When studying homomorphisms of modules over a fixed ring, we often encounter sequences like this:
where each is a homomorphism of modules. This sequence may terminate (on either end) or it may continue indefinitely.
Note on indices: usually the index of the map takes its source; thus maps for instance.
The sequence is said to be exact at if
The whole sequence is said to be exact if it is exact at each term except the ends. A short exact sequence is an exact sequence of the form
Having a short exact sequence above is equivalent to:
is injective, is surjective, .
Note that there is no harm in replacing by and by as the inclusion of a submodule. Then can be replaced with .
Why do we care about short exact sequences? In studying an A-module M, one trick is to decompose it as , then study the components separately. But unlike vector spaces, modules do not decompose so easily: if is a submodule, we do not have in general. This is already clear for , by taking and .
In many cases, decomposing into and is good enough.
1. If is an exact sequence of vector spaces, then M is finite-dimensional if and only if N and P are, in which case .
2. If is an exact sequence of abelian groups, then M is finite if and only if N and P are, in which case .
3. If is an exact sequence of A-modules, and N, P are both finitely generated, then so is M. [Proof: exercise.]
We will see many more examples later.
Note that in example 3, the converse is not true since a submodule of a finitely generated submodule is not finitely generated in general.
Decomposing an Exact Sequence
Suppose we have the following exact sequence of modules:
Define additional modules . Then we obtain multiple short exact sequences
In diagram form, this gives us:
where the diagonals give us short exact sequences.
Suppose we have the following long exact sequence of finite abelian groups.
Breaking this up as above gives us short exact sequences
Thus . The upshot: if we know the orders of all terms but one, we also know the order of the remaining term.
Through the remaining of this article, we will look at functors
for fixed rings A and B. Recall that is an A-module.
A covariant F is said to be additive if, for any ,
is a homomorphism of additive groups.
We only require for . Since the base and target rings are different, we cannot specify A-linearity.
1. Let B = A and . Any A-linear map induces an A-linear , .
2. Suppose is a homomorphism (i.e. B is an A-algebra). Each B-module M becomes an A-module via . This gives a functor
3. Let be an ideal and . We saw that is canonically an -module. This gives a functor since an A-linear induces a B-linear .
4. Let B = A. For any A-module M, we have a functor . This is additive since for , any give
Let F be an additive functor.
- Prove that F takes the zero module (over A) to the zero module (over B).
- Prove that for any A-modules M and N, we have .
[Hint for direct sum: let , , and be the obvious maps. Find relations among them.]
An additive functor F is said to be exact if, for any exact sequence ,
is also exact.
1. Prove that we always have if F is additive.
2. Among the four examples above, decide which are exact.
An additive functor F is exact if and only if it takes a short exact sequence to a short exact sequence
(⇒) Obvious: just take 3 terms at a time.
(⇐) Suppose is a 3-term exact sequence. Let . Break the sequence into:
Since the LHS are short exact sequences, so are the RHS. Also since is obtained by composing on the left, is obtained by composing on the right. Same goes with . Piecing the short exact sequences then gives an exact . ♦
Implications of Exact Functor
There are many reasons why it is desirable for F to be exact.
If is injective, so is .
Indeed, injectivity of is equivalent to exactness of . Hence, without loss of generality, for a submodule , we identify as a submodule of .
Under this identification, if is a submodule, .
Indeed since is a short exact sequence, so is:
where is identified as a submodule of . Thus .
If is surjective, so is .
Indeed surjectivity of is equivalent to exactness of .
For , we have , as submodules of .
Indeed, we have the exact sequence
which upon applying F gives us an exact sequence
which says that is the kernel of .
If are submodules, then as submodules of .
From we get . Similarly and thus . To show equality, since is surjective, so is
where the isomorphism was proven in exercise A above. Hence is generated by and .
1. For an A-linear , the cokernel of f is defined to be . Prove that .
[ Hint: show that the sequence is exact. ]
2. For an A-linear , prove that as submodules of .
3. Prove that for submodules , we have , as submodules of .
It is not true that or for any collection of submodules . Counter-examples are a bit hard to find at the moment, but we will encounter one later.