Homological Dimension

Homological Dimension David E V Rose April 17, 29 1 Introduction In this note, we explore the notion of homological dimension After introducing the basic concepts, our two main goals are to give a proof of the Hilbert syzygy theorem and to apply the theory of homological dimension to the study of local rings 2 Elementary results from homological algebra Assume throughout that R is a commutative ring with identity We assume some familiarity with the basic concepts from homological algebra, including but not limited to the following: chain complexes and double complexes, projective modules and resolutions, Tor, and Ext We state a number of elementary results without proof, all of which can be found in [4], that will be useful in the duration The first result gives a method for constructing projective resolutions: Lemma 21 Let P 1 P 1 P P M i π M M 1

be a diagram of R-modules where the column is exact and the rows are projective resolutions We can complete this diagram to give the commutative diagram P 1 P M i 1 P 1 i P i M π 1 P 1 π P π M where P i = P i P i give a projective resolution of M and the columns are exact with the canonical inclusion and projection maps Our next result, which follows directly from the long exact sequence for Ext i R(M, ) associated to a short exact sequence of R-modules, characterizes projective R-modules Proposition 22 Let M be an R-module, then the following are equivalent: 1 M is projective 2 Ext i R (M, N) = for all i > and all R-modules N 3 Ext 1 R(M, N) = for all R-modules N 3 Basic concepts Let M be an R-module The projective dimension of M, denoted pd R (M) is the smallest natural number n so that there exists an R-module projective resolution P n P M If M does not admit a projective resolution of finite length, then we set pd R (M) = Our first result gives several useful conditions characterizing projective dimension: Lemma 31 Let M be an R-module, then the following are equivalent: 2

1 pd R (M) n 2 Ext i R (M, N) = for all i > n and all R-modules N 3 Ext n+1 R (M, N) = for all R-modules N 4 Whenever K P n 1 P M is an exact sequence with each P i projective, K is projective Proof Since Ext i R(, N) is the right derived functor of Hom R (, N), we have that item 4 implies item 1 implies item 2 implies item 3 Now, if A P B is an exact sequence with P projective, the long exact sequence for Ext i R(, N) gives that Ext i R (A, N) = Ext i+1 R (B, N) for i 1 By interlacing long exact sequences with short exact sequences, this in turn implies that if K P n 1 P M is exact with each P i projective, then Ext n+1 R (M, N) = Ext 1 R (K, N) Thus, if item 3 holds then Ext 1 R(K, N) = for all R-modules N Proposition 22 then gives that K is projective In particular, item 2 gives that if pd R (M) < then pd R (M) = min{i : Ext i+1 R (M, N) = for all R-modules N} Proposition 31 now gives that for a ring R the following numbers are the same 1 sup{i : Ext i R (M, N) for some R-modules M and N} 2 sup{pd R (M) : M R-mod} Definition 32 This common number is the global dimension of the ring R, denoted gldim(r) 3

4 The Hilbert syzygy theorem Our goal in this section is to give a proof of the Hilbert syzygy theorem The strongest form of this statement is the following: Theorem 41 Let k be a field and suppose that M is a finitely generated k[x 1,, x n ]-module If K F n 1 F M (41) is an exact sequence with each F i free and finitely generated, then K is free Our approach is as follows: We will first deduce that gldim(k[x 1,, x n ]) = n (42) Since each F i is projective, Lemma 31 implies that K is projective, and K is finitely generated as k[x 1,, x n ] is noetherian The result then follows from the Quillen-Suslin theorem [3, Chapter XXI, 4, Theorem 37]: Theorem 42 Any finitely generated projective k[x 1,, x n ]-module is free We now aim to show (42) We follow the presentation given in [4, Section 43], elaborating on the proofs given there The result will follow from a sequence of change of rings results In order to prove the first of these results, we need the following preparatory lemma Lemma 43 Let {M i } i I be a collection of R-modules, then pd R ( i I ) M i = sup {pd R (M i )} i I Proof Since the direct sum of a family of projective modules is projective, given projective resolutions P,i M i for each M i we have a projective resolution P,i M i which gives pd R ( i I ) M i sup {pd R (M i )} i I Next, if pd R ( M i ) = then the result follows, so suppose pd R ( M i ) = n < For each i, consider an exact sequence K i P n 1,i P,i M i with each P m,i projective This gives an exact sequence i I K i i I P n 1,i i I P,i i I M i and Lemma 31 gives that K i is projective Since direct summands of projective modules are projective, it follows that each K i is projective Lemma 31 then gives pd R (M i ) n and the result follows 4

