Fret not if you’re unfamiliar with the term functor; it’s a concept in category theory we will use implicitly without delving into the specific definition. This topic is, unfortunately, a little on the dry side but it’s a necessary evil to get through.
Let R be a ring, and M, N be left R-modules. The set of R-module homomorphisms M→N is denoted by HomR(M, N) or just Hom(M, N) when the base ring is clear. This forms an abelian group under addition, where adding f, g : M → N just gives:
When R is commutative, in fact Hom(M, N) is an R-module since given r∈R, f : M → N, we can define (r·f) : M → N to be the map This fails when the base ring is non-commutative because
By functoriality, we mean the following:
Lemma. If g : N → N’ is a homomorphism of R-modules, then composing gives us group homomorphisms:
Furthermore, if h : N’ → N” is another R-module homomorphism, then composing gives:
Finally, given any module N, the identity map 1N : N → N gives the identity maps for the following:
The proofs for the above are quite easy and thus omitted. Hence, we need to bear in mind that the Hom functor is direction-reversing in the first term and direction-preserving in the second. [ At worst, just memorise it; that certainly beats getting confused by the direction all the time. ]
Under category theory, direction-reversing (resp. -preserving) is given the term contravariant (resp. covariant). It is standard notation in literature to denote a contravariant functor by and covariant one by Breaking this convention will result in unnecessary confusion!
To motivate the above definition, let us look not at modules but at rings instead. Consider the following ring R:
Any ring homomorphism f : R→S then corresponds to picking elements w, x, y, z ∈ S such that wz – xy = 1, i.e. a 2 × 2 matrix of determinant 1. Indeed, for such an f, we just take the images f(a), f(b), f(c), f(d) ∈ S. Furthermore, if g : S→T is a ring homomorphism, then the map:
corresponds to mapping elements w, x, y, z ∈ S such that wz – xy = 1 to g(w), g(x), g(y), g(z) ∈ T, which also satisfy the same equation.
When M or N is endowed with additional structure, it is possible to form a module out of Hom(M, N):
Definition. Let R, S be rings. An (R, S)-bimodule is an abelian group M equipped with a left R-module and right S-module structure such that both commute, i.e. for all r∈R, s∈S, m∈M, we have (rm)s = r(ms).
One way of visualizing this is via an expression:
which is associative, i.e. independent of our choice of bracketing.
- If R is commutative, any R-module is automatically an (R, R)-bimodule.
- Any ring R is an (R, R)-bimodule.
- More generally, if R→S is a ring homomorphism, then S is an R-algebra and S is an (R, S)-, (S, R)- and (R, R)-bimodule.
- If M is a left R-module, we can turn it into a right Rop module by defining mr’ (r’ ∈ Rop) to be rm where r∈R is the element corresponding to r’∈Rop. Does this turn M into an (R, Rop)-bimodule? [ Answer: no because the left and right actions don’t commute. ]
When either M or N is a bimodule, the resulting Hom(M, N) gets a module structure.
Proposition. Let M, N be R-modules.
- If M is an (R, S)-bimodule, then HomR(M, N) is a left S-module, via
- If N is an (R, S)-bimodule, then HomR(M, N) is a right S-module, via
We will only check the commutativity of the action of S. For the first case,
For the second case,
Here’s one way to remember this: as we mentioned earlier, Hom reverses direction in the first component by turning the right S-action into a left one, and preserves direction in the second.
We know that HomR(R, M) is uniquely determined by the image of 1∈R. Thus we get a bijection M ↔ HomR(R, M). This is clearly an isomorphism of abelian groups. On the other hand, R is an (R, R)-bimodule so HomR(R, M) also becomes a left R-module. Check that the bijection is an isomorphism of left R-modules.
Definition. Let M be a left R-module. From the (R, R)-bimodule structure of R, we get a right-module structure onHomR(M, R). This is called the dual of M and denoted by M*.
Linear algebra over division rings is mostly identical to that over fields, but the dual of a left vector space becomes a right vector space.
Hom Functor is Left Exact
The main result we’d like to show is:
Theorem. Suppose is an exact sequence of modules. Then for any module N, the following sequence is exact as well:
There are several parts to this result:
- is injective: if h : N → M’ is such that fh : N → M is zero, then since f is injective, h = 0.
- Since gf = 0, we also have Thus
- Conversely, suppose so h : N → M is a map such that gh : N → M” is the zero map. Then im(h) ⊆ ker(g) = im(f). Since f : M’ → M is injective, this gives and ♦
The following result is sometimes useful too:
Proposition. If is a sequence of modules such that
is exact for all N, then the sequence is exact.
Sketch of Proof
Let N = R, the base ring. Then we recover the exact desired sequence. ♦
Let us briefly explain why the above results are of interest. Ideally, we would like to have a functor F such that whenever M’ → M → M” is exact, the resulting F(M’) → F(M) → F(M”) is exact too. This is tremendously convenient, since F then commutes with kernel, image, cokernel etc. However, the above result says Hom(N, -) is only “left-exact”. For example, consider the surjective map of Z-modules, Z → Z/2 → 0. Taking N = Z/2 gives us Hom(Z/2, Z) → Hom(Z/2, Z/2) → 0 which is not exact since the first term is 0.
Suppose we would like to prove that some functor F is left-exact. Here’s one way:
- Represent F using the Hom-functor, i.e. find an N such that F(M) = Hom(N, M).
- From the above result, we immediately see that F is left-exact.
Prove the above for Hom(-, N) as well: i.e. suppose is an exact sequence. Then for any module N, we get an exact sequence:
Conversely, if the above sequence is exact for all N, then is exact.
Thus, Hom(N, -) takes left-exact sequences to left-exact sequences and Hom(-, N) takes right-exact to left-exact sequences too. In both cases, the final outcome is left-exact and one often says in short, the Hom functor is left-exact.