Commutative Algebra 33

Snake Lemma

Let us introduce a useful tool for computing kernels and cokernels in a complicated diagram of modules. Although it is only marginally useful for now, it will become a major tool in homological algebra.

Snake Lemma.

Suppose we have the following diagram of A-modules and homomorphisms, where the rows are exact.

snake_lemma_diag_1

Let K_i = \mathrm{ker} h_i and C_i = \mathrm{coker} h_i for i = 1, 2, 3. Then we get a long exact sequence:

0 \to \mathrm{ker } f_1 \to K_1 \to K_2 \to K_3 \to C_1 \to C_2 \to C_3 \to \mathrm{coker} g_2 \to 0.

In diagram, the long exact sequence is drawn in green:

snake_lemma_statement

Proof

The proof is rather tedious so we will only prove the existence of the “snake map” K_3 \to C_1.

  • Let m_3 \in K_3 so that h_3(m_3) = 0.
  • Since f_2 is surjective there exists m_2 \in M_2, f_2(m_2) = m_3.
  • Let n_2 = h_2(m_2).
  • Then g_2(n_2) = g_2(h_2(m_2)) = h_3(f_2(m_2)) = h_3(m_3) = 0.
  • Hence n_2 = g_1(n_1) for a unique n_1 \in N_1.
  • We let m_3 \mapsto image of n_1 in C_1.

This map is well-defined since if we pick another m_2' \in M_2 in the second step, then m_2' - m_2 \in \mathrm{ker} f_2 = \mathrm{im } f_1. Thus m_2' - m_2 = f_1(m_1) for some m_1 \in M_1 so h_2(m_2' - m_2) = h_2(f_1(m_1)) = g_1(h_1(m_1)) and thus the new n_1' \in N_1 in the above construction is n_1' = n_1 + h_1(m_1). So n_1 and n_1' have the same image in \mathrm{coker} h_1.

The rest of the proof is left as an exercise. ♦

Note

The above process is called diagram-chasing, and it is often far more effective to have it shown to you live. Here is the first part of the proof in video form.

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Finitely Presented Modules

Now let us compare \mathrm{Hom}_B(M^B, N^B) and \mathrm{Hom}_A(M, N) for A-modules M and N. Since any A-linear map f: M\to N induces a B-linear map f^B : M^B \to N^B we have a map \mathrm{Hom}_A(M, N) \to \mathrm{Hom}_B(M^B, N^B). This is clearly A-linear so it induces a B-linear map

[\mathrm{Hom}_A(M, N)]^B \longrightarrow \mathrm{Hom}_B(M^B, N^B), \quad b \otimes f \mapsto b\cdot f^B,

For this to be an isomorphism, we need to introduce a new concept.

Definition.

An A-module M is finitely presented if it is finitely generated, and for some surjective map A^n \to M, its kernel is also finitely generated.

Note

In other words, M is finitely presented if and only if there is an exact sequence of the form

A^m \to A^n \to M \to 0.

Intuitively, this means M can be described by a finite set of generators subjected to a finite set of relations.

Proposition 1.

If M is finitely presented, then the kernel K of any surjective map A^n \to M is finitely generated.

Proof

Since M is finitely presented there is an exact sequence of the form F_1 \to F_2 \to M \to 0 where F_1, F_2 are finite free modules. Since F_1, F_2 are free and hence projective, there exist f:F_2 \to A^n, g:F_1 \to K making the diagram commute.

proof_for_fp_modules

By the snake lemma we have an exact sequence

0 = \mathrm{ker} 1_M \to \mathrm{coker } g \to \mathrm{coker } f \to \mathrm{coker } 1_M =0.

Thus \mathrm{coker } g \cong \mathrm{coker } f is finitely generated. Furthermore, since F_1 is finite free \mathrm{im } g is finitely generated. Since \mathrm{im }g and K/\mathrm{im } g are finitely generated so is K. ♦

Exercise A

1. Find a module M over a ring A which is finitely generated but not finitely presented.

2. Prove that a finitely generated projective module is finitely presented.

3. Prove that in a short exact sequence of A-modules 0\to N \to M \to P \to 0, if N and P are finitely presented, so is M.

Optional Extra

The concept of finitely presented modules can be generalized further.

Definition

We say that an A-module M is of type FP_n if there is an exact sequence

F_n \longrightarrow \ldots \longrightarrow F_1 \longrightarrow F_0 \longrightarrow M \longrightarrow 0

where each F_i is finite free.

Thus saying M is FP_0 means it is finitely generated and saying M is FP_1 means it is finitely presented. One can show more generally that if M has type FP_n then the kernel of any surjection F\to M has type FP_{n-1}. However, this is outside our scope of discussion.

This has applications in non-commutative algebra, where we study (say) left modules over a non-commutative ring and consider their cohomological properties. For details, see GTM 87, Cohomology of Groups, by Kenneth S. Brown.

