# 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. 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: 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. # 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. 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. # 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. ♦ # 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. ♦ This entry was posted in Advanced Algebra and tagged , , , , , . Bookmark the permalink.