## Exact Sequences and the Grothendieck Group

As before, all rings are not commutative in general.

Definition. An exact sequence of R-modules is a collection of R-modules $M_i$ and a sequence of R-module homomorphisms:

$\ldots \stackrel{f_{i+2}}{\longrightarrow} M_{i+1} \stackrel{f_{i+1}}{\longrightarrow} M_i \stackrel{f_i}{\longrightarrow} M_{i-1} \stackrel{f_{i-1}}{\longrightarrow}\ldots$

such that $\text{ker}(f_i) = \text{im}(f_{i+1})$ for all i.

Examples

1. The sequence $0\longrightarrow N \stackrel f\longrightarrow M$ is exact if and only if f is injective.

2. The sequence $M \stackrel g\longrightarrow P\longrightarrow 0$ is exact if and only if g is surjective.

3. The sequence $0 \longrightarrow N \stackrel{f}{\longrightarrow} M \stackrel{g}{\longrightarrow} P\longrightarrow 0$ 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 $0 \to N \to M \to M/N \to 0$ 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 RZ. 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:

$0 \longrightarrow M_4 \stackrel{f_4}{\longrightarrow} M_3\stackrel{f_3}{\longrightarrow} M_2 \stackrel{f_2}{\longrightarrow} M_1 \stackrel{f_1}{\longrightarrow}M_0 \longrightarrow 0.$

Since $N_i := \text{ker}(f_i) = \text{im}(f_{i+1})$, this allows us to break it into a collection of short exact sequences.

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

$|M_4| = |N_3|,\ |M_3| = |N_3| + |N_2|, \ |M_2| = |N_2| + |N_1|, \ |M_1| = |N_1| + |M_0|,\ |M_0| = |M_0|$

which gives the alternating sum $|M_4| - |M_3| + |M_2| - |M_1| + |M_0| = 0.$ 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

$\sum_{i=1}^k a_i [M_i],\quad a_i \in \mathbf{Z}, M_i \in \Sigma$

which forms an abelian group freely generated by $\{[M_i] : M_i \in \Sigma\}.$ 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:

$[M] - [N] - [P]$ if there is an exact sequence $0\to N\to M \to P \to 0.$

The resulting quotient is called the Grothendieck group G(∑). Letting NMP = 0,  we have [0] = [0] + [0] in G(∑), so [0] = 0 as it should be. Next if MM’ ∈ ∑ 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:

$0 \longrightarrow M_n \stackrel{f_n}{\longrightarrow} M_{n-1}\stackrel{f_{n-1}}{\longrightarrow} \ldots \stackrel{f_2}{\longrightarrow} M_1 \stackrel{f_1}{\longrightarrow}M_0 \longrightarrow 0$

then gives the following relation in the Grothendieck group:

$[M_0] - [M_1] + \ldots + (-1)^{n-1}[M_{n-1}] + (-1)^n [M_n] = 0.$

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 ZΦ be the abelian group freely generated by elements of Φ. We define a group homomorphism ω : G(∑) → ZΦ as follows:

$[M], M\in \Sigma\ \mapsto\ [N_0] + [N_1]+ \ldots + [N_k] \in \mathbf{Z}\Phi$

where $N_0, N_1, \ldots, N_k$ 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 : ZΦ → 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.

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