As before, all rings are not commutative in general.
Definition. An exact sequence of R-modules is a collection of R-modules and a sequence of R-module homomorphisms:
such that for all i.
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.
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  =  +  in G(∑), so  = 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.
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:
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 : 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.