Given a morphism of rings R S, the first result relates the projective dimension of an S-module M over S to its projective dimension over R Theorem 44 If R S is a morphism of rings and M is an S-module, then pd R (M) pd R (S) + pd S (M) Proof We can assume that pd S (M) = n < and pd R (S) = d < or else there is nothing to prove Let Q n f n f 1 Q f M be an S-module projective resolution Choose R-module projective resolutions of M = imf and kerf and use Lemma 21 to construct an R-module projective resolution P, of Q Now construct R-module projective resolutions for each Q i via the same procedure, using the resolution of imf i = kerf i 1 used to construct the projective resolution of Q i 1 and choosing a projective resolution of kerf i We can then patch these short exact sequences of projective resolutions together to obtain the commutative diagram: Q n P,n P 1,n Q 1 P,1 P 1,1 Q P, P 1, where the columns are chain complexes and the rows are R-module projective resolutions Now, since each Q i is a projective S-module, it follows that for each i there exists an S-module N i so that Q i N i is a free S-module F i (S) Lemma 43 then implies that pd R (Q i ) pd R (F i (S)) = pd R (S) = d 5

We hence may construct the commutative diagram (43) Q n P,n P d,n Q P, P d, where the columns are chain complexes and the rows are R-module projective resolutions by taking P m,i = P m,i for m d and P d,i = P d,i / ker(p d,i P d 1,i ) and using Lemma 31 A standard argument now gives that the complexes Q and Tot(P, ), the total complex associated the double complex P, obtained from truncating the Q i s in the rows of (43), are quasi-isomorphic (ie have the same homology) Since Tot(P, ) is a complex of projective R-modules of length at most n + d, the result follows The proofs of the next two change of rings theorems proceed via induction on pd R (M), so the following lemma aids in these considerations Lemma 45 If the sequence of R-modules A B C is exact, then pd R (B) max{pd R (A), pd R (C)} Moreover, if we have strict inequality, then pd R (C) = pd R (A) + 1 Proof The first statement follows from Lemma 21 The second follows considering the characterization provided in Lemma 31 Indeed suppose that pd B (B) = n < max{pd R (A), pd R (C)} The long exact sequence for Ext i R(, N) gives that Ext i R(A, N) = Ext i+1 R (C, N) for i n + 1 and all R-modules N This gives the result provided pd R (A) n + 1 or pd R (C) n + 2 The remaining case is when pd R (C) = n + 1 and pd R (A) < n + 1 If pd R (A) = n then the result holds and otherwise the Ext i R (, N) long exact sequence implies that Extn+1 R (C, N) = for all R- modules N, contradicting the fact that pd R (C) = n + 1 6

Given a ring R and an element r R that is not a zero divisor, consider an R/rR-module M Such a module is an R-module that is annihilated by r The next result gives the projective dimension of M over R/rR in terms of the projective dimension of M over R This is particularly relevant to our present consideration, since taking R = S[x] for some ring S and r = x, we will be able to deduce information about gldim(s[x]) from gldim(s) Theorem 46 Let r R be a non-zerodivisor, M be a non-zero R/rR-module, and suppose that pd R/rR (M) <, then pd R (M) = 1 + pd R/rR (M) Proof First, note that pd R (M) Indeed, this would mean that M is a projective R-module, implying that M is a direct summand of a free R-module and contradicting the fact that M is annihilated by r We hence have that pd R (M) 1 Next, suppose that M is a projective R/rR-module, then since R r R R/rR (44) gives a projective resolution of R/rR of minimal length, Theorem 44 gives pd R (M) pd R (R/rR) + pd R/rR (M) = 1 so the result holds It now suffices to consider the case when pd R (M), pd R/rR (M) 1 and we proceed via induction on n = pd R/rR (M) < Consider an R/rR-module projective resolution P n f n f 1 P f A of minimal length Taking K = kerf, we have that K P M is exact and pd R/rR (M) = 1 + pd R/rR (K) Lemma 45 implies that either or 1 = pd R (P ) = max{pd R (K), pd R (M)} (45) Since the induction hypothesis gives that pd R (M) = 1 + pd R (K) (46) pd R (K) = 1 + pd R/rR (K) provided pd R (K) 1, we are done if (46) holds or if (45) holds and pd R (K) =, since pd R (M) 1 7

