Global Dimension and Regular Local Rings

Let R be a ring and let A be a left R-module. If there is a finite projective resolution 0 →Pn →· · · →P1 →P0 →A →0, then we write pd(A) ≤n. If n ≥0 is the smallest integer such that pd(A) ≤n, then we say that A has projective dimension n; if there is no finite projective resolution of A, then pd(A) = ∞.

Let E• = 0 →B η −→E0 d0 −→E1 d1 −→E2 →· · · be an injective resolution of a module B. If n ≥0, then the nth cosyzygy is ℧n(B, E•) = coker η if n = 0 coker dn−1 if n ≥1.

Let F• = · · · →F2 d2 −→F1 d1 −→F0 ε −→A →0 be a flat resolution of a module A. If n ≥0, then the nth yoke is Yn(A, F•) = ker ε if n = 0 ker dn if n ≥1. The term yoke is not standard; it is a translation of the Greek συζυγ ια (syzygy).

If M is an R-module over a commutative ring R, define M[x] = i≥0 Mi, where Mi ∼= M for all i. The R-module M[x] is an R[x]-module if we define x i ximi = i xi+1mi. In Lemma 9.55, we proved that if V is a free R-module over a commutative ring R, then V [x] is a free R[x]-module. The next result generalizes this from pd(V ) = 0 to higher dimensions.

If R is a commutative ring, then its prime-chain dimension is dim(R) = sup{ht(p) : p ∈Spec(R)}; that is, dim(R) is the length of a longest prime chain in R. If R is a one-dimensional integrally closed ACC domains, then dim(R) = 1, for every nonzero prime is a maximal ideal; if R is a domain, then dim(R) = 0 if and only if R is a field. The next proposition characterizes the ACC rings of prime-chain dimension 0.

If R is a commutative ring, then an associated prime ideal of a nonzero R-module B is a prime ideal of the form ann(b) for some nonzero b ∈B.

Let x1, . . . , xs be a sequence of elements in a commutative ring R. The exterior-multiplication complex M(x1, . . . , xs)• is defined as follows. M(x1, . . . , xs)p = <p(F), where F is the free R-module with basis {e1, . . . , es}. The differentiations Dp : ;p(F) → ;p−1(F) are defined by D1 ( s i=1 ciei ) = s i=1 ci xi, where ci ∈R for all i (so that D1(ei) = xi), and, for p > 1, Dp(ei1 ∧· · · ∧ei p) = p r=0 (−1)r−1xrei1 ∧· · · ∧ eir ∧· · · ∧ei p. If A is an R-module, the exterior-multiplication complex M(x1, . . . , xs, A)• is defined by M(x1, . . . , xs, A)• = A ⊗R M(x1, . . . , xs)•. We leave the straightforward calculation that Dp−1Dp = 0 to the reader; it is similar to that in the proof of Lemma 10.114. Thus, the exterior-multiplication complex really is a complex. Note that ;0(F) = R and that im d1 = I, where I = (x1, . . . , xs). In general, the exterior-multiplication complex is not acyclic; that is, it is not an exact sequence. However, if x1, . . . Observe that the pth term of M(x1, . . . , xs, k)• is, by definition, k ⊗R ;p(F). Since F is free of rank s, we know, from Theorem 9.140, the binomial theorem, that ;p(F) is free of rank (s p ) , and so k ⊗R ;p(F) is a vector space over k of dimension (s p ) . Thus, if we denote ;p(F) by Mp, as in Lemma 11.186, then k ⊗R ;p(F) is M p. If x1, . . . , xs is a minimal generating set for m, then Proposition 11.165 says that x∗ 1, . . . , x∗ s is a basis for m/m2, where x∗ i = xi + m2. If M is an R-module, then there is an isomorphism mM/m2M →(m/m2) ⊗R M, given by i xivi + m2M ↦ i x∗ i ⊗vi, where vi ∈M. If ϕ : M →M′ has the property that im ϕ ⊆mM′, then ϕ(u) = i xiv′ i, where v′ i ∈M′. Composing Δϕ : M/mM →mM′/m2M′ with the isomorphism above allows us to regard Δϕ : M/mM →(m/m2) ⊗R M′: Δϕ : u + mM ↦ϕ(u) + m2M′ = i xiv′ i + m2M′ ↦ i x∗ i ⊗v′ i. Regular Local Rings

If A is a set, let P(A)# denote the family of all its nonempty subsets. The axiom of choice states that if A is a nonempty set, then there exists a function β : P(A)# → A with β(S) ∈S for every nonempty subset S of A. Such a function β is called a choice function. Informally, the axiom of choice is a harmless looking statement; it says that we can simultaneously choose one element from each nonempty subset of a set. We now show that the axiom of choice is equivalent to a statement we would hate to be false. Proposition A.1. The axiom of choice holds if and only if the cartesian product i∈I Xi of nonempty sets is itself nonempty.1

For every n ≥1, for all left R-modules A and B, and for every projective resolution P• of B, there is an isomorphism Extn+1 R (A, B) ∼= Ext1 R($n−1(A, P•), B).

For all left R-modules A and B, for all n ≥0, and for any injective resolutions E• and E′• of B, there is an isomorphism Ext1 R(A, ℧n(B, E•)) ∼= Ext1 R(A, ℧n(B, E′•)).

For every n ≥1, for all right R-modules A and left R-modules B, and for every flat resolution F• of A, there is an isomorphism TorR n+1(A, B) ∼= TorR 1 (A, Yn−1(B, F•)).

Suppose that Extn R(A, B) = {0} for all left R-modules A and B. Then TorR n (C, D) = {0} for all right R-modules C and all left R-modules D.

