A COURSE IN HOMOLOGICAL ALGEBRA

Chapter **: Syzygies, Projective Dimension, Regular Sequences, and Depth

E. L. Lady (April 5, 1997)

Syzygies, Torsionless Modules, and Reflexive Modules

In this section, modules are usually assumed to be finitely generated.

Definition. A Λ-module A is called an 0^{th} syzygy if it is isomorphic to a submodule of a projective Λ-module. We now define the concept of an n^{th} syzygy recursively. A is an n^{th} syzygy if there exists a projective module P such that A can be embedded in P in such a way that P/A is a (n − 1)^{st} syzygy. In other words, A is an nth syzygy if there exists an exact sequence

0 ⟶ A ⟶ P_{n} ⟶ … ⟶ P_{1} ⟶ P_{0} ,

where the P_{i} are all projective Λ-modules.

The fascinating thing is that zeroth syzygies and first syzygies have an intrinsic significance in terms of the duality functor A ↦ A^{∗} = Hom_{Λ}(A, Λ). Namely, a left Λ-module A is a first syzygy if and only if it is the dual of some right Λ-module, and is a zeroth syzygy if and only if natural isomorphism A → A^{∗∗} is monic.

Another interesting fact is that A^{∗} = 0 if and only if A ≈ Ext_{Λ}^{1}(B, Λ) for some right Λ-module B. We will now proceed to prove these results.

References. Bass, Trans. Amer. Math Soc. 95(1960), 466-88. Jans, Trans. Amer. Math. Soc. 106(1963),330-40. Jans, Rings and Homology. (Auslander, “Coherent Functors,” in the La Jolla Conference on Categorical Algebra.)

Notation. If A is a left Λ-module, we set A^{∗} = Hom_{Λ}(A, Λ). A^{∗} is a right Λ-module. We define δ_{A}: A → A^{∗∗} by

δ_{A}(a)(φ) = φ(a) for φ ∈ A^{∗}.

Lemma. δ is a natural transformation. Furthermore, δ_{P} is an isomorphism if P is finitely generated projective.

proof: For the second statement it suffices to prove that δ_{Λ} is an isomorphism. ✓ Lemma. δ_{A^{∗}}: A^{∗} → A^{∗∗∗} is split monic.

proof: We claim that (δ_{A})^{∗}δ_{A^{∗}} = 1_{A^{∗}}. In fact,

A^{∗} ⟶^{δ_{A^{∗}}} A^{∗∗∗}  A^{∗∗∗} ⟶^{δ_{A}^{∗}} A^{∗}

and for φ ∈ A^{∗},

(∀a ∈ A) (δ_{A})^{∗}(δ_{A^{∗}}(φ))(a) = δ_{A^{∗}}(φ)(δ_{A}(a)) = δ_{A}(a)(φ) = φ(a)

so that (δ_{A})^{∗}(δ_{A^{∗}}(φ)) = φ and so (δ_{A})^{∗}δ_{A^{∗}} = 1_{A^{∗}}. ✓

Theorem. Let Λ be a left and right noetherian ring (not necessarily commutative) and let A be a finitely generated left Λ-module such that A^{∗} = 0. Then there exists a finitely generated right Λ-module B such that A ≈ Ext_{Λ}^{1}(B, Λ).

proof: Suppose that A^{∗} = 0 and consider a projective resolution

P_{1} ⟶^{∂} P_{0} ⟶ A ⟶ 0 .

Let B = Coker ∂^{∗}, so that

0 → A^{∗} = 0 ⟶ P_{0}^{∗} ⟶^{∂^{∗}} P_{1}^{∗} ⟶ B ⟶ 0

is exact. Since P_{i}^{∗∗} ≈ P_{i}, this yields an exact sequence

0 → B^{∗} ⟶ P_{1} ⟶^{∂} P_{0} ⟶ Ext_{Λ}^{1}(B, Λ) ⟶ 0 = Ext_{Λ}^{1}(P_{1}^{⋆}, Λ) ,

showing that A = Coker ∂ ≈ Ext_{Λ}^{1}(B, Λ). ✓

Definition. We say that a Λ-module A is torsionless if δ_{A} is a monomorphism and reflexive if δ_{A} is an isomorphism.

Theorem. Let Λ be right noetherian and let A be a finitely generated left Λ-module. Then A is torsionless if and only if A is a 0th syzygy.

proof: To say that δ_{A} is monic is to say that for every a ∈ A there exists φ ∈ A^{∗} such that φ(a) ≠ 0.

( ⇐ ): Easy. ( ⇒ ): Let F ↠ A be a surjection with F finitely generated and free. This induces A^{∗} ↣ F^{∗}. Thus A^{∗} is finitely generated since Λ is right noetherian. Thus there exists a finitely generated free right Λ-module G and a surjection G ↠ A^{∗}. Thus if δ_{A} is monic there are monomorphisms

A ↣ A^{∗∗} ↣ G^{∗}.

