# 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 and for . Then we get a long exact sequence:

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” .

- Let so that .
- Since is surjective there exists , .
- Let .
- Then .
- Hence for a unique .
- We let image of in .

This map is well-defined since if we pick another in the second step, then . Thus for some so and thus the new in the above construction is . So and have the same image in .

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 and for *A*-modules *M* and *N*. Since any *A*-linear map induces a *B*-linear map we have a map . This is clearly *A*-linear so it induces a *B*-linear map

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

Definition.An A-module M is

finitely presentedif it is finitely generated, and for some surjective map , 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

.

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

anysurjective map is finitely generated.

**Proof**

Since *M* is finitely presented there is an exact sequence of the form where are finite free modules. Since are free and hence projective, there exist , making the diagram commute.

By the snake lemma we have an exact sequence

.

Thus is finitely generated. Furthermore, since is finite free is finitely generated. Since and 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 , if *N* and *P* are finitely presented, so is *M*.

## Optional Extra

The concept of finitely presented modules can be generalized further.

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

where each is finite free.

Thus saying *M* is means it is finitely generated and saying *M* is means it is finitely presented. One can show more generally that if *M* has type then the kernel of any surjection has type . 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

**Proof**

Pick an exact sequence of *A*-modules: . Since Hom is left-exact, for any *A*-module *N* we have an exact sequence of *A*-modules

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

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

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

But we have natural isomorphisms

which commute with . Hence this map is an isomorphism. ♦

# F.P. ⇒ (Projectivity is Local)

Since localization is exact and naturally isomorphic to , as a special case we obtain the following.

Corollary 1.If

Mis finitely presented, then .

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 is -projective for all maximal .

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

**Proof**

(⇒) Let *M* be *A*-projective. For a maximal ; we wish to show is -projective. Let be a surjective -linear map. Now *f* is *A*-linear, and since *M* is *A*-projective

is surjective. Since *M* is finitely presented and , we get a surjective:

(⇐) Suppose is -projective for all maximal . To prove *M* is *A*-projective, let be a surjective *A*-linear map. We need to show

is surjective; equivalently, we need to show is surjective for all maximal . But this is just

which is surjective because is -projective. ♦