It remains to consider the case when pd R (M) = 1 = pd R/rR (M) and we shall now see that this cannot happen Let P 1 P M be an R-module projective resolution of M Applying the functor R R/rR gives the exact sequence Tor R 1 (M, R/rR) P 1 R R/rR P R R/rR M of R/rR-modules Since P 1 R R/rR and P R R/rR are projective R/rRmodules, Lemma 31 implies that Tor1 R (M, R/rR) is projective Computing Tor1 R (M, R/rR) by applying M R to the R-module projective resolution (44), we find that Tor R 1 (M, R/rR) = M which implies pd R/rR (M) =, a contradiction The next result examines the case where r R is neither a zero divisor on R nor on M Theorem 47 Let M be an R-module and r R be neither a zero divisor on R nor on M, then pd R/rR (M/rM) pd R (M) Proof Since the result is trivial if pd R (M) =, we proceed via induction on pd R (M) < If pd R (M) = then M is projective, hence M/rM = R/rR R M is projective, ie pd R/rR (M/rM) = and the result holds Now suppose n = pd R (M) 1 and let K F M (47) be an exact sequence of R-modules with F free Lemma 31 gives that pd R (K) = n 1 and the induction hypothesis gives Applying R R/rR to (47) gives pd R/rR (K/rK) n 1 Tor R 1 (M, R/rR) K/rK F/rF M/rM and a calculation gives that Tor R 1 (M, R/rR) =, so K/rK F/rF M/rM is exact Lemma 45 now gives that either = pd R/rR (F/rF) = max{pd R/rR (K/rK), pd R/rR (M/rM)} 8

so pd R/rR (M/rM) = or that which gives the result pd R/rR (M/rM) = pd R/rR (K/rK) + 1 Corollary 48 Let M be an R-module, then (n 1) + 1 = n pd R (M) = pd R[x] (R[x] R M) Proof Since x is a non-zerodivisor on R[x] and R[x] R M, Theorem 47 gives pd R (M) pd R[x] (R[x] R M) Now, if P M is an R-module projective resolution, then applying R[x] R gives an R[x]-module projective resolution R[x] R P R[x] R M since R[x] is a flat R-module This gives pd R (M) pd R[x] (R[x] R M) We now arrive at the main theorem in this section Theorem 49 Let R be a ring, then gldim(r[x]) = gldim(r) + 1 Proof If gldim(r) = then Corollary 48 implies that gldim(r[x]) = and the result holds We now assume that gldim(r) = n < If N is an R-module, then N is an R[x]-module (let x act trivially) and Theorem 46 gives pd R[x] (N) = 1 + pd R (N) and hence gldim(r[x]) n + 1 Now, let M be an R[x]-module A computation shows that the sequence of R[x]-modules with R[x] R M ϕ R[x] R M ψ M ϕ(f m) = xf m f xm ψ(f m) = fm 9

is exact Lemma 45 then gives that either pd R[x] (M) = 1 + pd R[x] (R[x] R M) or pd R[x] (M) pd R[x] (R[x] R M) This gives pd R[x] (M) 1 + pd R[x] (R[x] R M) = 1 + pd R (M) 1 + n where we have used Corollary 48 This implies gldim(r[x]) n + 1 which completes the proof This result immediately implies the following corollary Corollary 41 Let R be a ring, then gldim(r[x 1,, x n ]) = gldim(r) + n Our proof of the Hilbert syzygy theorem follows directly from this result, noting that if k is a field, then gldim(k) = since all k-modules are free 5 Local Rings For the duration, we consider a noetherian local ring R with maximal ideal m Recall that such a ring is called a regular local ring provided dim(r) = dim k (m/m 2 ) where k = R/m is the residue field, dim( ) denotes Krull dimension, and dim k ( ) denotes vector space dimension Such rings arise naturally in algebraic geometry, where they characterize nonsingularity of varieties It is a standard fact that if R is a noetherian local ring then dim(r) is finite and in fact dim(r) dim k (m/m 2 ) [1, Chapter 11] In this section we aim to prove the following result [4, Corollary 4418]: Theorem 51 If R is a regular local ring and p R is a prime ideal, then R p is a regular local ring It is interesting to note that from the statement of this theorem, one would not suspect that homological algebra would play any role in its proof As in the last section, we will prove this result using a sequence of preliminary results We again follow the presentation given in [4, Section 44] We begin with an additional change of rings theorem, a strengthening of Theorem 47 in the case that M is finitely generated Theorem 52 Let R a noetherian local ring with maximal ideal m If M is a finitely generated R-module and r m is neither a zero divisor on R nor on M, then pd R/rR (M/rM) = pd R (M) 1

