILII element fi = Vi aij V j . Since the V,: are free, we see that )'11/+1, '" ,fn generate a free submodule N* of N. On the other hand, N contains the relation module N** of VI, •.. , Vm , and N** is the I'olation module of lYI. We are to prove that N is the direct sum of N** and N*. It is obvious that N* N** = O. Therefore it is sufficient 1,0 show that N is generated by N* and N**. Let g = L biV i be 1111 arbitrary element of N. Then 9  L;':+1 b,ji is in RV1 RVm , whence the element is in N**. Thus N = N* + N** and the /tssertion is proved.
L:7'
n
+ ... +
[91]
'1'111'1 '1'1111111111' 1)/1'
By Vil'Llln of (~(;'I), WIII'Ii ,,~ iH
/I,
1\,,~,\'IIIil:r\
IOI'/I,II'ill/l: lVI' 1~/l.1I "1'lilll' 1.111' lIoLioli
of lJyzygie8 of fini(.c modl/II~ Hi"! I'ollowi"!: Let R be a local ring and let M hl~ a filliLn NIIIOd"II'. Tlan mil i"!Y:'.yJ.!;Y of M is M itself; when the ith syzygy Zi of M iH ddillnd, Lito (i + I) Lia syzygy of M is the relation module of Z i. The iLh Hy:'.ygy of 111 i,.., denoted by syzi M; if we want to express that M is an RmoduIC', WI' write syz~ M. Even if R is not a local ring, we can give the following notion of a weak syzygy sequence of a finite Rmodule JJ![: A sequence of finitl~ Rmodules Zi (i = 0, 1, 2, '" ) is a weak syzygy sequence of M if Zo = M and if each Zi( i 2:: 1) is the relation module of a finite basis for Zil . Zi is called a weak ith syzygy of M. (26.2) If M = Zo, ZI, ... , Zn, .. is a weak syzygy sequence of a finite module M, then, with an arbitrary, multiplicatively closed subset S of R such that ~ S, Zo ® Rs , ZI ® R s , .. , , Zn ® Rs, ... is a weak syzygy sequence of M ® Rs over Rs . Proof: Let 11,1, •• , , Un be a basis for Zm such that the relation module of the Uj is Zm+l . Let N be the relation module of the
°
Ui
® 1
(1 E R,s).
It is obvious that Zm+l ® Rs is contained in N. In order to prove the converse inclusion, let La~ Ui be an el(ement of N, i.e.,
Since the elements of S are mapped to units, we may assume that a; are images of elements ai of R. Then L a:(u ® 1) = implies that L aiUi is in the kernel K of the natural map from Minto M ® Rs , whence there is an element s of S such that L a;sui = 0, and L Ui E Zm+l ® Rs . Thus N = Z",+1 ® Rs , and the assertion is proved. (26.3) When R is a local ring, any weak itk syzygy Z of a finite module M is the direct sum of the ith syzygy of M and a free module, hence the relation module of Z coincides with syzi+l M. The proof is immediate from (26.1). From now on, let R be a local ring. A finite module M is said to be of homological dimension n (over R) if n + 1 is the smallest i such that syzi M = 0. In symbols, we write hd M. = n; when we want to express "over R" explicitly, we write hd R • The definition implies that: (26.4) Let M be a finite module over the local ring R. Then: (1)
°
a;
if fll/d (llIluif III
1111 III
0,
(~)
lid M
() '(/' (t'Iul only if M
(:l) if M ~ 0 and if F is a free //lOr/II/I', 1/11'11 II(I M = hd(M E9 F), (4) hd(syzi M) = max(l, lid III  ',;) anti (Ii) ~f Z is a weak ith syzygy of M then hd Z is equal /(1 hd M  i for i ::; hd M; it is equal to either or 1 for other i. III
II ,1'1'1'1' 1I1I1//ltll' tll)il'/'('II/ fl'/I/II, Z(:/'II,
°
(:W.Ii) THEOHEM: If a submodule N of a finite module M is given, Ihl'//', for each n, there is a weak nth syzygy Zn of M and an isomorphism If from syzn N into Zn such that ZnlCT(syzn N) is isomorphic to
syzn (MIN). I'roof: By the definition of syzygies, we have only to prove the case wltnrc n = 1. Let nl , ... ,nr be a minimal base of N and let m~ , ... , , .. , m: be a minimal base of MIN. Let mi be an clement of M such l,hnL m: = mi modulo N. Let Zl be the relation module of the base nl, ... , n r , ml, .. , , m., of M. L aiMi E syzl (MIN) if and only II' L a,m: = 0, that is, L aim, E N, or equivalently, L aimi + bjnj = with some lij E ii, whence syzl (MIN) is the image of :11 by the map ¢ such that ¢( aiMi + L bjNj) = L a;lYf; . The Iwmel of ¢ is the set of clements of Zl of the form L bjNj , which is obviously the relation module of N, i.e., syzl N. Thus the proof is I'omplete.
2:
°
L:
(26.6) COHOLLAHY: Among the homological dimensions of N, M, and M IN, one of the following must hold: (1) hd N ::; hd M = hd( MIN), (~) hd 111 ::; hd N = hd(MIN)  1, or (3) hd(MIN) < hd M lid N. Proof: Let Z" be a weak syzygy of lYI sueh that Zn/syz n N ~'yzn (MIN). Case I : ASimmethat hd N is minimal among homologieal dimensions of M, N, MIN, and let n be hd N. Then Zn+l is isomorphic to syzn+l (MIN). Therefore, either hd M = hd (MIN) 01' ~'yzn+1 AI = 0. In the latter ease Zn+l is a free module, and therefore we see that, in ease I, either (1) or (2) holds. Case II: Assume that hd M is minimal among the homological dimensions of M, N, MIN, nnd set m = hd M. Since syzm M is free, syzm N is the relation module oj' a base of syzm (M)N), henee either both syzm (MIN) and syzm N nrcfree and different from zero or hd (syzm N) = hd (syzm+l (MIN)), nnd therci'ore either (1) or (2) holds. Case III: The remaining case iH where hd (MIN) is smaller than both hd Nand hd M. Set n' = hd (MIN). Then Zn'+l/syzn'+l N = 0, i.e., Zn'+l = syzn"I'1 N. By
III
'1'11111 '1'1111111111' 1111' IIH,I'IIIIIIII
11111' II,HHI II II pl,ioll
wn ItIl.V(~
!.ImL hd (II/ IN)
iH 1111111,\\1'1'1,\11\.11\1111,,,
lid N /1.1111 ltd nI,
(:1). '/'ItII14 I,hl' prolll' iH 1·lIllIpll'I",.
11;~(II:ItI'INI':H: LI1L
U 111,11, N"dhnrill,lI l'illJl; II.lld 1'''"Hid,,1' Iillil.n 11i",IiIl"H OVIll' ,,~. modliin Ai iH II, '1II'IIjIW/'il'l: 'lllIIt/III" il' M Iiii N", iH 1'1'111, 1'01' IUlY maximal idlJILl 111 or u. Th" hOl//,o[o!/,£I:/J,1 tlhll{;II,,~io/(, of II IIIOdlll" M iH d"fillnd 1.0 lie Lhe maximum (or tho HII\lrmnllm) Ill' 1i0IllOlogil::d R' is exact. Prove that Zo 18> R', Z, 0 R', ... ,Zn 0 R', '" is a weak syzygy sequence of M 0 R'. .
W"
HHy LIla" 1J,
27. Change oj Rings (27.1) Tm]OREM: Let M be a finite module over a local ring (R, m) and let x be an element oj m. Assume that x is not a zero divisor with respect to both Rand M. Then for every natural number n, there is a natural isomorphism between (syz n M) I x (syz n M) and SYZ~/XR (MlxM).
Proof: Using the induction on n, we have only to prove the caSe where n = 1. Let Ul, ... , Ur be a minimal base of M. Then u' = Ui modulo xM form a minimal base of MlxM. L a:U i E syzl (MlxM) if and only if L a:u~ = 0, i.e., L aiUi E xM with ai such that a: = ai modulo xR, namely L aiUi = L xbiui. Thus the map ¢ such that ¢( a,U i ) = (ai modulo xR) U i is a homomorphism from syzl M onto SYZ1/XR MlxM. The kernel of ¢ is the set of elements xaiUi such that xaiUi = 0, whence aiUi = because x is not a zero divisor with respect to M. Th us the kernel of ¢ is x (syz 1 lrf), and the assertion is proved. We say that a sequence of elements Xl, ... ,X r of the maximal ideal m of a local ring R is an M sequence, M being an Rmodule, if Xi is not a zero divisor with respect to M/L~l xjM for each i = 1, '" , r. Then we have:
L
L
L
L
L
°
(27.2) COROLLAIty: Let M be aflnite module over a local ring (R, m). If Xl, '" , Xs is an I11sequence and at the same time an Rsequence, then hd R }J![ = hdR/oM I aM with II = L xiR.
