ON THE NUMBER OF GENERATORS OF MODULES OVER POLYNOMIAL AFFINE RINGS

Preprint enviat per a la seva publicacio en una revista cientifica: Mathematische Zeitschrift. 1991, Vol. 208, p. 11-21. [https://doi.org/10.1007/BF02571506]

On the number of generators of modules over polynomial affine rings RICARDO GARCÍA LÓPEZ 0.INTRODUCTION In thís paper we give sorne bounds for the minimal number of generators of a finitely generated module Mover a polynomial ring B = A [Xi, ... ,Xn], where A is an affine ring satisfying sorne regularity conditions. For n = l , A.Sathaye and N.M.Kumar proved in [S] and [MK] that if Bis commutative and noetherian, then: µ(M)::; max { µ(Mp) + dim B/p I dim B/p < dim B } de/ ee (M) where µ denotes minimal number of generators and p runs over the prime ideals of B, solving a conjecture of D.Eisenbud and G.E.Evans Jr.
For A= k, an infinite field, G.Lyubeznik has proved in [Ly] the following bound: (0.1) µ(M)::; max { µ(Mp) + dim B/p I p E SpecB such that Mp is not free} It is easy to see that thís bound turns out to be especially sharp when the primes of B at which M is not locally free determine a closed subset of SpecB of small dimension. As a consequence of our main result (Thm. 2.1), we prove that if A is a regular affine algebra over an infinite field and Mp is a free Bp-module for all p E SpecB excepta finite number of maximal ideals m 1 , ... , ms, then: µ(J..1)::; max { dimA+r, µ(A1mi), ... ,µ(Mm.)} where r = rank (M), thus generalizing for this type of modules the bound (0.1) (see Prop. 3.9).
When n=l we prove that if AJ is a torsion free E-module and A is a regular affine ring ( or is an affine domain such that SpecA has at worst isolated singularities ), then: µ(AJ) :' .S: max { dimA + r , µ (M/ lMlvf)} where IM = ideal of definition of the set of primes of B at which Mis not locally free. We recall that the ideal lM can be effectively calculated if a presentation of M is given (see (Br]) and that if A is a domain the Eisenbud -Evans bound can be written as ee (M)  On the other hand, if M = I is an ideal, one knows that µ( I / 1 2 ) :S: µ( I) :S: µ( I / 1 2 ) + 1 and one is interested on knowing when the inequality on the left side becomes an equality. If E is commutative and noetherian and I ~ E is an ideal which contains a monic polynomial, S.Mandal has proved in [M] that if µ(I/1 2 ) ~ dimE + 2, then µ(I) = µ(I/1 2 ). But, to the extent we know, the only result which holds for general I ~ B is that the equality is attained when µ(I/1 2 ) ~ dimB (see [MK]). In Corollary 3.2 we prove that if A is a regular affine ring and I ~ A[X 1 , ... ,Xn] is an ideal such that ht(I) ~ 2n and µ(I/1 2 ) > dimA, then µ(I) = µ(I/ 1 2 ). This result follows from the main result in [M] in case I contains a monic polynomial (possibly after a change of variables). However, this may not be so if ht(I) :S: dimA (see Remark 3.3).
Thanks are due to J .M.Giral for his help on the preparation of this paper and to G.Lyubeznik for sending me the paper [Ly] and sorne other material when still was in pre-print form.