Proof We have pd R/rR (M/rM) pd R (M) by Theorem 47 so we are done if pd R/rR (M/rM) = Otherwise, we may proceed via induction on n = pd R/rR (M/rM) If n = then M/rM is projective and it follows that M/rM is free since R/rR is local and projective modules over local rings are free We now claim that this implies that M is free, ie pd R (M) = Indeed, let x 1,, x m M be elements mapping to a basis for M/rM and let (x 1,,x m ) denote the R-submodule they generate We have M = (x 1,, x m ) + rm = (x 1,, x m ) + mm so Nakayama s lemma implies (x 1,, x m ) = M To see that these elements are independent, suppose there exist s 1,, s m R so that s i x i = This gives that s i x i = in M/rM where x i denotes the image of x i under the canonical quotient map For each i we have, s i = r s i, so r ( s i x i) = and since r is not a zero divisor on M this implies s i x i = We may iterate this procedure and obtain an increasing sequence of ideals where s (j) i s (l) i R = s (l+1) i s i R s i R s(j) i R = rs (j+1) i, which must stabilize since R is noetherian This gives that R for some l and hence there exists a R so that ars (l+1) i = s (l+1) i Rearranging, this gives (1 ar)s (l+1) i is a unit [1, Proposition 19] This implies s (l+1) = and since r m we have that 1 ar = so s i = r l+1 s (l+1) i = for all i Having handled the n = case, we now easily finish the proof Let K F M be an exact sequence of R-modules with F free, so pd R (K) = pd R (M) 1 Applying the functor R R/R gives the exact sequence K/rK F/rF M/rM since Tor R 1 (M, R/rR) = if r is not a zero divisor on M Since F/rF is free, pd R/rR (K/rK) = n 1, and since M is finitely generated, we can suppose that F is as well As R is noetherian, this implies that K is finitely generated The induction hypothesis then gives that pd R (K) = n 1 and so pd R (M) = pd R (K) + 1 = n We now need to introduce further machinery for analyzing local rings Let M be an R-module, then a regular sequence on M is a sequence r 1, r d m so that r 1 is not a zero divisor on M and r i is not a zero divisor on M/(r 1,, r i 1 )M We let G(M) denote the length of the longest regular sequence on M Now recall the following result from the dimension theory of noetherian local rings [1, Corollary 1118]: i 11

Proposition 53 Let R be a noetherian local ring and r m not a zero divisor, then dim R/rR = dimr 1 This shows that G(R) dim(r) and R is called Cohen-Macaulay if we have G(R) = dim(r) Proposition 54 A regular local ring R is Cohen-Macaulay Moreover, any set r 1,,r d m mapping to a basis of m/m 2 is a regular sequence on R Proof It suffices to show that dim(r) = dim k (m/m 2 ) G(R), and we proceed via induction on d = dim(r) The case d = is trivial, so assume d 1 and let r 1,, r d m map to a basis of m/m 2 Since regular local rings are integral domains [1, Lemma 1123], r 1 is not a zero divisor and hence, by Proposition 53, dim(r/r 1 R) = d 1 Let r 2,, r d denote the images in R = R/r 1 R of r 2,, r d and note that r 2,, r d map to a basis of m/ m 2 where m = m R, hence R is regular The induction hypothesis implies that r 2,, r d is a regular sequence on R, and thus r 1,,r d is a regular sequence on R We now consider a result describing noetherian local rings R with G(R) = Note that this condition is equivalent to all elements of m being zerodivisors Lemma 55 Let R be a noetherian local ring with G(R) =, then if M is a finitely generated R-module either pd R (M) = or pd R (M) = Proof Suppose G(R) = and that pd R (M) = n, There then exists an exact sequence P n f n P f M with each P i projective and finitely generated Let K = imf n 1, which is a finitely generated syzygy of M with pd R (K) = 1 Let m 1,, m t M be elements which map to a basis for M/mM and hence generate M We now have an exact sequence P R t K where R t K is the obvious map, and Lemma 31 gives that P is projective, hence free Since mr t is the kernel of the map R t K K/mK, it follows that P mr t Now, since m consists only of zerodivisors, it is contained in the union of the associated primes of R and as it is maximal, it must be one of the associated primes (by prime avoidance) By definition, this gives that there exists s R so that m = {r R rs = } This gives that sp =, contradicting the fact that P is free We state the next two results, Theorem 422 and Corollary 4412 from [4] without proof, as the proofs would necessitate the introduction of a number of additional concepts (in particular injective dimension, flat dimension, and Tor-dimension [4, Section 41]) 12