('III 1"1'1'111 I V
(~7,:11 'l'1111:IIIII':M:
/,rl ill III' (/ ./illill' IIlIIdll/I'
(/1/1//1'1,/'111' II//I'!I'IIII'III
Id
III /I'hil'" iN
/101 ill
III'/'I' /I /lIl'Itll'ill(/
(/t', 1111
III", ,lssllI/lI'.I'III'I!lI'l'lIllil'I'I!Jn{
/11111/ '::/'I'II dipislIl' ':/1. /i', Ihol. M i8 l'onlrt.i/II'r!':1I I/, .1'/'1'/1 U/.oduk F ,,~{ Ii ( {I" ' , , , (I,. Iwi/l(/ I/, fl'(,(, 111/,81 1 I~( /1') IUUf, thai, ,rF ~ JI,l ~ 111/1'. '/'lil' II , fol' I'lli'll. I/I/./I/i/,(/./ 1/.1/./11111"1'1/., HYY." ill m'fl, hi: imlwddr:d in a Iter: II/Udllk /1'" '£1/. SI/.(lh (!, 'W(/,I! Uw,/. :r/I'n C HYII" A;j C lIl!?n and such that (HYY." JI/) /.d l '" '/:s nalu,raUy isomorphic to SYII{;lxR (M / xF). Pl'oof: lJHill1!; illdudion on n, we have only to prove the case where I/. "= I. Lei; a~, ... ,a; be a minimal basis for M/xF and let aI, ... ,as
,/' iN
L
1)(1 elements of M sueh that a~ = ai modulo xF. Then aI, .. , , as, .rl1, , .. , , xUr is a basis for M. They really form a minimal basis for M, since otherwise there would be a relation Clal + .,. + csa s + d,:tU l + ... + drxU r = 0 with some Ci EE m or some d j EE m. If some (:,: ~: )11, then this contradicts the fact that the a~ form a minimal basis for M/xF; if some d j EE m, then the coefficient of U j in L Ciai is not ill )112, whence some Ci EE m (because M emF) which is not the case. Thus the ai and the xU j form a minimal basis for M. syzl M is therefore the relation module of aI, .. , , a r , x UI , .. , , X Us. L ciA i + L bjVj E syzl M if and only if L Ciai + L bjxU j = O. Let (J be Lhe map from syzl Minto FI = L RA i such that (J( L CiAi + L bj V j) = L ciA i . This (J is obviously a homomorphism. Element of the kernel of (J is bjVj with bjxUj = O. Since x is not a zero divisor and since Uj are free, we have bj = 0 and therefore (J is an isomorphism. Since the ai and the xU j form a minimal basis, we have (J(syzl M) C mFl . Since xai E xF, we have xAi E (J(syzl M). Thus :lPI ~ (J(syzl M) C mF I . L c;A i E syzlAxR (M/xF) if and only if L cA i E xF with Ci such that c; = Ci modulo xR, or equivalently L ciAi E (J(syzl M). Thus we see that
L:
syzAlxR (M/xF)
L:
=
(J(syzl M)/xFl,
and the assertion is proved completely. (27.4) COROLLARY : With the same notation and assumptions as in (27.3) above, we have hd R M = 1 hd RlxR M/xF, except for the case
+
where F = O. Proof: If both hd M and hd M / xF are infinite, then the assertion is obvious. Therefore, we assume that one of them is finite and we prove Lhe corollary by induction on the finite homological dimension. hd RlxR 111 /xF = 1 if and only if M /xF = 0, i.e., M = :rF, or equivnlently hd M = 0 because, in the notation in the proof of (27.3) the
II(I (/, Hlld
'1'1111\ '1'111,1(111\' 1. Then the union of the sets (If zero divisors with respect to M I xlM and M I ylM does not cover 111, whence there is an element z such that both Xl, Z and Yl, z form MHnquences. Let Xl, Z, ZR, " . , Z t and Yl, Z, W3, ,." w" be maximal Msequences. Considering M IxdYI, we have r = t by induction, ConHidering M/zM, we have t = u, and, considering Mly1M, we have 'I/, = 8. Thus r = s, which completes the proof. EXERCISES: Let M be a finite module over a Noetherian ring and let a be an illeal of R. Then the notion of Msequence in a is defined simibrly using a inHtead of m in the previous definition. 1. Confirm that, under the assumption that M laM T" 0, (27.6) and (27.8) (Jan be genemlized to such a case. 2. Prove (27.2) when R is a semilocal ring and when the Xi are in the Jacob~on mdical. 3. Adapt (27.4) to the case of semilocal rings.
11K
'1'111'1 'I'll 1':01/ \
11/" i1 \ 'f,\ 1111'11 \
2M. U('lllllm' IO('t11 1'I1I1l.\' Wn Il('gili wiLli /I, 11'111111/1,: (::lKI) I,d. (N, 11/) 1)(: a {,I/(:(/,( 'rill!!. If 0: III / 0, /Ju:n hd M ,D fll/' eveTY finite mod'ute M which i8 'not II'I'(:. Proof: Assume that hd M = n < 00 amI Chat n ?:. I. 'I'lwll :..;yll /I III is free and different from zero. Since syz n M i::; j,he l'elaLiol\ modltin or syznl ].II, there is a free module F such that syzn M ~ lllf?, wh(~II(',n (O:m)(syzn M) = 0, which is a contradiction. Now we prove: (28.2) THEOREM: Let (R, m) be a local Ting. If hd m is finite, then R is TegulaT. ConveTsely, if R is a TegulaT local Ting, then faT eveTY finite Rmodule M diffeTent fTom zeTa, we have hd M =c altitude R  s, wheTe s is the length of a maximal M sequence. Proof: Assume that hd m is finite. We prove the regularity of R by induction on r = altitude R. We begin with the following remark: If O:m ,L 0, then m must be free by (28.1), whence m = 0 because 0: m ,L O. Therefore, R is a field in this case. Now, if l' = 0, then 0: m ,L 0, and the assertion is proved already. Assume that l' > O. Then by the above remark, we have 0: m = O. Therefore, there is an element x of m which is not in m2 such that x is not a zero divisor. Then hd m = hd R / xR m/xR 1 by (27.4), whence RfxR is regular by induction. Therefore we see that R is also a regular local ring by (9.11). Conversely, assume that R is a regular local ring and let M be a finite Rmodule different from zero. Let Xl , ... , Xs be a maximal Msequence. We may assume, by (27.7), that Xs ~ m2 if 8 ?:. 1. We prove the equality hd M = l'  S (1' = altitude R) by induction on 1'. If l' = 0, then R is a field, and M is free (8 is obviously zero). Thus this case is obvious. Assume that l' ?:. 1. If s ?:. 1, then hd M = hd R / xsR M/::c M by (27.2), whence by induction hd M = (1'  1) (8  1) = l'  s. If s = 0, then M is not free, whence syzI M ,L 0 and hd M = 1 hd syzl M. Since syzl M is contained in a free module, y E m (y ,L 0) forms an (syzl M)sequence; if it is maximal, then by the above proof, we have hd (syzl M) = l'  1, and hd M = r. Thus, it is sufficient to prove that every element z of m is a zero divisor with respect to (syzl M)/y(syzl M). Let F be a free module such that F/(syzl M) "" M. Since z is a zero divisor with respect to M, there is an element a of F which is not in syzl M such that za E syzI M, whence yza E y(syzl M). But ya ~ y(syzl M) because y is
+
8
+
('11111"1'11111, IV 1111(, II, Y,PI'I)
divir4ol' II.lld
111'1'/1,111'1(11/
II HY If, , II!,
'l'IH'I'PI'OI'!', .:: iH
/I,
~,(II'O IIi
ViNOI' willi IW1(1('1'I, 1,0 (N.v~,1 III )/!I(H.v~,' /1/ l, 1I,lId LlHI limo\' iH "Olilpil·l,ll,
(:.lH,:I) (!O/l.Old,AI1I': IJ' p /8 (/,fII'/I/I/' il!('(/l /~r /I /'I'(/lIlu/, lul'/l.!, '/'/11(/ h', h\, ':8 //, 'I'I'l/n{(I./' lo/'/r!, '1"1:11(/, "I'OO\': lid p iH 1;lIil.(', wlH,tll:n 1111/"" ~>Iip iH lillil.n I,.v (~o,:.n, wllidl
IIM'/I.