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Hom Functor and Induced Module

Now our main result is as follows.

Proposition 2.

If M is finitely presented and B is A-flat, we have an isomorphism

[\mathrm{Hom}_A(M, N)]^B \stackrel \cong \longrightarrow \mathrm{Hom}_B(M^B, N^B), \quad b \otimes f \mapsto b\cdot f^B.

Proof

Pick an exact sequence of A-modules: A^m \stackrel f\to A^n \stackrel g\to M\to 0. Since Hom is left-exact, for any A-module N we have an exact sequence of A-modules

0 \longrightarrow \mathrm{Hom}_A (M, N) \stackrel {g^*} \longrightarrow \mathrm{Hom}_A(A^n, N) \stackrel {f^*} \longrightarrow \mathrm{Hom}_A(A^m, N).

And since B\otimes_A - is exact, we get an exact sequence of B-modules

0 \longrightarrow [\mathrm{Hom}_A (M, N)]^B \stackrel {(g^*)^B} \longrightarrow [\mathrm{Hom}_A(A^n, N)]^B \stackrel {(f^*)^B } \longrightarrow [\mathrm{Hom}_A(A^m, N)]^B.

Since tensor product is right-exact, we also get an exact sequence of B-modules

B^m \stackrel{f^B}\longrightarrow B^n \stackrel{g^B}  \longrightarrow M^B \longrightarrow 0.

Again since Hom is left-exact, we get an exact sequence of B-modules

0 \longrightarrow \mathrm{Hom}_B(M^B, N^B) \stackrel {(g^B)^*} \longrightarrow \mathrm{Hom}_B(B^n, N^B) \stackrel {(f^B)^*} \longrightarrow \mathrm{Hom}_B(B^m, N^B).

But we have natural isomorphisms

[\mathrm{Hom}_A(A^n, N)]^B \cong (N^n)^B \cong (N^B)^n \cong \mathrm{Hom}_B(B^n, N^B)

which commute with [\mathrm{Hom}_A(M, N)]^B \longrightarrow \mathrm{Hom}_B(M^B, N^B). Hence this map is an isomorphism. ♦

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F.P. ⇒ (Projectivity is Local)

Since localization is exact and naturally isomorphic to S^{-1}A \otimes_A - , as a special case we obtain the following.

Corollary 1.

If M is finitely presented, then S^{-1}\mathrm{Hom}_A(M, N) \cong \mathrm{Hom}_{S^{-1}A}(S^{-1}M, S^{-1}N).

Now we are ready to prove the following result.

Proposition 3.

If M is a finitely presented A-module, then M is A-projective if and only if M_{\mathfrak m} is A_{\mathfrak m}-projective for all maximal \mathfrak m\subset A.

In summary, for a finitely presented module, projectivity is a local property.

Proof

(⇒) Let M be A-projective. For a maximal \mathfrak m\subset A; we wish to show M_{\mathfrak m} is A_{\mathfrak m}-projective. Let f :N_1 \to N_2 be a surjective A_{\mathfrak m}-linear map. Now f is A-linear, and since M is A-projective

\mathrm{Hom}_A(M, N_1) \stackrel {f_*}\longrightarrow \mathrm{Hom}_A(M, N_2)

is surjective. Since M is finitely presented and (N_i)_{\mathfrak m} \cong N_i, we get a surjective:

\mathrm{Hom}_{A_{\mathfrak m}}(M_{\mathfrak m}, N_1) \cong \mathrm{Hom}_A(M, N_1)_{\mathfrak m} \longrightarrow \mathrm{Hom}_A(M, N_2)_{\mathfrak m}\cong \mathrm{Hom}_{A_{\mathfrak m}}(M_{\mathfrak m}, N_2).

(⇐) Suppose M_{\mathfrak m} is A_{\mathfrak m}-projective for all maximal \mathfrak m. To prove M is A-projective, let f:N_1 \to N_2 be a surjective A-linear map. We need to show

\mathrm{Hom}_A(M, N_1) \stackrel {f_*} \longrightarrow \mathrm{Hom}_A(M, N_2)

is surjective; equivalently, we need to show (f_*)_{\mathfrak m} is surjective for all maximal \mathfrak m. But this is just

\mathrm{Hom}_{A_{\mathfrak m}}(M_{\mathfrak m}, (N_1)_{\mathfrak m}) \cong \mathrm{Hom}_A(M, N_1)_{\mathfrak m} \longrightarrow \mathrm{Hom}_A(M, N_2)_{\mathfrak m} \cong \mathrm{Hom}_{A_{\mathfrak m}}(M_{\mathfrak m}, (N_2)_{\mathfrak m})

which is surjective because M_{\mathfrak m} is A_{\mathfrak m}-projective. ♦

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