Proposition 56 A ring R is semisimple (ie every ideal is a direct summand) iff gldim(r) = Proposition 57 Let R be a noetherian local ring with residue field k, then gldim(r) = pd R (k) We now arrive at our main theorem, which characterizes regular local rings via their global dimension Theorem 51 will follow as an easy corollary Theorem 58 A noetherian local ring R is regular iff gldim(r) Moreover, if R is regular then dim(r) = gldim(r) Proof We first assume that R is regular and show gldim(r) <, proceeding via induction on d = dim(r) If d = then m = m 2 so Nakayama s lemma implies that R is a field and the result holds Now suppose d >, then Proposition 54 gives G(R) = d so there exists r m so that r is not a zero divisor on R Since R = R/rR is regular and dim( R) = d 1, we have that gldim( R) = d 1 by the induction hypothesis Now, we compute gldim(r) = pd R (R/m) = pd R ( R) + pd R(R/m) = 1 + pd R( R/ m) = 1 + gldim( R) = d where m = m R is the maximal ideal of R and we have used Proposition 57 and Theorem 44 We now assume that gldim(r) < and deduce that R is regular; again, we proceed via induction Suppose that gldim(r) =, then R is local and semisimple (by Proposition 56) so R is a field, hence regular Now, let gldim(r), so by Lemma 55 there exists r m which is not a zerodivisor Infact, we can even choose r m m 2 ; indeed, if m m 2 consists only of zero divisors, then m m 2 p 1 p m where p 1,, p m are the associated primes of R (it is a standard fact that the union of these ideals precisely equals the zero divisors in R) This gives m m 2 p 1 p m and the strong form of prime avoidance [2, Lemma 33] shows that either m p j for some j or m = This implies that all elements of m are zero divisors Hence let r m m 2 be a non-zerodivisor Consider R = R/rR, which we now aim to show is regular We have exact sequences m R R k (51) and rr/rm m/rm α m R (52) 13

The Tor long exact sequence associated to (51) gives that rr/rm = Tor R 1 (R/rR, k), and a standard computation shows that Tor R 1 (R/rR, k) = k Now let r 2,, r n m be such that the images of r, r 2,,r n in m/m 2 form a basis (this is possible since r / m 2 ) Let n = (r 2,, r n )R+rm and note that n/rm surjects onto m R via α Since kerα = rr/rm is isomorphic to a field and contains r + rm (which is not in n/rm) we have that (rr/rm) (n/rm) = which implies n/rm = m R It then follows that m/rm = k m R as R-modules We now compute gldim( R) = pd R(k) pd R(m/rm) = pd R (m) = pd R (k) 1 = gldim(r) 1 where we have used Proposition 57, Lemma 43, and Theorem 52 The induction hypothesis now implies that R is regular This in turn implies that R is regular Indeed, let d = dim(r) and note that dim( R) = d 1 Since R is regular, Proposition 54 gives that there exist elements r 2,, r d m whose images r 2,, r d m R generate m R and give an R-regular sequence It follows that r, r 2,,r d give an R-regular sequence and generate m This shows dim k (m/m 2 ) G(R) and since we always have G(R) dim(r) dim k (m/m 2 ), this implies R is regular We now explain how this result implies Theorem 51 Suppose R is a regular local ring and p R is a prime ideal Let M be an R p module, then viewing M as an R-module, there is a projective resolution P n P M where n gldim(r) < Applying the functor R p R then gives an exact sequence R p R P n R p R P M of R p modules, since R p is a flat R-module and R p R M = M Each R p R P i is a projective R p -module, so this implies gldim(r p ) gldim(r) < By Theorem 58, R p is regular 14

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