PI'OV(!H LlI(: Il.HH(!I'I.iOlI.
Lnl. (/(" Ill) h(: a I'n~I\III,J' IO(Ial J'illp; :l,lullnL 71 h(: 1.11(: cilal':I.(,i,(!I·iHI.iI' oi' iH (:allnd an 'Il'f/,r'(un'!j,:/,r/ '/'c(J'I1.lllr local r"';n.(j if nil.l\(\l· f{, ('OI\(,II,iIIH II,
U/lli.
n
P q:
2
• Namely, H iH 1I1lmmilicd if H.l\d only if H/p!?' iH I' 0, we see Chat hd syzl M + 1 = t, and hd M = t, which completes the proof. 1'~XERCISES:
1. Let a be an ideal of a regular local ring R. Prove that if
p, , .. , , lJn are all of the prime divisors of a, then hd a is not greater t.han the
IIItLximum of the depth lJi + 1. 2. Let R be a Noetherian ring. Prove the equivaJence of the following condiI,ions: (1) hd M :::; n for every finit.e module M, (2) R is a regular ring such that IIHitude R ;;; n. 3. Assume that M is a finite Riamodule (a being an ideal of a local ring Il). I't'Ove that hd Rla M + hd R Ria = hd R M, provided that each of them is tillite.
29. Syzygies of graded modules Since there i8 a very close relation between graded ring::; and semilocal rings or between graded rings over loc:al ring" and local rings, we Itdd some remarks on syzygies of graded aodule". Let R = L Rn he a graded Noetherian ring sueh that Ro is a 10c:a1
III',',
'11111\ '1'111 11(111\' 1111'
~\\1,\1111'1I\
L"
I'illg willi 11111.\illlll.l idl'/I,I 1111), HI'I III 1111) I I h'", III iN oill'iollNly 11, Ill/I,xil1\1d ilil'/I,\. 1.1'1 M 1)(' H lillill' /l;I'HIIi'd 11111111111', WI~ d('(illl' Ii, NIP/lllli ,~/'ql//'I//'/' or III /1,,'4 rollnw,'4: TII(~ lil'HL 1111'1111)('1', wllidl IIlay 1)/· d/'liol.l~d I,y HYY,I) M, iN 11/ il.NI'IL WIII'II Clip ",1,11 1111'111111'1' NYX" I 11/, whidl iN a gl'/Likd [illil.l~ m()dllll~, iH ddilll'd, WI' ddill(\ !.Iw
+
('II. I )NC Il\mnl>I~1' NYX" 11/ aN folioWN: Ldo 'UI , '" , U r IH~ a Illillimal IIaNiN for NyxlI I M (',onNi,,('ing only of homogeneollN nli~In r • • . , br ' _ p "[Kpnl(1 . II, . . . , b) r''K  ] _ p' p (n1)·,. . 1'1lllS'
WI~
prove (31.~). We note that (:n.2) implies that every element of K is expressed as pn II. polynomial in elements of B with coefficients in K in such a way Chat the degree of the expression is less than pn for each member of B /l,lId that such an expression is uniquely determined by the element or K. Next we prove another lemma: (31.3) Let p be a prime number and assume that an ideal m of a ring
pn  b Em, then a Il" E m n+1. Conseq1lently, if, furthermore, m t = 0 and if m is a maximal 'I:deal, then the map ¢ such that ¢( a) = a P' with s 1 2: t induces a oneone map from the field Rim into R. q q 1 Proof: Set c = b  a, q = pn. Then bq = a qa  c q (;) aq"c r c . If r = ptr', (p, r') = 1, then a simple calcula/?, contains p (i.e., the pfold of the identity).
If a
+
+
+ '" +
+ .. , +
Lion shows that (;) is a multiple of pnt, whence (;) E m,,t. Since
> t, c" E mt+\ and therefore the assertion is proved. Now we proeeed with the proof of (31.1). Let B* be a pbase of f( = Rim and fix a set B of representatives of B* in R (i.e., we take only one b E R for each b* E B* such that b* = b modulo In, and B 7)1
IOH
'1'1111:(11/1
1111'
I'II~II'I"'I'I'II:
1,111'\1, IIINIIII
iN 1.111' Hi'1. 01' 11111,1111),1 1'01'1'11.1'11 I 111,1.11 I'HI 1111111111'1' 1/ 11'1. 'i'" III' 1.111' IIlnp 1'1'11111 NIIII" illl.o il.HI,11' giVI'1i I,y 'i'" (0. 1 (/,I!''', '1'111'11 'i'" illillll'I'H I/. lilli' IllIn Il\ap 1'/'11111 NIIII illl.o h~/III" h.v (;\1.;\); I.Ili:: lilli' olin Iliap iN ill'III1LI'iI hy ITn' Wn (kIlIlLI~ by II n 1.111' illlngl' III' H/llI by (/" ' 1.1'1, 8" III~ Lill' NI'I, 01' polynomialN ill einnll'nLN III' /) I.aknll \l\lIdulo III" wiLli codli('.i(,IIL:: ill /1" such that the degree of 1,he polynomial ill n11.M (;}~rd~ )1\11 i8in pHlJn
by induction because CM, d M are in Sn. Therefore L pi(aM
+ h)M =
L pi(CM
+
dM)aJV[
+
piC Lr,M
(;)C~rri~)M
is in piJn. Thus, the above assertion ii:> proved, and we see that I n is a ring. Obviously J n ml 111 n = pJ" , whence pJn is a maximal ideal of I n • The oneone correspondence between Sn and Rim now induces the natural isomorphism between JnlpJ n and Rim. We want to show
n
I'IIA 1"1'1'111 V
lOll
II(\~L LlIII.I, LlII' 1l/I,i,III'n1110IIIOllllll'pltitHIl II" 1'1'0111 U/III" Olli,o "~/I"" I ill 11,1111,1.111'11,1 hOlllOlllOl'pliiHl1i 1'1'0111 .I" 0111.0.1" I, II, iH ohvioliH UmL 1r,,( III ) l'I'I'ol\ll'H /1.11 ('11'11\('111, 01' ,','" I ,111'111"(' 7r,,( 8,,) C:.I" I, whieh implil'H LIm!. '71'" iH H h01\1011IOl'phiH1l1 1'1'0111 .1/1 illto .1/11. On the other hnlld, iL iH ol,violiH hy Lh" nOIlHLI'llId,ioli !.ImL (~adl element of 8 n  1 is in 71",,(,'1,,), whidl implinH Lh:1.L 71",,(.1,,) l,onLaintl '/nI' Therefore 7I"n in"III'('M a lIaLlIl'al hOIllOIlIIlI'phiHIlI 1'1'0111./" onto '/nl' Now, let {an} be a M(\II'II'III'(: of finite over R, by (3,1), Thus we may assume Lhat L = L', Let q = [L:Kl (q is a power of p) and set I* = Ilia, q i/.,: = x~lq, R* = I*[[Yl, '" , Yrll, If a E R', then a E R, whence U' C R*, which implies that every element a of R' is uniquely expressed as a power series in the Yi with coefficients in I*, and in that Hense we define the leading form of an element of R', We want to show first that when aI, ". , as are elements of R' such that the leading forms /I, .,. , I" or these elements are linearly independent over R, then aI, ... , as are linearly independent over R. Indeed, if ~ aio.i (Ii i E R) is a nontrivial linear combination, then the leading form of L aib i it:: a nontrivial linear combination of the Ii which cannot be lIero because of their linear independency. Thus aI, ... ,as are linearly independent. Since L is finite over K, we see that there are elements aI, ". , as of R' with leading forms iI, .,. , is such that if a is an dement of R', then the leading form i of a is linearly dependent on II, .. , , is . Let Cl, . , ' , Ct be the set of coefficients of 11, .. , , Is . Then the eoefficients of 1 are in I(cl, ... , Ct). Therefore, if do = 1, dl , . , . , du is a linear base of I (Cl' . , . , C t) over I and if mo = 1, ml, .. , , mv is the set of monomials in the Yi of degree less than rq, then 1 is in the module L Rmid j , whence the module M generated by leading forms of elements of R' is a finite Rmodule; let gl , ... , gw be a base of M such that each gj is the leading form of an element bj of R' and let R" = R[b 1 , • , • , bwl. R" is a finite Rmodule, whence it is complete. Therefore, R" is a subspace of both R* and R' by (30.2). Let d be an element of R', We want to show by induction on n that there is a sequence I dn} of clements of R" such that d  dn has leading degree not less than n, We may start with do = O. If do, ... , dn are already defined, let f be the leading form of d  d71 . We can write I = L hjg j with hj E R, and we may assume that h j are homogeneous forms, whence d* = L h)}j has the same leading form f, and Lhe1'e
III
'1'111'1111/\ 01" l'IIMI'",q'I'I'! 1,111' II, IIINIIi!
