As before, all rings are not commutative in general.

Definition. Anexact sequenceof R-modules is a collection of R-modules and a sequence of R-module homomorphisms:such that for all i.

**Examples**

1. The sequence is exact if and only if *f* is injective.

2. The sequence is exact if and only if *g* is surjective.

3. The sequence is exact if and only if *f* is injective, *g* is surjective, and im(*f*) = ker(*g*). This is called a **short exact sequence**.

## Short Exact Sequences

Note that the short exact sequence is equivalent to the following assertions:

*N*is isomorphic to a submodule*N’*of*M*;*P*is isomorphic to the quotient*M*/*N’*.

Hence, given a short exact sequence we may assume it is of the form for a submodule *N* of *M*. Short exact sequences are important because knowledge of *N* and *M*/*N* often tell us something about *M* itself. Here are some examples.

1. *M* is of finite order if and only if *N* and *M*/*N* are, in which case we have |*M*| = |*M*/*N*| × |*N*|.

2. Suppose *R* = **Z**. Then *M* is finitely generated if and only if *N* and *M*/*N* are, in which case rk(*M*) = rk(*M*/*N*) + rk(*N*).

3. Suppose *R* is a division ring. Then *M* is finite-dimensional if and only if *N* and *M*/*N* are, in which case dim(*M*) = dim(*M*/*N*) + dim(*N*).

4. Over any ring, *M* is of finite length if and only if *N* and *M*/*N* are, in which case we have *l*(*M*) = *l*(*M*/*N*) + *l*(*N*). More generally, if we let CF(*M*) denote the formal sum of its composition factors, then CF(*M)* = CF(*M*/*N)* + CF(*N)*.

In general, it is hard to recover *M* from *M*/*N* and *N*; this is called the *extension problem* and is classified by something called the “Ext functor”, which is another story for another day.

## Breaking Up Long Exact Sequences

Suppose we have the following long exact sequence:

Since , this allows us to break it into a collection of short exact sequences.

For example, if all the terms are finite, we obtain:

which gives the alternating sum We could repeat the same argument for the ranks of the terms, etc, but instead let’s have a common framework.

## Grothendieck Group

Let ∑ be any set of *R*-modules, including the zero module. Take the set of all formal sums

which forms an abelian group freely generated by Note that we specifically said ∑ is a set to avoid running into set-theoretic paradoxes. Take the quotient of this group by the subgroup generated by relations:

if there is an exact sequence

The resulting quotient is called the **Grothendieck group** *G*(∑). Letting *N* = *M* = *P* = 0, we have [0] = [0] + [0] in *G*(∑), so [0] = 0 as it should be. Next if *M*, *M’* ∈ ∑ are two isomorphic modules, the short exact sequence 0 → *M* → *M’* → 0 → 0 then gives us [*M*] = [*M’*] in the Grothendieck group, as we would expect.

Suppose ∑ is closed under submodules and quotients. From the above observation, a long exact sequence of the form:

then gives the following relation in the Grothendieck group:

Now we have a convenient language to describe the above.

**Examples**

1. Let ∑ be the collection of *isomorphism classes* of *R*-modules of finite cardinality. Taking the order then gives a group homomorphism *G*(∑) → **Q***, where each [*M*] → |*M*|. [ We need to take the isomorphism classes so that ∑ forms a set instead of a proper class. This will be implicit from now onwards. ]

2. Let ∑ be the collection of finitely generated **Z**-modules. Taking the rank then gives a group homomorphism *G*(∑) → **Z**, where each [*M*] → rk(*M*).

3. When *R* is a division ring, taking the dimension for finite-dimensional vector spaces also gives a group homomorphism *G*(∑) → **Z**.

4. Let *R* be any ring and ∑ be the collection of *R*-modules of finite length. Let Φ be the set of isomorphism classes of simple modules, and **Z**Φ be the abelian group freely generated by elements of Φ. We define a group homomorphism ω : *G*(∑) → **Z**Φ as follows:

where are the composition factors of *M*. Recall that the composition factors of *M* are independent of the choice of composition series, up to isomorphism and permutation. Hence ω is a well-defined map. It is a group homomorphism since we can concatenate composition series of *N* and *M*/*N* to obtain one for *M*. It is clearly surjective since ω[*M*] = [*M*] for any simple module *M*.

On the other hand, consider the map ν : **Z**Φ → *G*(∑) via extending [*M*] → [*M*] for any simple *M*. Clearly νω is the identity on *G*(∑) since [*M*] gets mapped to the sum of its composition factors in *G*(∑), which is equal to [*M*] in *G*(∑). Thus, ω is injective. In conclusion:

Theorem. Let ∑ be the collection of all modules of finite length. The Grothendieck group G(∑) is an abelian group freely generated by [M], for all simple modules M.If R is artinian, there are only finitely many such M so G(∑) is finitely generated.