Since G^{∗} is free, A is a 0th syzygy. ✓ Lemma. Let A be a submodule of the left Λ-module P. Define

A′ = {φ ∈ P^{∗} | φ(A) = 0}
A′′ = {p ∈ P | (∀φ ∈ A′) φ(p) = 0}.

Then there is an exact sequence

0 → A′ → P^{∗} ⟶^{ρ} A^{∗} .

Furthermore, if P/A is torsionless then A′′ = A.

proof: A′ is clearly the kernel of the map P^{∗} ⟶^{ρ} A^{∗} induced by the inclusion A ↪ P (since ρ(φ) is the restriction of φ to A). This justifies the first assertion.

Now let φ′ ∈ (P/A)^{∗}. Then φ′ is induced by a map φ ∈ P^{∗} such that φ(A) = 0, i. e. φ ∈ A′. Then for p ∈ A′′ ⊆ P,

δ_{P/A}(p + A)(φ′) = φ′(p + A) = φ(p) = 0,

by definition of A′′. Thus if P/A is torsionless, so that Ker δ_{P/A} = 0, then A′′/A = 0 so

that A′′ = A. ✓

Theorem. Let Λ be both left and right noetherian and let A be a finitely generated left Λ-module. Then A is a first syzygy if and only if A ≈ C^{∗} for some right Λ-module C.

proof: ( ⇒ ): Suppose that A is a first syzygy. Then there is an exact sequence

0 → A → P_{1} → P_{0} ,

where P_{0} and P_{1} are finitely generated projective. Let A′ ⊆ P_{1}^{∗} and A′′ ⊆ P_{1} be defined as in the previous lemma. Then

A′′ = {p ∈ P_{1} | δ_{P_{1}}(p)(A′) = 0} ,

so that there is an exact sequence

0 ⟶ A′′ ⟶ P_{1} ⟶ (A′)^{∗} ,

where the right hand map is induced by δ_{P_{1}} , so that there is a commutative diagram

0 ⟶ A′′ ⟶ P_{1} ⟶ (A′)^{∗}
δ_{P_{1}} ↓^{≈}  ∥
0 ⟶ (P_{1}^{∗}/A′)^{∗} ⟶ P_{1}^{∗∗} ⟶^{ρ} (A′)^{∗} ,

where ρ is induced by the inclusion A′ ↪ P_{1}^{∗}, so that Ker ρ = (P_{1}^{∗}/A′)^{∗}, as indicated. Thus A′′ ≈ (P_{1}^{∗}/A′)^{∗}. But P_{1}/A is a zeroth syzygy since P_{1}/A ⊆ P_{0} and is thus torsionless, so that by the preceding Lemma A′′ = A. Thus A ≈ C^{∗} with C ≈ P_{1}^{∗}/A′.

( ⇐ ): Suppose that A = C^{∗}. Let P_{1} be a projective which maps onto C, so that we have

\begin{matrix}0 & ⟶ & K & ⟶ & P_{1} & ⟶ & C & ⟶ & 0 \\ 0 & ⟶ & C^{∗} & ⟶ & P_{1}^{∗} & ⟶ & K^{∗} . \end{matrix}

But K^{∗} is a 0th syzygy since δ_{K^{∗}}: K^{∗} → K^{∗∗∗} is monic (in fact, split monic). Thus C^{∗} is a first syzygy. ✓ Theorem. Let Λ be left and right noetherian and let A be a finitely generated torsionless left Λ-module. Then there exists a torsionless right Λ-module B such that there are exact sequences

0 ⟶ A ⟶^{δ_{A}} A^{∗∗} ⟶ Ext_{Λ}^{1}(B, Λ) ⟶ 0
0 ⟶ B ⟶^{δ_{B}} B^{∗∗} ⟶ Ext_{Λ}^{1}(A, Λ) ⟶ 0.

proof: Starting with a short exact sequence 0 → K → P → A → 0 with P projective, one gets

(1)  0 ⟶ A^{∗} ⟶ P^{∗} ⟶ B ⟶ 0,

with B = P^{∗}/A^{∗}. Since A = P/K, this can be written

0 ⟶ K′ ⟶ P^{∗} ⟶ B ⟶ 0.

Since P^{∗} is projective, (1) induces

(2)  0 ⟶ B^{∗} ⟶ P^{∗∗} ⟶ A^{∗∗} ⟶ Ext_{Λ}^{1}(B, Λ) ⟶ 0.

As above, let

K′′ = {p ∈ P | (∀φ ∈ K′) φ(p) = 0}.