tI" I t/'" ill LI", l'I''IlIil'l'd ,,1, ' 111,'111, '1'111111, 1,111' 1'~ild"'III'I' IIi' 1.111' 111'11\'1'11. ~illl'(' It'" ill /I, IlltlllIPHI'(' IIi' 11"" 1.111' HI'IIiII'III'I' iH if, (!/I,I/('hy HI'ql/('III',y ill It'''. HilWI' U" ill ('111111'1('1.1', LlII' H('(II/I'III'I' 1111.:1 lI. liJlli!. tI" ill /(,". tI 1/" IW('II.lIHI' U" iH II. HI IItHIIII.I',I' III' h~', Whi('11 illllllil'H !.Im!. U' ~ U", :1.111\ N' iH:I, lilli!.n 1(lIllIillIll'. Wn NlI.y LIm!. a HI'milo(',al rill/!: R iN wnn{lIli(:u,Uy '11:n;/,(/;It/,'/jil'd ir (,\II~ ('.olllpll'Lioll N* IIi' N linK ])0 nilpoLcl\L e1mncl\i; (,xenpL )ICI'O; all i(i 1, and set B = I[xd/(q I[Xl]) , ~' = 111 B. Then altitude B~, + trans. degI/wnI) B/~' = altitude I(~'nI) + trans. deg[ B by the case n = 1, and altitude R + trans. degB/p' R/m = altitude B p' + trans. degB R. Since m 1 = ~' I, we prove the formula. Thus (35.5) is proved completely. As an application of (35.5) to general Noetherian integral domains, we can prove the following theorem:
n
n
n
n
(35.6) THEOREM. Let A be an affine ring over a Noetherian integral domain I and set r = trans. degI A. If ~ is a prime ideal of I and if ~' is a minimal prime divisor of ~A such that ~ = ~' n I, then trans. degI/~ A/~'
2:
r.
Proof. Let S be the complement of ~ in 1. Then considering Is and As, we may assume that (I, ~) is a local ring. We prove the assertion by double induction on height ~ and the number of generators of A over I. If height ~ :::; 1, then by our assumption, I is a Macaulay ring, whence by the altitude formula we prove the assertion in this case. Therefore, we assume that height ~ ~ 2. If height ~' T'" 1, then let q' be a prime ideal of A sueh that q' c ~', q' I F 0 and such that height q' = 1. Sinc:e ~' is a minimal prime divisor of ~A, we see that q' n I F ~. Sinec height q' = 1, q' ii:l a minimal prime divisor of (q I)A, whence by our induction on height ~, applied to q' I,
n
n
n
1'1111"1'1111/ 1'1
It'/q'
by 0111' illlilll',l.illll H,c,HI'I'l,illll iH 1.1'111' ill 1,liiN l'nNI', 11'1' 1I~lclllllll' LlmL IlI'iglil, ~l' I, LI'C ,1'1, , , , , ,I'", !In a H(~L of gl'III'l'al.ol'~ oi' ,I OVI'I' /, Ii' ,Ihl' iH 1101. algl'l>l'ai(~ OV(H' /Ip, Hay if ,(;, IIloilltio p' iH IIO!, algidH'ail" OVI~I' I Ip, U)(~II we eonHider A" = I[xI) Hlld ~I" ~I' n /1", Hilwe :1:1 modulo p" is transcendental over lip, W(~ HI~n I.haL ~l" ~c, pll" and Chat height p" :::;: height p by the altitude Lllnol'(~rn of Krllll :llld by the fact that a system of parameters of I~ gnncml.cH an ideal which is primary to p"A"~" in A"~", whence by our indudion on n applied to A over A" with the prime ideal p", we Hne i1wt trans, degA"w Alp' 2: trans, degA" A, Since trans, degrr~ II" Ip" = 1 and since trans, degr A" :::;: 1, we prove the assertion in Clii::; case. Therefore, we assume that Alp' is algebraic over lip. It Huffices to prove now that A is algebraic over I. Let A * be the derived Ilormal ring of A and let p* be a prime ideal of A * which lies over p', Then height p* = 1 by the goingup theorem, Since A * is a Krull I'ing by (;33,10), A:, i:,; a Noetherian valuation ring. Let K be the field of quotients of I and set B = A:, K, B is a Noetherian valuation ring by (33.7), Set C = B[A); this last ring is an affine ring over B, Set furthermore p** = p*A;. n Band p" = p*A;. n C, and let C* be the derived normal ring of C. Then obviously A * c C* C A;. , whence is a ring of quotients of C*, that is, = C: with r = p* A;, C*. It follows that r is a minimal prime divisor of p**C* and height r = 1. Since B contains I, Blp** contains lip, whence A;,/p*A;, is algebraic over Blp**. Therefore C*/r is algebraic over Blp**, whence C* is algebraic over B by (35,5) and by the fact that B is a Macaulay ring, which proves that A is algebraic over I. Thus the proof is complete, 1\'1' 111'(' 11111,1, 11'11.1111. dI'IJ:'il'l'III)
npplil'd III 'l'III'I'I'i'III'I',
P,'(
q' () /),
11'1' ~II'I' /1111.1
I'. '1'111'1'1·1'111'1',
/111'
n
A:.
n
A:.
36. Pseudogeometric rings We say that a ring R is a pseudogeometric ring if R is Noetherian and if, for every prime ideal p of R, Rip satisfies the finiteness condition for integral extensions. We note first that: (36,1) If R is a pseudogcometric ring, then every homomorphic image of R, every ring of quotients of R, and every ring which is a finite module over R arc pscwiogeometric rings, The proof i:,; :,;tmightfol'ward. (::)6.2) Let R be a semilocal integral domain and assume that p is a
I :I:~ Will/I' i"I'II/ I!!, /iI'i(/h/ I ill II' NIII'1i 111111 /i'p ill /I /11/11/1/1/111/ I'illfl· If P /N al/.oJy/.im.lly 1/1/.1'/('lIlIjil'" 1/.1/." if pl' i.~ II IIli/li/l/1l1 11,.//1/1' dillisor I!f' pH'f, R* being tlw (;mnJltl'l'ill/l, I!!, N, thl'//' h~I;' /8 1/ /111///(/,/.//11/. rin!/. Proof. Let w be an dement or \J whidl iH IIO!, iii p~N~\ . 'I'h(~11 pNp
wR~ , whence p*R;. = wR:• . Since w i" not a J1pr() diviHOt' ill N* hy (18.1), we see that is a valuation ring by (12.1). (36.3) Let R be a semilocal integral domain and let x (~O) be an element oj the Jacobson radical of R. Assume that xR has no imbedded prime divisor and that, for every prime divisor P of xR, P is analytically unramified and R~ is a valuation ring. Then R is analytically unramified. Proof. Let R* be the completion of R and let the prime divisors of xR be PI , ... , pr . By our assumption, and by (36.2), if the prime divisors of p;1l* are p7j (j = 1, ... ,n(i)), then eachp7j contains a prime divisor lj3;j of zero which is the kernel of the natural homomorphism from R* into the valuation ring R:. i i . Let n be the intersection of all the l.l3;j . Sinee the Pi are all the prime divisors of xli, we sec that the p7j are all the prime divisors of xR* by (18.11). Since 1j37j is contained in any primary ideal belonging to p7j , we see that n is contained in xR*. Since x is in the .Jacobson radical of R* and since n:xR*= n, we sec that 11 = by (4.3), which proves that the zero ideal of R* is semiprime, i.e., R is analytically unramified.
R:.