1.SOME NOTATION AND LEMMAS
Throughout the paper all rings will be assumed to be commutative and noetherian and all modules will be finitely generated. By an affine ring we will always understand a finitely generated k-algebra, k a field.
If Bis a ring and M a E-module, the rank of Mis defined as: We recall that given a E-module, the set of primes of E at which M is not locally free determines a closed subset of SpecE. We denote by IM the (radical) ideal which defines this set. Deflnition 1.1 Let E be a ring, M a E-module and J ~ E an ideal. We will say that Mis free out of J if Mp is a free Ep-module for all p E SpecE such that p ~ J. It is clear that Mis free out of J if and only if J ~ IM. The lemma which follows is close in spirit to sorne avoidance lemmas proved in [MK] and [M], although its proof and the one of Proposition 1.5 are inspired in the proof of [Ly,Theorem 1]. We keep the same notations as in [Ly], i.e., if E is a ring, p E SpecE, N is a E-module and J ~ E is an ideal, put:  for 1 :::; i :::; n + l. Then we can find mt+ 1 E M such that: .. ,X,-1 l(M¡/ < m 1 , ... , mt+I >):::; b 1 (X 1 ,···,X,_ 1 J(M¡) -tl for 1 :::; i:::; n + l PROOF: Set, for 1 :::; i :::; n + l N¡ = M¡/ < m1, ... , mt > As it is observed in [Ly], it follows from the proof of the main theorem of [F) that, in order to get (0.2), it is enough to make mt+I basic for N¡ at certain primes Pi,t, ... , Pi,j, of A[X 1 , ••• ,X¡_ 1 ], no one of them containing J(X 1 , ... ,X¡_ 1 ]. But we have that if q E SpecA[X 1 , ... , X¡-1] and m E Mis basic for Nn+I at ( q,X¡, ... , Xn), then mis basic for N¡ at q (One only needs to check that if m E q(N¡)q, then m E ( q, X¡, ... , Xn)(Nn+i)(q,X,, ... ,Xn))-Therefore, we have to make mt+1 basic for N n+I at a finite set oí primes: no one oí them containing J[X 1 , ... ,Xn]-For this purpose, we make the following claim: Claim: Let N be a finitely generated E-module and { P 1, ... , P s} ~ SpecB. If x, y E M and y is basic for M at each p¡, then there is a e E B such that x + cy is also basic for M at all the primes p¡.
PROOF(OF THE CLAIM): We can assume that P; is minimal in {p¡, ... ,P;} and we make induction on s, the cases = l being trivial. By induction hypothesis there is a e' E B such that x+c'y ft. p¡Mp, for 1:::; i:::; s-l. Suppose x+c'y E PsMp, and take a En::: p¡-Ps• If we put e= e'+ a, then x + cy is basic for M at P1, ... , Ps• 1 Now take y E M basic at all p E P (this is possible by [F,Hilfssatz] where the M¡ are defined as in lemma 1.3. Now from our hypothesis on A and this inequality we get that ht(J¡) > dimA, so we are done. 1 Remark 1.6 It is clear from the proof that the proposition above remains valid under much weaker assumptions on the ring A (for example, it would be enough to assume that it is an universally catenary equicodimensional Jacobson ring). However, the later use will involve only affine domains.
Finally we need a lemma from [Li2] which will be important on proving the main result.
We give a different proof from the one in [Li2]. i} L s' can be assumed to be extended. Put S 1 = 1 + As. S1 ~ As is a multiplicative system and the regularity hipothesis on A implies that S 1 1 As is a regular domain. Let m be a maximal ideal of S 1 1 As. Then (S 1 1 L)m will be a stably free (S  From this and Quillen's extendability criterion we get that S 1 1 Lis extended. Let s 1 E S 1 be such that L 81 is extended. Replacing s 1 by s1s 1 we have (s, s 1 s 1 ) = A and all above works the same. In what follows we assume that this replacement has already been made.
-----ii) L~ can be assumed to be extended. 1 Ms)m is also free (it is projective and free after inverting a monic polynomial) and then we are through.

Remark 3.3
Suppose that A is a ring which satisfies the requirements of corolary 3.2. and J ~ A[X1,X 2 ] has height 4. If dimA = 5 and µ(J/J 2 ) = 6 from 3.2 one gets µ(J) = 6. But dimA >ht(J) , so in general J will not contain a monic polynomial (not even after a change of variables) and the main theorem of [M) could have not been applied.
It is known (see [F]) that µ(N) $ f(N) and for general B, N this is the best possible bound. However, there are a number of situations where it can be sharpened.
For the proof of (0.6) observe that: and from this he deduces (see [Ly,Theorem 2]) that: µ(M) $ max{ µ(Mp) + dimB/p I p E SpecB such that l\1p is not free} def r¡ (M) He also points out that if dimB ~ 2, then every module is the surjective image of a projective module of rank 77(M) (so it is 17(M)-generated if B-projectives are free).
We study now the case dimB = 3, B a polynomial ring. As a consequence of our result, we get that the bound µ(M) ~ 17(M) also holds for modules over Fp[X 1 ,X 2 ,X 3 ] (Fp a finite field ). c. ht( J) = 3. lf d > r + 1 then d ~ r + dimA and we can use again 3.4. If d = r + 1 one can find s,s' E A[X] such that Ms is projective ofrank r and M 8 , is (r+l)-generated, and then apply the last proposition of [Ly]. 1 Finally, we are·going to give a bound for the number of generators of a finitely generated A[X 1, ... , X n]-module which is locally free everywhere except at a finite number of maximal ideals. First we have to observe that the main result in [Ly] (Theorem 1) can be extended to rings of the form ( k 1 x ... X kr )[X 1 , ... , X n] where the k¡ are extensions of an infinite field k.
The only diffi.culty could appear on the geometric argument used to prove InA+JnA = (1) (in the notations of [Ly]). But one has the following easy lemma: Remark 3.9 Using Prop. 3.7 it is easy to see that if n ~ 3, the result above remains true when k is a finite field.