Then the image of B^{∗} in P^{∗∗} is δ_{P} (K′′) and K′′ = K by the Lemma above, since by hypothesis P/K = A is torsionless. Thus we have

0 ⟶ K′′ ⟶ P ⟶ A ⟶ 0
δ_{K} ↓^{≈}  ^{δ_{P}} ↓^{≈}  ↓^{δ_{A}}
0 ⟶ B^{∗} ⟶ P^{∗∗} ⟶ A^{∗∗}

where δ_{A} is monic since A is a zeroth syzygy. Thus A ≈ P^{∗∗}/B^{∗}. Combining this with (2) one has

(3)

0 ⟶ P^{∗∗}/B^{∗} ⟶ A^{∗∗} ⟶ Ext_{Λ}^{1}(B, Λ) ⟶ 0
↓^{≈}  ∥  ∥
0 ⟶ A ⟶^{δ_{A}} A^{∗∗} ⟶ Ext_{Λ}^{1}(B, Λ) ⟶ 0 ,

as required.

Applying the same reasoning to (3) we get

0 ⟶ B ⟶ B^{∗∗} ⟶ Ext_{Λ}^{1}(A, Λ) ⟶ 0 . ✓

Corollary. Λ has the property that every finitely generated Λ-module is reflexive if and only if Λ is left and right self-injective.

proof: ( ⇐ ): If A is a finitely generated Λ-module, consider a projective resolution P_{1} → P_{0} ↠ A. Now if Λ is injective as a left and a right Λ-module, then the functor A ↦ A^{∗∗} is exact, so the rows in the following diagram are exact:

P_{1} ⟶ P_{0} ⟶ A ⟶ 0
δ_{P_{1}} ↓^{≈}  ^{δ_{P_{0}}} ↓^{≈}  ^{δ_{A}}↓
P_{1}^{∗∗} ⟶ P_{0}^{∗∗} ⟶ A^{∗∗} ⟶ 0 ,

showing that δ_{A} is an isomorphism, so A is reflexive.

( ⇒ ): If every finitely generated Λ-module is reflexive, then a fortiori every finitely generated A is torsionless. Thus by the Theorem there exists a Λ-module B and an exact sequence

0 ⟶ B ⟶^{δ_{B}} B^{∗∗} ⟶ Ext_{Λ}^{1}(A, Λ) ⟶ 0 .

Since by hypothesis δ_{B} is an isomorphism, Ext_{Λ}^{1}(A, Λ) = 0. Since this is true for all finitely generated A, Λ is an injective module. ✓

Corollary. Let Q denote the injective envelope of Λ. Then all finitely generated torsionless modules are reflexive if and only if Q/Λ is injective (i. e. inj dim Λ ≤ 1).

proof: ( ⇒ ): By the same reasoning as in the proof of the previous corollary, if all finitely generated torsionless modules are reflexive, then the theorem implies that Ext_{Λ}^{1}(A, Λ) = 0 whenever A is torsionless, i. e. whenever A is a zeroth syzygy. Now let M be any finitely generated Λ-module and let P be a projective module which maps onto M, giving a short exact sequence

0 ⟶ K ⟶ P ⟶ M ⟶ 0 .

Then K is a zeroth syzygy and thus torsionless, so Ext_{Λ}^{1}(K, Λ) = 0. But then

Ext_{Λ}^{2}(M, Λ) ≈ Ext_{Λ}^{1}(K, Λ) = 0.

On the other hand, from the sequence 0 → Λ → Q → Q/Λ → 0 it follows that

Ext_{Λ}^{1}(M, Q/Λ) ≈ Ext_{Λ}^{2}(M, Λ) = 0 .

Since this is true for every finitely generated M, Q/Λ must be injective. ✓

Corollary. Let Λ be a commutative noetherian ring and let T be its total quotient ring. Then Λ has the property that all finitely generated duals A^{∗} are reflexive if and only if T/Λ is an injective Λ-module.

proof: T is the injective envelope of Λ, since it is a maximal essential extension

of Λ. ✓

Projective Dimension

Schanuel’s Lemma. Let

0 → K → P →M → 0
0 → K_{1} → P_{1} →M → 0

be short exact sequences such that P and P_{1} are projective. Then

K ⊕ P_{1} ≈ K_{1} ⊕ P.

Definition. Projective Dimension Theorem. The projective dimention of a Λ-module A is the smallest n such that Ext_{Λ}^{n+1}(A, ) = 0, or ∞ if there is no such n. Theorem. If Λ is left noetherian, then proj. dim A is the smallest n such that Ext_{Λ}^{n+1}(A, B) = 0 for all finitely generated B. proof: It suffices to see that K is projective if and only if Ext_{Λ}^{1}(K, B) = 0 for all finitely generated B. To see this, take B to be a zeroth syzygy for K. ✓ Theorem. If R is commutative then proj. dim_{R} A = sup_{𝔪} proj. dim_{R_{𝔪}} A_{𝔪}. Theorem. Let

0 ⟶ A ⟶ B ⟶ C ⟶ 0

be an exact sequence. If any two of these modules have finite projective dimension, then so does the third. Furthermore, either

proj. dim B < proj. dim C = proj. dim A + 1

or

proj. dim B = max{proj. dim A, proj. dim C}.