°
(36.4) THEOREM. A pseudogeometric semilocal integral domain R is analytically unramified. Proof. We prove the assertion by induction on I' = altitude R. Let R' be the derived normal ring of R. Then R' is a finite Rmodule by our assumption, whence R is a subspace of R' by (16.8). Therefore, it is sufficient to show that R' is analytically unramified. Since R' is pseudogeometric, we may assume that R is normaL If r = 0, then the assertion is obvious, and we assume that r ;::: 1. Let x (~o) be an element of the .Jacobson radical of R. Then, by (36.3) applied to this x, we see that R is analytically unramified, which proves the assertion. (36.5) THEOREM. If R is a pseudogeometric ring, then every ring which is of finitely generated type over R is a p8e1ldogeometric ring. Proof. By the definition, and by (:36. J ), we have only to prove that if R is a pseudogeometric integral domain, then every affine ring A over R is a pseudogeometric ring. Using induction on the
1'11.11"1'1':11 1'1 111I/lillI'l' of J!;I'IIPI'II,i,ol'f: of ,I, WI' 1111.\'1' 11111,1' 1,0 /11'(11'1' 1.111' l'II.fIl' ",111'1'1' ,I /"1,1'1 IVilll 1111 1'11'1111'111 ,/, of ,I, II Ilidlil'I'H III "I,ow Lila!. if q iH :f, /l1'illl(' iill'I'" or ,I, 1.111'11 ,I /q ,~/I,Li:4Iil',cl 1111' liIIiLI'III'HH l'olliliLillll fill' illll' j.t;1'1I.1 I'XI.I'IIt!ioIlH, ~illl'I' /,'1 ( q n N) it! /lH(,lIlIo gI'IlIIII'LI'il', WI' Ilal'I' Olily (,(1 /11'111'(' Lllal. ,I II.H 11.11111'1' HII.liHlil',Y 1.111' lilliLI'III','iH I'ollilil.illil 1'111' ilil,I'gl'al I'XI.I'IiHioIlH, If ,I' iH 1,l'aIlNI'I'IIIII'IIL:I,1 01'1'1' N, 1,111'11 I.II(~ aHHI'I'l.ioll follolV,e; fl'lllli (;i!'i,~), 11,1111 WI' aNHtlllw Ihat. ,I' iH aigillll'ail: ov('l' U. 'I'hl'lI, 1'11/1Hiilcrill!!: a HltiLablc filii\'(' illl.q.!;I'nl l.xLl'lif'doll of fl, whidt iH PHI'tlilo gl:ome',l'ie by (:1(LI ), WI' may aHHllme that x is in the field of lIl/olil'l 11,'; J( of R. Let L be an arbitrary finite algebraic extensioll (If /\', III order to prove the finiteness of the integral closure A' of A ill I), Hilll'I' the integral closure of R in L is a finite Rmodule, we may :I,,'iHIIIIII' that L = K and that R is a normal ring. Thus it is sufficient to PI'II"I' the finiteness of the derived normal ring of A (assuming that U iH normal, and that x E K). By virtue of (35.3), it suffices to show t.hal. if lJ is a prime ideal of A, then the derived normall'ing of A~ is a finite module over A~. Obviously, we have only to prove it in the case where ~ is maximal. Since Rlpn R) is a pseudogeometric local ring, we may assume furthermore that R is a local ring with maximal ideal R. Since ~ is maximal, x modulo ~ is integral over Rim, m = ~ If x E R, then there is nothing to prove. Assume that x EE R. Let f(X) be an irreducible monic polynomial over R which is irreducible modulo m and such that f( x) E ~. Considering a suitable, finite integral extension of R, we may assume furthermore that f(x) is of degree 1, hence that x E ~. Let F be the set of pairs (a, b) of elements of R Ruch that ax = b. Then the set of aX  b generates a prime ideal q of the polynomial ring R[X] such that R[X]/q is naturally isomorphic to R[x] by (11.13). Then the set b of the b has no imbedded prime divisor by (11.13), and by its proof, and AI xA is isomorphic to Rib, which implies that xA, hence xAp , both have no imbedded prime divisors. If ~' is a prime divisor of xA~ , then ~' n R is a prime divisor of b, hence we see that RWnR) is a valuation ring and x is in the valuation ring. Thus we see that A~ is analytically unramified by virtue of (36.3), whence the derived normal ring of Ap is a finite A~modtlle by (32.2), which completes the proof of (36.5).
n
(36,6) COROLLARY. If A is an affine ring over a pseudogeometric integral domain, then the derived normal ring of A is a finite A module.
I: II
(;1(1.',') .1 1II'IIIi 111('(1/ 1'/1/(/ h' /.4 /11111/11111'/1111/ /I/lI'/I//lljil'tI If 1 III /S (1111111/1/('(11//1 IW I'll IIIlji/'(1 fill' ('1 ('1'/1 111(1.1'/111111 /(1('(/1 III (~r 'l'lin PI"OO!' is illllll('dill.l.n hy (17.7).
i/ U
/11111 11111.1/
u.
(;W.H) '/'I1I':()I{,II:M. I,d .1'1, ..• , :1'1/ II(' 111(/1'111'11/('(11//1 JIlt/(·II(·lliI('III. I'!(' ments O'I!(~r (t sem:iiow/ 'I"inl/ N. 11881/,/11(' /,//,(/,1, N /8 (Ulldlli/mlll! 11I/1'(/'llIlji('d. If a semilocal rino H' is a rino (~r (Jlwl'il'n/'8 I(/' 11,(, jllltY/WIII/ill.! I'ill!! R[x] = R[XI , ... , Xn], then Il' is analytically unmll/:~/it:d. In PW'I'I:('It.iIll', 1j p is a prime ideal of R, then R~ is analytically unram'~/i('d. Proof. By virtue of (36.7), we may assume that R' is a loeal rillg. Let q be the prime ideal of R[xj such that R[x]q = R' and set p = q R. Let R* be the completion of R and let p* be a minimal priffi(~ divisor of pR*. Then the theorem of transition holds for R~ and R:. and R~ is a subspace of R:•. Since R* is a pseudogeometric ring by (32.1), we see that is a pseudogeometric ring by (36.1), which implies that the completion of has no nilpotent elements except zero, whence the same is true for its subspace R~ . Thus Rp is analytically unramified and we may assume that R is a local ring with maximal ideal p. Since R*[x]/pnR*[x] is isomorphic to R[x]/p"R[x], qR*[x] is a prime ideal of R*[x] (because R*[xJlqR*[x] = (R*[xJl pR*[x]) / (qR*[xJlpR*[;r])), and furthermore the theorem of transition holds for the local rings R' and R*[X]qR'[x] by virtue of (IS.S) and (19.1), whence R' is a subspace of R*[X]qR*[X] . Therefore we may assume that R = R*, whence R' is pseudogeometric by (32.1) and (36.5), and the assertion is proved by (36.4). We add, here, a result on the normality of pseudogeometrir: semilocal integral domains. We first prove it in a more geneml case:
n
R:.
R:.
(36.9) THEOREM. Let R be a Noetherian integral domain and assume that a ( ~ 0) is an element of the Jacobson radical of R. A 8sume f urthermore that: (1) aR has only one minimal prime divisor p, (2) aR~ = pR~ , and (3) R/p is a normal ring. Let R' be the derived normal ring of R. If R' is a finite Rmodule and if every (minimal) prime divisor p' of aR' lies over p, then R itself is normal and p = aR. Proof. R' /p' is integral over R/p. Since R~ is a valuation ring by our assumption (2) and by (12.1), we see that R' is contained in R~. Therefore R'/p' C R~/pR~, which implies that R'/p' and R/p have a common field of quotients. Therefore that R/p is normal implies that R'/p' = R/p. On the other hand, R' C R~ implies that p' R' pR~, whence p' is unique, and that R;, = Rp , whence aR;, =
n
I!II A l"I'llllt V I
,
p'h~""
fL roliowH LlmL flH' )1'. 'l'IH'I'I'/'OI'(I WlaN' Nip, whidl illlpli"H LlI/I.L h~1 (/'I~' I N. Nilll~n H' j~ /I. IilliLn 1~lI\odlll(~ alld !:lineo ((. iH ill LlII' .1n""ilHoll 1'11."i':/l.1 or u, w(~ H('(l LllaL U = "~' by LIw lemma, of 1\I'ldl,AYIIIIIII.yll., wialllH:n (J."~' "" pi implieH that all = p. Thus the pl'ool' iH I:oll\plnl,n. Now Wp apply j,}w above result to the case of pseudogeometric illLegml domaiuH. (36.10) TllIWRlDM. Assume that a psewlogeometric integral domain U is a homomorphic image of a locally Macaulay ring, and that aI, ... , ... , ar are elements of R such that height (L aiR) = r and such that they are in the Jacobson radical of R. If L aiR has only one minimal 71rime divisor p, if the ai form a regular system of parameters of Rp and 'zf Rip is a. normal ring, then R is norma.l and p = LaiR, and furthermore every RIL{ aiR is a normal ring. (LEMMA OF HmONAKA) Proof. We prove the assertion by induction on r. If r = 1, then (36.9) a,nd (34.9) imply the validity of the assertion, and we assume Chat r > 1. Let q be a minimal prime divisor of arR. If r is a minimal L aiR, then height r/q :::;: r  I, whence, by prime divisor of q the validity of the chain condition for prime ideals, height r ::; r, and p = r. Therefore q c p. Since the ai form a regular system of parameters of Rp we see that q is unique and that arR q = qRq. Consider R" = Rlq. Then, applying the induction to R" with elements ai modulo q, we see that p/q = (q L aiR)/q and that RI(q L{ aiR) is normal for each j = 0, I, ... , r  1. In particular, Rlq is normal. This, the uniqueness of q and the equality arR q = qRq imply that R is normal and that q = arR by the case where r = 1. Therefore the proof is complete.
