Module morphisms and submodules

There is a tight connection between module morphisms and submodules.

Kernels


Suppose f:MN is an R-module morphism. Let ker(f) denote the usual kernel of f as a morphism of abelian groups, i.e., ker(f)={mMf(m)=0N}. One can easily verify this set is not only a subgroup of M (when viewed as an abelian group), but also a submodule of M. It is still called the kernel of the morphism f. One can also give a definition of the kernel without reference to any elements using the zero morphism (see here).

For future reference, let's cryptically note that the kernel of the module morphism f:MN should really be the "inclusion" morphism i:KM, where K=ker(f) is the submodule defined above.

Images


Similarly to the kernel, let im(f) denote the usual image of f (as a set map or group morphism). As with the kernel, this set is not just a subgroup of N (when viewed as abelian group) but also a submodule of N. It is still called the image of the morphism f.

As with the kernel, we should really define the image of f:MN as a certain "special" module morphism j:IN through which f "factors," but we'll hold off on that until we study Yoneda's Lemma and maybe even abelian categories.

Hom-sets? More like hom-modules!


For each pair of R-modules M and N, we can consider the set HomR-Mod(M,N) of all R-module morphisms from M to N. (You might also see this set sometimes denoted HomR(M,N).) This set has a natural (!) structure of an abelian group, where addition of morphisms is defined by "addition of outputs"; i.e., (f+g)(m)=f(m)+g(m).

Of course, to be careful one should verify that if f,g:MN are R-module morphisms, then f+g:MN defined in this way is indeed an R-module morphism. And then we should verify that this operation on HomR-Mod(M,N) does indeed give the set the structure of an abelian group. (What's the additive identity? Can you describe the additive inverse of a morphism f?) We will skip those details here (at least for now), but I promise there are no surprises. Everything checks out as directly as you might hope.

Along the same lines, when R is commutative the set HomR-Mod(M,N) even has the structure of an R-module. Specifically, define a left action of R on the hom-set through the R-action on the outputs, i.e., using the R-action in N. In other words, for each R-module morphism f:MN and rR define a map rf:MN by (rf)(m)=rf(m). As above, technically we should check that rf is indeed a module morphism MN. It's in the compatibility with the R-actions that the commutativity of R is required (or at least enough to guarantee the result we need). Indeed, for each sR and mM we must verify that (rf)(sm)=s(rf)(m). On the left-hand side we have

(rf)(sm)=rf(sm)=r(sf(m))=(rs)f(m).

On the other hand, we have

s(rf)(m)=s(rf(m))=(sr)f(m).

If R is commutative, then rs=sr and so (rs)f(m)=(sr)f(m), as desired.

When N=M, one can check that the set HomR-Mod(M,M) has a natural structure of a ring (with unity). It is called the endomorphism ring of M and is sometimes denoted EndR-Mod(M); when R is commutative, the ring EndR-Mod(M) has a natural structure of an R-algebra.

Enriched categories

Discoveries like these, namely of categories in which hom-sets naturally have additional structure, inevitably lead one to the concept of an enriched category. A quick, informal definition of an enriched category is that it is a category in which the hom-sets have an algebraic structure (e.g., are abelian groups), together with the requirement that the composition operation respects that structure (e.g., define morphisms between abelian groups).

Suggested next notes


Quotient modules
The Isomorphism Theorems for Modules