Regular M -sequences and Depth

Reference. Matsumura, Commutative Algebra.

In the rest of this chapter, R will denote a commutative ring. Definition. Ass M. Regular M-sequence. Theorem. If R is noetherian then an R-module M is trivial if and only if Ass M = ∅. Furthermore, if x is a zero divisor on an R-module M, then there exists a prime ideal 𝔭 ∈ Ass M with x ∈ 𝔭.

Proposition. If 𝔭 is a prime ideal in R, then Ass R/𝔭 = {𝔭}.

Theorem. If R is noetherian and M is finitely generated, then Ass M is finite.

proof: Since M is noetherian, there is a maximal submodule M′ of M such that Ass M′ is finite. If M′ ≠ M, one easily gets a contradiction. ✓

Lemma. For N ⊆ M,

Ass M ⊆ Ass N ∪ Ass(M/N).

Definition. The support of an R-module M consists of the set of primes 𝔭 such that M_{𝔭} ≠ 0.

Proposition. Every element of Ass M belongs to Supp M. Furthermore, the primes which are minimal in Supp M belong to Ass M.

Lemma. If an ideal is contained in a finite union of prime ideals, then it is contained in one of those primes.

Corollary. Let R be noetherian and M a finitely generated R-module. If 𝔞 is an ideal consisting of zero divisors on M, then 𝔞 is contained in some associated prime 𝔭 for M.

Definition. Regular M-sequence. Depth_{𝔞}M.

Theorem. Let R be a noetherian ring and M a finitely generated R-module and 𝔞 an ideal such that 𝔞M ≠ M. If r is the smallest integer such that Ext_{R}^{r}(R/𝔞, M) ≠ 0, then every regular M-sequence in 𝔞 has length r. Thus r = depth_{𝔞} M.

proof: Clear if r = 0, since if Hom_{R}(R/𝔞, M) ≠ 0 then no element of 𝔞 is regular on M and conversely, if no regular M-sequences exist in 𝔞 then 𝔞 consists of zero divisors on M and consequently 𝔞 ⊆ 𝔭 for some 𝔭 ∈ Ass M by the preceding Corollary, and Hom_{R}(R/𝔭, M) ≠ 0 (since M contains a submodule isomorphic to R/𝔭), so also Hom_{R}(R/𝔞, M) ≠ 0.

Now let r ≥ 1. Then by the preceding paragraph, we can choose x_{1} ∈ 𝔞 which is regular on M. Now use induction, since clearly x_{2}, …, x_{r} is a maximal regular sequence of M/x_{1}M if and only if x_{1}, …, x_{r} is a maximal regular M-sequence. ✓

Lemma. Let (R, 𝔪) be local and suppose that depth_{𝔪} R = 0 (i. e. 𝔪 consists of zero divisors). Then any R-module M with finite projective dimension is in fact projective.

proof: It suffices to prove that all modules M with proj. dim M ≤ 1 are projective. If M is such a module, take a projective resolution

0 ⟶ P_{1} ⟶ P_{0} ⟶^{ε} M ⟶ 0

such that ε induces an isomorphism from P_{0}/𝔪P_{0} onto M/𝔪M. Then P_{1} ⊆ 𝔪P_{0} . Since 𝔪 consists of zero divisors, 𝔪 ⊆ 𝔭 for some 𝔭 ∈ Ass R. Since 𝔪 is maximal, thus 𝔪 ∈ Ass R. Thus there exists x ∈ R with x𝔪 = 0, and consequently xP_{0} = 0. Since x can’t be invertible, x ∈ 𝔪, so by Nakayama’s Lemma, P_{1} = 0, so that ε is an isomorphism and M is projective. ✓ Lemma. Let (R, 𝔪) be local, let M be an R-module with finite projective dimension, and let x ∈ M be regular on M. Then proj. dim(M/xM) = 1 + proj. dim M.

proof: Since x is regular on M, the sequence

0 ⟶ M ⟶^{x} M ⟶ M/xM ⟶ 0

is exact. The induced long exact Ext sequence looks like

… ⟶ Ext_{R}^{k}(M, ) ⟶^{x} Ext_{R}^{k}(M, ) ⟶ Ext_{R}^{k+1}(M/xM, ) ⟶ Ext_{R}^{k+1}(M, ) ⟶ …

Now if Ext_{R}^{k}(M, ) ≠ 0, then Ext_{R}^{k}(M, X) ≠ 0 for some finitely generated X, and in this case Ext_{R}^{k}(M, X) is finitely generated. Since x ∈ 𝔪, multiplication by x cannot be surjective on Ext_{R}^{k}(M, X) by Nakayama’s Lemma. It then follows the Ext_{R}^{k+1}(M/xM, ) ≠ 0.