+
+
+
EXERCISE. Prove that a pseudogeometric semilocal integral domain R is unmixod if and only if R satisfies the second chain condition for prime ideals. GenernJi7.o (36.10) to the case whor~ R it! a pseudogeometric unmixed local illtOgl'al dOIll~dn.
37. Analytical normality We I:la,y that ul:lemilocal integral domain R is analytically irreducible if the completion of R is an integral domain. R is called analytically normal if the completion of R is a normal ring; in this case, R itself must be a normal ring by (18.4). We say that a prime ideal p of a semilocal ring R is analytically irreducible if Rip is analytically irreducible.
1:\11
(lIiiIlMII)'I'I/.1(1 I.(II'ill. IIINIlII
"1'1
(;i7.I) U III' 1/ 'I'il/(/ (I//d (/IlIlI//I1f' Iho! I, ." ( fa: (//'(' I. ·,:s '/I./I! It Z(·'/'o···dilli8U'1' ,:'" N /UU{ (:!) '''~:I/U IH.!J /' /8
N/II'II
IIII/l ( I )
(/./1. 1'//'/1/('//1 /(/
the total quotient ri'/l.(f 4 H 8U,(:/i. UUI.l ll! ((.nd ·W) nrc i", N, !hl'''' ·0 '/N ":/1. fa'. Proof. Since tuv E tU, we have tv C tH:'uN = lU, wl\l~I\(:c LlII'I'I~ i.'j an element v' of R such that tv = tv'. Since t is noL a ;!'(:I'O divi:iOl', W(' have v = v' E R. (37.2) Let R be a normal semilocalring and let R* be its compl/:/i/ll/.. A SSUme that t is an element of R which is neither zero nor unit in N such that every prime divisor of tR is analytically unramified. If v ·is an element of the total quotient ring of R* such that v is integral over HI and such that tv E R*, then v is in R*, Proof. Let the prime divisor of tR be ~1 , • • • , ~r • They are of heigh (, 1 by (12.9). Let S be the intersection of the complements of ~i in U. Then Us is a semilocal Dedekind domain with maximal ideals ~iR,~ , whence ~iRs is a principal ideal for every i by (28.9). Let Xi be an element of ~i such that ~iRs = xiRs for each i, and let ei be natural numbers such that IRs = X~l ... x;r Rs . Since tR*: sR* = tR* for any s of S by (18.1), it is sufficient by virtue of (37.1) to show that there is an s of S such that tvs E x?··· x;rR* (observing that X~l ... x~r ( tR). Let ~jj (j = 1, .. ' ,n(i)) be all the prime divisors of ~iR* and let Wij be the valuation of the field of quotients of R:' ii with R~ij as valuation ring and such that Wij(Xi) = l. Let q,ij be the natural homomorphism from R* into R:. i i . Let fl , ... ,fr be nonnegative integers satisfying the following condition: tvs is in X{l .. . X;rR* for some s of S but for any s of S and for any i, tvs is not in X{l .. ·x~r. xiR*. Then it is sufficient to show that fi 2: ei. Assume the contrary, for instance that fl < el. We take an element s of S such that tvs is in x{l .. . X;rR* and let z be an element of R* such that tvs = x{l ... x{rz. We may regard q,ij as a homomorphism from the total quotient ring R** of R* into the field of quotients of R:. i j • Then, :,lince v is integral over R*, q,ij( v) is in R:. i; , whence Wi/( q,ij( v )) 2: 0, Since WJj(q,lj(t)) = el > 11 = Wlj(q,lj(X{l .. 'x~r)), we have Wlj(q,lj(Z)) 2: 1. This shows that q,lj(Z) iH in q,lj( ~~j). Since the kernel of q,l; is contained in ~~j, it follows that Z is in ~ij . Since x 1U; = j ~ij R,~ , Z is in x1R; and therefore there is an element s' of S such that ZS' E xlR*. Thus, with s" = ss' which is in S, tvs" E Xl·x{l .. 'x~rR*, which is a contradiction and we have ei :::; fi for every i, Thus the proof is complete. (37.3) A ssume that a normal local ring R is analytically normal. Let
n
I'll
11"1' 1'111 1'1
1',rll'll,ql"l/ I~( 1/11' ,1il'ld I~r 1IIIIIIil'lIl,q II I~r Ii' 111111 11'1 //1/1'(/1'111 I'lmilll'l' 1(1' Ii' ill /" ,\ ,%'/11111' l/il/I 1'1'1'1'/1 111'ill/I' itll'lll )I 1(1' /it'il/hi I ill I iN 111I1I1,l/lil'llll,ll 1I11I'1//llIjil,tI, '1'/11'11 1/11' 1'11111/1/1'110 /1 I" I~I' I IN illll'tll'lIlI!! 1'11I,o.;I'd, IItIII 1,0.;, fill' 1'/'1'1'11 11I1I,rllllll! itll'l/I III IIf I, I", is 111111 1/11//'((,11/1 /l.III'IIIIIJ,
" III'
1/
I /11'
l/il'
./il/III' 81'llltl'l/llll'
(,11'1\\('111, or I cHlI'h LhaL I, 1\(1/.) alld 1('(, dill' il'l'('dlll'ibll~ IlIOlli(, (loIYllollli:t.i OV('I' N \Vliil,li bn,.; /1, HH a root.. Ld Ii"' lin Lhl~ (lolllpidioll of It nlld ltd. 1** I )I~ (,he ill" Lnp;m[ e/OHlll'l, of /* ill i(.,y Lotal quotillld, rillp;, Hi II I'.i\ / iH a {illiLlr u~ I\lOdlde hy (10, J(;), t.he ill (,egral I'lmmre of n*[aJ in itH LoL:d (jlloLi('11 L I'illg I'oineide::; wiLh 1** by (J7,8), Therefore we see that dJ** C 1(,*1111
P('OO\'. L(,(,
II, 1)(' :1.\1
(.1\1' di:;('l'illlill:l.IIL or Lill'
hy (10,15), whieh implies that 1** = J* by (37,2), and therdOl'(I 1* iH illtegrally closed, If b is a nilpotent element of J* and if x iH lIoi, :\, y.W'O divisor, then bj:r; is integral over 1* and is in J**, and b E :r1** ~,~ :r/*, which implies that b E mnJ* with Jacobson radieal 111 of R alld with an arbitrary natural number n, Therefore we have b = 0 by (4,2), whieh proves that 1* has no nilpotent elements except J1(1J'(), i,I)" the zero ideal of J* ilS tlw intersection of a finite number of prim(r ideals, say pi , ' , , , The idempotents of t.he total quot.ient rillg of J* are integral over J*, and t.herefore they are in J*, and we HCO that J* is the direct. sum of 1*/p; . Sinee 1* is integrally closed, c[l,(Ih /* / p; mURt, be normal, and the proof if; (~omplcte,
P: '
(:i7,4) THEom;J\L Assume that a normal local ring (R, nt) is analytically irreducible, Let K be the field of quotients oj R, Assume that a local ring (R', m') satisfies the following three conditions: (1) R ::s; H' c K, (2) RI/mR' is a finite R/111~mod1lle, and (:~) altit.ude R' = altitude R. Then R' coincides with R, Proof. Let R* and R ' * be the completions of Rand R ' , reRpectively. R = m, we see t.hat 111' C m,i R for every i, Therefore Since m' there exists a natural homomorphism ¢ from R* into R ' * and ¢(R*) becomes the elosure of R in R ' *, By assumption (2), we see that R '* is a finite ¢(R*)~module by (30,6), whenc:c altitude R ' * = altitude ¢(R*) by (10,10), Since altitude R ' * = altitude R' = alt.itude R = altitude R* by (17,12), we have :t11.iLllilc R* = altitude ¢(R*), Since R* is an integral domain, it. follow:: that ¢ is an isomorphism, Thus we sec that. R is a subspace of W :,,"1 !t,'* is integral over R*, Let alb (a, b E R) be an arbitrary nl(rlll('I\1, or R ' , Since alb is iutegral over R*, there are elements ci , ", , c~ of Hoi' such that (a/b)n + ci(a/b)nl c! = 0 :tlld Lhnl'dol'c an banlei
n
n
+ '" +
+
+ ,., +
(1I1\()~IIII'I'111i