If 0 →A′ →A →A′′ →0 is a short exact sequence, then pd(A′′) ≤1 + max{pd(A), pd(A′)}.

allows us to compute global dimension as the supremum of projective dimensions of cyclic modules. When R is a local ring, there is a dramatic improvement; global dimension is determined by the projective dimension of one cyclic module: the residue field k.

says that there are only finitely many minimal prime ideals. Since dim(R) = 0, every prime ideal is a minimal prime ideal (and a maximal ideal). We conclude that R has only finitely many prime ideals, say, p1, . . . , pn. Define N = p1 · · · pn ⊆p1 ∩· · · ∩pn = nil(R), so that N m = (p1 · · · pn)m = {0}. Let M be a finitely generated R-module, and consider the chain M ⊇p1M ⊇p1p2M ⊇· · · ⊇N M. The factor module p1 · · · pi−1M/p1 · · · pi M is an (R/pi)-module; that is, it is a vector space over the field R/pi (for pi is a maximal ideal). Since M is finitely generated, the factor module is finite-dimensional, and so the chain can be refined so that all the factor modules are simple. Finally, repeat this argument for the chains N j M ⊇p1N j M ⊇p1p2N j M ⊇· · · ⊇N j+1M. Since N m = {0}, we have constructed a composition series for M. Conversely, if every finitely generated R-module has a composition series, then the cyclic R-module R has a composition series; say, of length ℓ. It follows that any ascending chain of ideals has length at most ℓ, and so R is ascending-chain-finite. To prove that dim(R) = 0, Regular Local Rings we must show that R does not contain any prime ideals p ⊋q. Passing to the quotient ring R/q, we may restate the hypotheses: R is a domain having a nonzero prime ideal as well as a composition series R ⊇I1 ⊇· · · ⊇Id ̸= {0}. The last ideal Id is a minimal ideal; choose a nonzero element x ∈Id. Of course, x Id ⊆Id; since R is a domain, x Id ̸= {0}, so that minimality of Id gives x Id = Id. Hence, there is y ∈Id with xy = x; that is, 1 = y ∈Id, and so Id = R. We conclude that R is a field, contradicting its having a nonzero prime ideal. • We are going to prove a theorem of W. prime-chain, the principal ideal theorem, which implies that every prime ideal in a ACC rings has finite height. Rees. We begin with a technical lemma.

If R is a ACC rings, then every prime ideal has finite height, and so Spec(R) has DCC.

Let (R, m) be a ascending-chain-finite local ring. If m can be generated by an R-sequence x1, . . . , xd, then R is a regular local ring and d = dim(R) = µ(m). Remark. We will soon prove the converse: In a regular local ring, the maximal ideal can be generated by an R-sequence. ◀

Let (R, m, k) be a ascending-chain-finite local ring of finite global dimension. If µ(m) ≤D(R) and D(R) ≤d, where d is the length of a longest R-sequence in m, then R is a regular local ring.

Let R be a commutative ring, let A and B be R-modules, and let x1, . . . , xn be a B-sequence in ann(A). If I = (x1, . . . , xn), then HomR(A, B/I B) ∼= Extn R(A, B). Regular Local Rings

Let (R, m, k) be a ascending-chain-finite local ring, let f : A →B be a map of finitely generated R-modules, and let f = f ⊗1k. (i) f is surjective if and only if f is surjective. (ii) If, in addition, both A and B are free R-modules, then f injective implies that f is a (split) injection.

If (R, m, k) is a ascending-chain-finite local ring of finite global dimension D(R), then µ(m) ≤D(R).

Let R be a domain, and let I be a nonzero projective ideal in R. If m is a maximal ideal in R, then Im ∼= Rm.

(i) A module A is projective if and only if pd(A) = 0. We may thus regard pd(A) as a measure of how far away A is from being projective. (ii) If R is a one-dimensional integrally closed ACC domains, then pd(A) ≤1 for every R-module A. By Theorem 11.101, every submodule of a free R-module is projective. Hence, if F is a free R-module and ε: F →A is a surjection, then 0 →ker ε →F ε −→A →0 is a projective resolution of A. This argument extends to left hereditary rings. ◀

Let k be a field, and let R = k[[x1, . . . , xr]] be the ring of formal power series in r variables x1, . . . , xr. Recall that an element f ∈R is a sequence f = ( f0, f1, f2, . . . , fn, . . .), where fn is a homogeneous polynomial of total degree n in k[x1, . . . , xr], and that multiplication is defined by ( f0, f1, f2, . . .)(g0, g1, g2, . . .) = (h0, h1, h2, . . .), where hn = i+ j=n fig j. We claim that R is a local ring with maximal ideal m = (x1, . . . , xr) and residue field k. First, R/m ∼= k, so that m is a maximal ideal. Now f /∈m if and only if f0 ̸= 0, and we now show that f is a unit if and only if f0 ̸= 0. If f g = 1, then f0g0 = 1, and f0 ̸= 0; conversely, if f0 ̸= 0, we can solve ( f0, f1, f2, . . .)(g0, g1, g2, . . .) = 1 recursively for gn, and f g = 1 if we define g = (g0, g1, g2, . . .). But R/(x1, . . . , xi−1) is a domain, because it is isomorphic to k[[xi, . . . , xr]], and so xi is a regular element on it. Hence, Proposition 11.168 shows that R = k[[x1, . . . , xr]] is a regular local ring, for x1, . . . , xr is an R-sequence. ◀ The next lemmas prepare us for induction. Regular Local Rings