But conversely, the long exact sequence shows that if Ext_{R}^{k}(M, ) = 0 (and therefore also Ext_{R}^{k+1}(M, ) = 0), then Ext_{R}^{k+1}(M/xM, ) = 0. ✓

Theorem. Let (R, 𝔪) be local and x_{1}, …, x_{r} a regular M-sequence. Then

proj. dim M/(x_{1}, …, x_{r})M = r + proj. dim M.

Theorem. Let (R, 𝔪) be local and let M be a finitely generated R-module with finite projective dimension. Then

proj. dim M = depth_{𝔪} R − depth_{𝔪} M.

proof: A previous lemma covers the case depth_{𝔪} R = 0 (i. e. 𝔪 ∈ Ass R).

Now use a double induction on depth_{𝔪} R and depth_{𝔪} M by choosing x ∈ 𝔪 regular on R and using the Lemma below. The difficult part concerns the case when depth_{𝔪} M = 0 and depth_{𝔪} R > 0.

Lemma. Let (R, 𝔪) be local, M finitely generated, and let x ∈ 𝔪 be regular on both R and M. Write R̄ = R/xR and M̄ = M/xM. Then

proj. dim_{R} M = proj. dimR̄ M̄.

proof: Consider a minimal projective resolution for M,

0 ⟶ P_{n} ⟶ … ⟶ P_{2} ⟶^{∂_{2}} P_{1} ⟶^{∂_{1}} P_{0} ⟶ M ⟶ 0.

(I.e. suppose that the induced maps P_{i}/𝔪P_{i} ⟶ ∂_{i}(P_{i})/𝔪∂_{i}(P_{i}) are all isomorphisms.) Then proj. dim M is the largest n such that P_{n} ≠ 0. Now since x is regular on M, Tor_{i}^{R}(R/xR, M) = 0 for all i > 0. Therefore the induced sequence

0 ⟶ P_{n}/xP_{m} ⟶ … ⟶ P_{2}/xP_{2} ⟶ P_{1}/xP_{1} ⟶ P_{0}/xP_{0} ⟶ M̄ ⟶ 0

is exact. But this is a minimal projective resolution for M̄ over R̄, so that proj. dimR̄ M̄ = n = proj. dim_{R} M. ✓ Addendum. If x_{1}, …, x_{r} is a regular M-sequence in 𝔞 and r = depth_{𝔞} M, then

Ext_{R}^{r}(R/𝔞, M) ≈ Hom_{R}(R/𝔞, M/(x_{1}, …, x_{r})M).

Proposition. Let R be noetherian, M a finitely generated R-module, N a submodule of M, and 𝔞 an ideal such that 𝔞M ≠ M, 𝔞N ≠ N, and 𝔞(M/N) = M/N. Then either

(*)  depth_{𝔞} M = min{depth_{𝔞} N, depth_{𝔞} M/N}

or

(**)  depth_{𝔞} M > depth_{𝔞} M/N = depth_{𝔞} N − 1 .

proof: If depth_{𝔞} M and depth_{𝔞} M/N are non-zero, choose an element x ∈ 𝔞 which is regular on both M and M/N. Since Tor_{1}^{R}(R/xR, M/N) = 0,

0 ⟶ N/xN ⟶ M/xM ⟶ (M/N)/x(M/N) ⟶ 0

is exact, making an induction possible.

Now it is easily seen that if depth_{𝔞} M = 0 then either depth_{𝔞} N = 0 or depth_{𝔞} M/N = 0, so that (∗) holds in this case.

Now if depth_{𝔞} M/N = 0 and depth_{𝔞} M ≠ 0, then there exists a coset m + N ∈ M/N such that 𝔞(m + N) = 0. There also exists an element x ∈ 𝔞 which is regular on M. Then xm ∉ xN but 𝔞xm ⊆ xN since 𝔞(m + N) = 0 ∈ M/N, so that 𝔞 consists of zero divisors on N/xN, and thus depth_{𝔞} N = 1 and (**) holds. ✓

Lemma. Let (R, 𝔪) be local and let k = R/𝔪. Let M be a finitely generated non-trivial R-module. Then proj. dim M is the smallest n such that Tor_{n+1}^{R}(M, k) = 0.

proof: By considering the nth syzygy in a projective resolution for M, it suffices to see that a finitely generated R-module N is projective if Tor_{1}^{R}(N, k) = 0. Consider an short exact sequence

0 ⟶ K ⟶ F ⟶^{ε} N ⟶ 0

such that the induced map F/𝔪F → N/𝔪N is an isomorphism. If Tor_{1}^{R}(N, k) = 0, then

0 ⟶ K/𝔪K ⟶ F/𝔪F ⟶ N/𝔪N ⟶ 0

is exact, so that K/𝔪K = 0. Therefore K = 0 by Nakayama’s Lemma, so that N is

free. ✓

Theorem. Let Λ be a ring which need not be either commutative nor noetherian. The following conditions are equivalent:

(1) proj. dim M ≤ n for all left Λ-modules M.