I 1,( I( ',I 1. 11/ N 1. Ld d be the discriminant of RI. Assume that d is a nonunit. Then th(~I'I' is a prime ideal q of height 1 in R which contains d. Let a be sueh ilil element as given by (41.4) applied to q. Then the discriminant d', of the irreducible monic polynomial f( x) such that f( a) = 0, is not ill q, As was remarked before (41.3), d' is the discriminant of R[aJ and iN in dR, which is a contradiction, whence d is a unit in R, which prOV(~H the assertion by virtue of (41.3). ( 41.6) Let (R, m) be a regular local ring with a regular system /~r parameter s Xl , . , , ,X r and let (V, n) be a valuation ring which dom'inates R. Set a = xlR xsR (2 ::::; s ::::; r), If XdXl , . , , ,xS/rl modulo n are algebraically independent over R/m and if R* is the dilatation of R by the ideal a with respect to V, then R is a subspace of R*. Proof. Set RI = R[xdxl' . , . ,Xs/XlJ. Then xlR' is a prime ideal and xlR I = aR'. Set R" = R:w . Then R" is a Noetherian valuation ring. Obviously R* is dominated by R" and RI! is a quadratic dilatation of both Rand R*. For an element f of R, f E mn if and only if f E x~ R", and, if we denote by m* the maximal ideal of R*, then fol' an element f of R*, f E m*n if and only if f E x~ RI!. Therefore m*n n R = m n and we complete the proof. We need one more preliminary:
n
+ ... +
(41.7) THEOREM. Let (R, m) be a complete regular local ring with a regular system of parameters Xl , ' •. ,Xr with r 2: 3, Let P be a normal local ring which is a finite separable integral extension of R. Then there is a finite number of elements al , . .. ,as of R such that, letting y be a transcendental element over R, and c an element of R, the local ring Pc = P(Y)/(X3Y  Xl  CX2)P(y) is analytically irreducible, whenever c is such that II (c  ai) ~ m. Proof. Let b be an element of P such that P is the derived normal
1'11.\ l"I'lilll
VI
I'ill~'; or 11'1111111111 11'1/1 .\' I III' III(' II'I'I'IIIII'illll' 111111111' pol,I'llolllinl 11\'1'1' Ii' 1 I qi and let Q: be the derived normal ring of Qi ' Since Qi i!'i complete, Q; is a normal local ring, By the induction assumption, if r is a prime ideal of P different from 11, then P r is unramified over R(rnR) , Applying this fact to those r containing qi, we have: (1) r/qi is unramified over RlxR; hence (Qi)r/qi is a regular local ring, and consequently (2) the conductor of Qi in Q; contains a power of the maximal ideal; and (3) is unramified over RlxR, In particular, the residue class field of is separable over RI m, whence Pin = Rim by our assumption made above. If QI 'F Q~, then L~ 'F Pin by ( 41. 7). Then, taking an element a' which generates over Pin = Rim, we extend both P and R so that their residue class fields become L~ by the method we used above. Namely, let f(x) be a monic polynomial over R such that f modulo m is the irreducible monic polynomial for a'. Set PI = P[x]lf(x)P[x], RI = R[xllf(x)R[x]. Since the discriminant of f is unit in R, it follows that PI and RI are normal and are unramified over P and R, respectively. mP 'F n if and only if mPI 'F nPl and therefore (RI, PI) is equivalent to (R, P), Furthermore, f(:1:) modulo ql is reducible over QI , and we see that qi splits into several prime ideals. Since the total number 8 of the qi does not exceed the degree of extension of the field of quotients L of P over the field of quotients K of R, we see that, after a finite number of steps, we come to the case where QI = Q~, whence QI = RlxR. Thus we may assume that there is an element x of m which is not in 2 m such that xP has a prime divisor ql with the property that Qlql = RlxR. Now let c, y and Xl, •.. ,X r be as in (41.6) and consider (R(y), P(y)), which is equivalent to (R, P). Set z = X3Y  Xl CX2. Then zP(y) is prime, and our observation for X can be applied to z and we have: (1') if r' is a prime ideal of P(y) such that z E r' c nP(y), then rlzp(y) is unramified over R(y)/zR(y) and consequently (2') the conductor of P(y)/zP(y) ill its derived normal ring P" contains a power of the maximal ideal nP(Y)/zP(y). Since P(y)/zP(y) is analytically irreducible, the completion of P" is an integral domain, which implies that P" is a local ring. By our induction assumption, we have (3') P" is unramified over R(y)lzR(y).
Q:
L; Q;
L;
10i'{
(j1'l(IMlli'I'IJ,I(! 1,11i'AI, IIINilil
WI: h:w(~ 0Ii1'y 1,0 Hhow I,!IHI, I'" 1'( /I )/'1'( 1/), 1,(11, 1.1\1' 11Il\,\illlid iil('al of I'" hn II". Ld r" hn a lIIillimal pl'illl(' di"iHo(' oi' //I/'(!/) I ,,'I'(!/), Then the COl 1(1 u ei,o I' 01' I'(U)/zl'(!!) ill I'" iH 1101, I'olll.nilll't! ill r"/zP(y), whence the derived lIol',"al rillg oi' 1'(1/)/1''' haH I\. ('I'Hidlll' class field which contains P"/n". On iJw oU\(~1' halld, ::il\('(\ 1'/,1,("
II.
III I ~"1.i(Oh IIIlp;en zwischen del' Idealen verschiedener Ringe, Math. AHll. !17 (I D27), PI" 490523. "",("')11,'1', Il. III (111(01' dioj,itcorie del' algebraischen Formen, Math, Ann, 36 (1890), pp, 471 1i:!,1. 1'21 0(,1\1' die vollen Invariantensysteme, Math. Ann. 42 (1893), pp. 31337:3,
II I III1NAI(A,
n.
III II 1I0t.e on algebraic geometry over ground ringsThe invariance of Hilbert. d'll,meLeristic function under the specialization process, Illinois J. Math. '2 (11158), pp. 355366. 1\11,111,[" W. III I'l'imidealketten in allgemeinen Ringbereichen, S.B. Heiderberg Akad. Wiss, 7 (1928). 121 IIllgemeine Bewertungstheorie, J. Reine Angew. Math. 167 (1931), pp. 160196. 1:\1 Ober die Zerlegung der Hauptideale in algemeinen Ringen, Math. Ann. 105 (1931), pp. 114. 1,11 Idealtheorie, Ergeb. der Math, 4, No.3, Jlllius Springer, Berlin, 1935. llil Beitrage zur Arithmet.ik kommutativer Integritatsbereiche, Math. Z. 41 (1936), pp. 545577. 16] Beitrage zur Arithmetik kommutativer Integritatsbereiche, IT, Math. Z. 41 (1936), pp. 665679. [7] Beitrage zur Arithmetik kommutativer Integritatsbereiche, III, Math. Z. 42 (1937), pp. 745766. IHI Beitrage zur Arithmetik kommutativer Integritatsbereiche V, Math. Z. 43 (1938), pp. 768782.
,,'1 
.,
IIII 1111".11111011111111'''111' III ~\I'i'''I'''l'ill~.I·",.1 H,·illl' !\IIV,,'w.I\111I11. 1',illlll:IHI, I'll. ~O I ~~~~I \ 1101 1.111' 'l'III'lIl'il' cI,,,' 1'''III1I1I1LnLivl'lI 11I1'''I':I·iLii.LHI)(!I'ni,dl'', .1. ((,·i"" i\1I1':"W. MaUl. 1\1~l 110[,1), PI'. ~:\O ~Ii~.
L I'll 'II,
(~.
III ()II I.III~ :l.HHOeiILi,ivi Ly formula for multiplicities, Arkiv. Math. 3 (1956), pp. :\I)I:~I
(I.
F. S. [1] Algebraic theory of modular systems, Cambridge Tract.s Mat.h., 19 Cambridge University Press, Cambridge, 1916.
MA':AlILAY,
S. [1] Modular fields. I, Duke Math. J. 5 (1939), pp. 372393.
MAcLANE,
MORI,
Y.
[1] On t.he integral closure of an integral domain, Mem. Coil. Sci., Univ. Kyoto
27 (195253), pp. 249256; Errat.a, Mom. Coli. Sci., Univ. Kyot.o 28 (19531954), pp. 327328. [2] On the int.egral closure of an int.egral domain, II, Bnll. Kyot.o Gakugei U niv. B7 (1955), pp. 1930. NAGA'rA,
M.
[11 On the struct.ure of complete local rings, Nagoya Math . .T. 1 (1950), pp.
12] [3] [4] [51 [6] [7J [8] [9] [10] [11] [12] [13] [14] [15]
6370; Enata, Nagoya Math. J. 5 (1953), pp. 145147. On the theory of semilocal rings, Proc. Japan Acad. 26 (1950) pp. 131140. Local rings (in Japanese), Sugaku 5 (195354) pp. 104114 and pp. 229238. On the theory of Henselian rings, Nagoya Math. J. 5 (1953), pp. 4557. On the theory of Henselian rings, II, Nagoya Math. J. 7 (1954), pp. 119. Some remarks on local rings, Nagoya Math. J. 6 (1953), pp. 5358. Some remarks on local rings, II, Mem. Coll. Sci., Univ. Kyoto 28 (195354), pp.109120. Note on integral closures of Noet.herian domains, Mem. Coli. Sci., Univ. Kyoto 28 (195354), pp. 121124. Note on complete local integrity domains, Mom. Coll. Sci., Univ. Kyoto 28 (195354), pp. 271278. Basic theorems on general commutative rings, Mem. Coll. Sci., Univ. Kyoto 29 (1955), pp. 5977. On the derived normal rings of Noetherian integral domains, Mem. Coll Sci., Univ. Kyoto 29 (1955), pp. 293303. An example of normal ring which is analytically ramified, Nagoya Math. J. 9 (1955), pp. 111113. A general theory of algebraic geometry over Dedekind domains, I, Am. J. Math. 78 (1956), pp. 78116. A general theory of algebraic geometry over Dedekind domains, II, Am. J. Math. 80 (1958), pp. 382420. A general theory of algebraic geometry over Dedekind domains, III, Am. J. Math., 81 (1959), pp. 401435.
'_" 'II
11111 '1'111' 1111''''1' "I III,IIllpl"'II.I' III ~y,"q,,1 1""111 111I~,iI, l'I',"'""dll'~,,, "I IIII' '"II'rlllIl,I""1I1 H,I'"I'''''IiIIlI, '1',,1'111 NII,I", 1ill,II ~1"'''"III,,' (""III"Ii "I ,1111'1111, '1'" I, .1''', 111[,1 I, PI' IIII '!:,~II 11','1 1111 1111' "llIdll 1'1',,1"""' "I' I'rillll, id,'" III, N "~,"I'II 1\11111, .J IIIII!I[,III, 1'1'. id II I I L~ I N OJ,l' 011 II, pll pl'I' 1)1' HIIIIIIWI ('OIIC'cq'lIi IIJl. JlIl,\' 1111 II 01 iI' 1I"qpPI'CJC'li III' iI/I'll III ~1I'1i1. (!"II.H,'i., (1IIiv.I\.I'"I,,;I() 11%111,',1, 1'". III[) 1'/1" 11111 1\ .JII,'ol,illll ""il,I'ri"" "I' {,ill""" poi II 1,". '"illl,i:,.1. 1\111111. I I 1%'iI. PI', 1:2',' 1:1:2_ 1:2111 1\11 mlllIl1>lI' "I' II, 11111'11111.1 101.1..1 rill/!: lVilil'l, i,y II.llltiyl,i"II.IIy !'I·dll",i!>I,·. ~I,,"I. (~oll, H"i,. (1IIiv, 1',YIII,II:\1 (IOIiX), PI>, X:I Xli, I~lll NoL,' Oil '" "I,,,,ill (lolidiLioli 1'01' I>rillll' idnalH, M"IlI. Coli. HI',i., lI11iv. 1\.1'''1,,) :1:2 (lilliO 1!lIiO) , PI>. Xli !IO. I~l~ll N"I,,, Oil l.o"f1i .. i(lIiL fiddH or ('lImplet(. llle:i.! I'ill/!:H, M()Il'. Coli. H"i .• lilliv. l\yoLo ;1:2 (IHIiB WHO), I>p. !11 H2. 1:1;11 (III 1,11(1 L111'Ol'y or """Holiall I'ill/,:H, "I, MOIll. ColI. Hei., lilliv. 1\'yoLil ;I:.l (IOIiO lOtiO), PI>. H:I 101. I~II ()II 1,1111 p'lriLy or !>mlleh loei ill t'('/!:lllal' I(H,:d l'illgH, [Ililloi" '/. MaLII. :\ (IDW), PI>. :128 :\:1:1. I~lil Oil t.h« el()Hedll()~H of "ingular lOlli, Puhi". MIlII!. IIIHt.. II alit.. ji;l.lId. Hl'i. 2 (IHGH), pp. 2936. NA"AI, Y. alld
NAGATA,
IV!:.
III I\I/!:i'brnie geometry (in Japanese). Ky6ritsu, Tokyo, 1957. NAIII'I'A, M. III (III Lhe "il'ucture of complete local rings, J. Math. Soc. Japan 7 (1955), Pl'· ,1:11i,14:1. I~ I (Ill lhe unique factorization theorem in regular local rings, Proc. JaptLiI I\WI!. :\5 (1959), pp. 329331. NI~III,
III
M, the dimension of local rings, Mem. ColI. Sci., Univ. Kyoto 29 (1955), pp. 79. ()Il
N OI,n'IlER, E, III Idealtheorie in Ringbereichen, Math. Ann. 83 (Hl21), pp. 2466. 121 Der Endlichkeits~atz del' Invarianten endlicher linearer Gruppen der Charakteristik p, N achr. Ges. Wiss. Giittingen, 1926, pp. 2835. [:\1 Abstrakter Aufbau del' Idealtheorie in algebraischen Zahl und Funktionenkarpel'll. Math. Ann. 96 (1926) pp. 2661. NOR'rHCOTT,
D. G.
III Hilbert function in a loeal ring, Quart .•T. Math. Oxford 4 (Hl53), pp. 6780. D, G. and Rl!lES, D. III Heduction of ideals in local rings, Proc. Cambridge Phil. Soc. 50 (1954), pp. 145158. 121 A note on reduct.ions of ideals wiLh an application to the goneralized Hilbert function, Proc. Cambridge Phil. Soc. 50 (1954), pp. 353359.
NORTHCOTT,
( ) h '\
III
I
~._
,~III' 1"11 1,1101'1101111
IIlIlilyl,i'lII""
.1 "1'1111 :1 I Iilid I. JlI', I( 1':1':11.
~~III
.I,.
1'II/IIi'''"'H vlIrillhle'H, VIII, .1, Malli, N"I',
:.l1,1,
I),
III 1\ lIIIII' 1111 VlllIIILi,ioIiH aHHoeiat.ed with a local domain, Proc. Cambridge Phil. HilI', 1)1 (1!11i1i), Pl'. 252253. [21 'l'wl) I:In.HHieal theorems of ideal theory, Proc. Cambridge Phil. Soc. 52 (I!lSI;), pp. 155157.
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P. [1] La notion de multiplicite en algebre et en geometrie algebrique, J. math. pures appl. 30 (1951), pp. 159274; These, Paris, 1951. [2] Algebre locale, Memorial Sci. Math. 123. GauthierVillars, Paris, 1953.
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H. [1] Some remarks on Zariski rings, J. Sci. Hiroshima Univ. 20 (19561957), pp. 9399. [2] A note on principal ideals, J. Sci. Hiroshima Univ, 21 (19571958), pp, 7778,
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J.P. [1] Sur la dimension homologique des anneaux et des modules noetherian, Proceedings of the International Symposium, TokyoNikko 1955, Scientific Council of Japan, Tokyo, 1956, pp. 175189. [2] Multiplicites d'intersection, mimeographed notes, 1955. [3] Geometrie algebrique et geometrie analytique, Ann. inst. Fourier 6 (19551956), pp, 142.
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Math. No,:. ii:\ (1\);1:1), PI'. ·IDO 5·1:.l. (ililll\l'lIli~,:d H"lIli 10,,11,1 I'ill/,:H, NIlIllIlla. BraHi!. Math. J (I!Hli), PI'. Iii\) IHii, '1'11/\ IIOIHiopL or /I, Himplo point of nil abstract alg;ehmie val'idy, TI'IIIIH. 11111. MaLh. No". G:.l (1!117), pp. 152. IIl1alyLil1ld irrntlueibility of normal varieties, Ann. Math. 1!l (III/IX), PI'. aii:.l :W 1. NIII' III, lIol'lllaliLe analytique des variete normales, Ann. inHt.. 1,'ollri,:r ~ (1!l50), pp. 161164. (III i,1!" pilrity of the branch locus of algebraic functions, Proc. N aL. A.lad. 1I. :;. 14 (1958), pp. 791796.
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AND SAMUEL,
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III COllllllutative algebra, 1, van Nostrand, New York, 1!J58.
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