(2) proj. dim M ≤ n for all finitely generated left Λ-modules M.

(3) inj. dim M ≤ n for all finitely generated left Λ-modules M.

(4) Ext_{Λ}( , ) = 0.

Definitions. Left global dimension.

A commutative local ring (R, 𝔪) is regular if 𝔪 is generated by a regular R-sequence.

A commutative ring is regular if R_{𝔪} is regular for every maximal ideal 𝔪.

height_{M} 𝔭.

The Krull dimension of M is the supremum of height_{M} 𝔭 for all maximal ideals 𝔭.

Definition. Let M be a finitely generated R-module and let 𝔞 be its annihilator. Then the grade of M is defined to be depth_{𝔞} R, in other words grade M is the length of the longest R-sequence consisting completely of elements that annihilate M.

Warning. Unfortunately, the words depth and grade have been used inconsistently in the literature. Kaplansky uses grade to mean what we have defined as depth. I guess some people feel rather strongly about this issue, or at least such was true many years ago. The definitions given are the ones that seemed to be most prevalent at the time these notes were written.

Theorem. If M is finitely generated, then grade M is the smallest integer n such that Ext_{R}^{n}(M, R) ≠ 0.

proof: If grade M = 0, then 𝔞 = ann M consists of zero-divisors in R, and so there must exist 𝔭 ∈ Ass R such that 𝔭 ⊃ 𝔞 = ann M. Since R then contains a submodule isomorphic to R/𝔭, in order to prove that Hom_{R}(M, R) ≠ 0 it suffices to prove that Hom_{R}(M, R/𝔭) ≠ 0. Now 𝔭M ≠ M, otherwise M_{𝔭} = 0 by Nakayama’s Lemma, contrary to the fact that 𝔭 ∈ Supp M since 𝔭 ⊃ 𝔞. It thus suffices to show that Hom_{R}(M/𝔭M, R/𝔭) ≠ 0. Therefore there is no loss of generality in supposing that M is an integral domain and 𝔭 = 0 ∈ Supp M. With this assumption, let K be the quotient field of R and choose a basis u_{1}, …, u_{r} for M_{𝔭}. Let α: M → M_{𝔭} be the canonical map and let s ∈ R be such that sα(M) ⊆ Ru_{1} ⊕ ⋯ ⊕ Ru_{r}. Then the composition of one of the projection mappings from Ru_{1} ⊕ ⋯ ⊕ Ru_{r} into R with sα yields a non-trivial homomorphism in Hom_{R}(M, R).

Conversely, if there exists φ ≠ 0 ∈ Hom_{R}(M, R) then for every a ∈ ann M, aφ(M) = 0 so that ann M consists of zero divisors and grade M = 0.

Now suppose that grade M ≥ 1 and let x ∈ 𝔞 = ann M be regular in R. It is easily seen that grade_{R} M = 1 + grade_{R/xR} M/xM. The theorem is therefore a consequence of the following lemma:

Lemma. Let x be a regular element in R such that xM = 0. Let C be the category of R/xR-modules, identified as the full subcategory of the category of R-modules consisting of those modules M such that xM = 0. Then for i ≥ 1, the restriction of Ext_{R}^{i}( , R) to C is naturally isomorphic to Ext_{R/xR}^{i−1}( , R/xR).

proof: {Ext_{R}^{i+1}}_{i≥0} is clearly an exact co-connected sequence of functors on C.

(1) From the fact that Hom_{R}( , R) = 0 on C we get

0 = Hom_{R}( , R) ⟶ Hom_{R}( , R/xR) ⟶ Ext_{R}^{1}( , R) ⟶_{0}^{x} Ext^{1}( , R) ⟶ …

for modules in C, so that

Hom_{R/xR}( , R/xR) ≈ Hom_{R}( , R/xR) ≈ Ext_{R}^{1}( , R)

for modules in C.

(2) Since proj. dim R/xR ≤ 1, Ext_{R}^{i+1}(R/xR, R) = 0 for i > 0. Thus Ext_{R}^{i+1}( , R) vanishes on free R/xR-modules. The theorem now follows from the characterization of derived functors in terms of universal properties. ✓

Corollary. If 𝔞 is an idea such that grade R/𝔞 ≥ 2, then any map φ: 𝔞 → R is given by mujltiplication by a unique element of R.

proof: This follows from the exact sequence

0 → Hom_{R}(R/𝔞, R) = 0 → Hom_{R}(R, R) → Hom_{R}(𝔞, R) → Ext_{R}^{1}(R/𝔞, R) = 0. ✓

Auslander-Buchsbaum, Annals 68(1958), pp. 625–57.

Lemma. If

0 → F_{n} → ⋯ → F_{1} → F_{0} → F_{−1} → 0

is an exact sequence of finitely generated free R-modules, then

∑_{−1}^{n} rank F_{i} = 0.

Proposition 6.2. If M has a finite resolution

0 → F_{n} → ⋯ → F_{1} → F_{0} → M → 0

where the F_{i} are finitely generated free R-modules, then M is faithful if and only if grade M = 0.

(In other words, if M has a finite free resolution then the annihilator of M is non-trivial if and only if it contains a regular element.)

proof: ( ⇒ ): Trivial.

( ⇐ ): Let 𝔭 ∈ Ass R. Localizing the above resolution at 𝔭, we see that M_{𝔭} has finite projective dimension over R_{𝔭}. But depth R_{𝔭} = 0, so that M_{𝔭} is free over R_{𝔭}. Thus

rank_{R_{𝔭}} M_{𝔭} = ∑_{0}^{n} rank F_{i},

so that rank_{R_{𝔭}} M_{𝔭} is the same for all 𝔭 ∈ Ass R. But if M has grade 0 then ann M consists of zero divisors, so there exists 𝔭 ∈ Ass R with ann M ⊆ 𝔭, so that M_{𝔭} ≠ 0. Thus M_{𝔭} ≠ 0 for all 𝔭 ∈ Ass R, i. e.

Ass R ⊆ Supp M.

For any 𝔭 ∈ Ass R, then, M_{𝔭} is a non-zero free R_{𝔭}-module. Thus if 𝔞 = ann M, we conclude that 𝔞_{𝔭} = 0 for all 𝔭 ∈ Ass R, i. e.

Supp 𝔞 ∩ Ass R = ∅.

But Ass 𝔞 ⊆ Ass R, so this implies that Ass 𝔞 = ∅. Thus 𝔞 = 0, so M is faithful. ✓

The Koszul Complex

Reference. Matsumura, Commutative Algebra, §18.D, p. 132.

Construction. Let R be a commutative noetherian ring and a_{1}, …, a_{n} ∈ R. Let E be the free exterian algebra over R on the symbols T_{1}, …, T_{n}. Thus E_{0} = R, E_{1} = RT_{1} ⊕ ⋯ ⊕ RT_{n}, E_{i} ≈ R^{\binom{n}{i}}, x^{2} = 0 for x ∈ E_{1} , and for x ∈ E_{i}, y ∈ E_{j}, xy = (−1)^{ij}yx.

Define a graded R-linear map d: E → E with degree 1 by defining d(T_{i}) = a_{i}, d(xy) = (dx)y + (−1)^{i}x(dy) for x ∈ E_{i}, y ∈ E_{j}. Note that d(R) = d(E_{0}) = 0 and

d(T_{i_{1}} ⋯ T_{i_{k}}) = ∑_{r=1}^{k}(−1)^{r−1}a_{i_{r}}T_{i_{1}} ⋯ T̂_{i_{r}} ⋯ T_{i_{k}}.

We define the Koszul Ring or Koszul Complex for R with respect to a_{1}, …, a_{n} to be this graded ring E, and denote it by R< T_{i}; dT_{i} = a_{i} >.

Lemma. (E, d) is a chain complex.

proof: Use induction on i to show that d^{2}(D_{i}) = 0. ✓ Lemma. Let Z = Ker d and B = d(E). Then Z is a ring and B an ideal in Z, so that H(E) = Z/B is a graded ring.

Lemma. Let 𝔞 = (a_{1}, …, a_{n}).

(1) H_{0}(E) = R/𝔞

(2) H_{n}(E) ≈ ann_{R} 𝔞

(3) H_{q}(E) = 0 for q < 0 or q > n.

(4) 𝔞H(E) = 0.

proof: (4) Since 𝔞 = d(E_{1}) ⊆ B and BZ ⊆ B, this is clear. ✓

Theorem. Let R be noetherian, let 𝔞 = (a_{1}, …, a_{n}), and let r = depth_{𝔞} R = grade R/𝔞. Let E = R< T_{i}; dT_{i} = a_{i} >. Then H_{n−r}(E) ≠ 0 and H_{q}(E) = 0 for q > n − r.

proof: By induction on r. If r = depth_{𝔞} R = 0 then H_{n}(E) = ann 𝔞 ≠ 0, and by construction H_{q} = 0 for q > n.

If depth_{𝔞} R > 0, let x_{1} be a regular element in 𝔞. Let R̄ = R/(x_{1}), Ē = E/x_{1}E. Since x_{1} is regular on E there is an exact sequence of complexes

0 ⟶ E ⟶^{x_{1}} E ⟶ E/x_{1}E ⟶ 0 .

Now let q be the largest integer such that H_{q}(E) ≠ 0. Then we get

0 = H_{q+1}(E) ⟶ H_{q+1}(Ē) ⟶ H_{q}(E) ⟶^{x_{1}} H_{q}(E) ⟶ · · · .

Now 𝔞H_{q}(E) = 0 so, since x_{1} ∈ 𝔞, this implies

H_{q+1}(Ē) ≈ H_{q}(E).

On the other hand, for p > q + 1, we have

H_{p}(E) = 0 ⟶ H_{p}(Ē) ⟶ H_{p−1}(E) = 0.

Thus H_{q+1}(Ē) ≠ 0 and H_{p}(Ē) = 0 for p > q + 1. Since depth𝔞̄ R̄ = r − 1, it now follows by induction that q + 1 = n − (r − 1) so that q = n − r. ✓

Theorem. Let R be noetherian, and suppose that 𝔞 = (a_{1}, …, a_{n}) ⊆ J(R). Let E = R< T_{i}; dT_{i} = a_{i} >. Then the following are equivalent:

(1) a_{1}, …, a_{n} is a regular R-sequence.

(2) H_{i}(E) = 0 for i ≥ 1.

(3) H_{1}(E) = 0.

(4) Any sequence of n elements generating 𝔞 is a regular R-sequence.

(5) grade R/𝔞 = 1.

proof: By the previous theorem, (1) ⇒ (5) ⇒ (2) ⇒ (3).

(4) ⇒ (1): Easy from the previous theorem and from (3) ⇒ (1).

(3) ⇒ (1): By induction on n. Let E′ = R <T_{1}, …, T_{n−1}; d T_{i} = a_{i} >. Consider the “chain maps” i: E′ ↪ E and j: E → E′, where j is defined by the conditions j(E′) = 0, j(T_{k_{1}} … T_{k_{s}}T_{n}) = T_{k_{1}} … T_{k_{s}}. This gives a short exact sequences of chain complexes

0 ⟶ E′ ⟶ E ⟶ E′ ⟶ 0 .

Since j has degree −1, we get a long exact sequence

⋯ ⟶ H_{1}(E′) ⟶^{δ_{1}} H_{1}(E′) ⟶^{i_{∗}} H_{1}(E) ⟶^{j_{∗}} H_{0}(E′) ⟶^{δ_{0}} H_{0}(E) ⟶ · · · .

We now compute δ_{1} . An element in H_{q}(E′) has the form [z], where z ∈ E′_{q} and dz = 0. Now z = j(zT_{n}) and δ_{q}[z] = [d(zT_{n})]. But

d(zT_{n}) = dz T_{n} + (−1)^{q}z dT_{n} = (−1)^{q}a_{n}z ∈ E′,

so that, up to sign, δ_{q} is just multiplication by a_{n}: δ_{q}[z] = (−1)^{q}a_{n}[z].

Now H_{1}(E) = 0 by hypothesis, so δ_{1} is surjective and δ_{0} monic. In other words a_{n}H_{1}(E′) = H_{1}(E′), and multiplication by a_{n} is monic on H_{0}(E′). Therefore H_{1}(E′) = 0 by Nakayama’s Lemma. Therefore by the induction hypothesis, a_{1}, …, a_{n−1} is a regular R-sequence. Since H_{0}(E′) = R/(a_{1}, …, a_{n−1}) and a_{n} is regular on H_{0}(E′), it follows that a_{1}, …, a_{n} is a regular R-sequence. ✓

Corollary. Suppose that R is noetherian and let a_{1}, …, a_{n} ∈ J(R). If a_{1}, …, a_{n} is a regular R-sequence, then every permutation of it is also a regular R-sequence.

This corollary may seem not very surprising. However it is not valid without the hypothesis that the sequence be contained in J(R).

Example [Kaplansky, Commutative Rings, §3.1, Exercise 7, p. 102]. Let R = K[X, Y, Z], where K is a field. The elements X, Y − XY , Z − XZ form a regular R-sequence, but in the order Y − XY , Z − XZ, X they do not.

Corollary. Suppose that R is noetherian. Let 𝔞 = (a_{1}, …, a_{n}) ⊆ J(R) and suppose that grade R/𝔞 = n. Then E is a projective resolution for R/𝔞. In consequence,

Tor_{i}^{R}(R/𝔞, R/𝔞) ≈ (R/𝔞)^{\binom{n}{i}} ≈ Ext_{R}^{i}(R/𝔞, R/𝔞).

proof: By construction, E_{q} is a free R-module for each q. And by the theorem, if grade R/𝔞 = n then E is exact in degrees larger than 0 and H_{0}(E) = R/𝔞. Thus E is a projective resolution for R/𝔞. Therefore Tor^{R}(R/𝔞, R/𝔞) is the homology of E/𝔞E = E ⊗_{R} R/𝔞. But since d(D) ⊆ 𝔞E, the differentiation on this complex is

trivial. ✓