Modules, The Ring of Endomorphisms

The Ring of Endomorphisms

Let M be a left R module. As you know, a module homomorphism from M into itself is called an endomorphism. Verify the steps below to show that these endomorphisms form a ring.

The endomorphisms form an abelian group via (f+g)(x = f(x)+g(x). In other words, show that f+g is a function that respects addition in M and scaling by elements of R.
The product of endomorphisms f and g is f followed by g. (Definition)
The product of f followed by g can be written f*g, or fg. (Convention)
The product fg satisfies the criteria for a module homomorphism. It is just the composition of two homomorphisms.
Multiplication is associative. Either way it's f followed by g followed by h.
Multiplication distributes over addition.
f*(g+h) = fg+fh and (g+h)*f = gf+hf.
The zero and identity homomorphisms correspond to 0 and 1 respectively.
The endomorphisms form a ring.
The module automorphisms form a nonabelian multiplicative group inside the ring of endomorphisms.
The left module M is also a right module, when acted upon by the ring of automorphisms. If x is in M, and f is an endomorphism, the "module function" maps xf to f(x). Note that x(fg) = fg applied to x = f(x) followed by g = (xf)g. Show that the other module criteria are satisfied.

Bimodule

Given two rings R and S, M is an rs bimodule if M is a left R module and a right S module, and (R*M)*S = R*(M*S).

An example of a bimodule is any R module M, where S is the ring of R endomorphisms of M, written on the right. By definition, (R*M)*S = R*(M*S). That's what we mean by an R endomorphism. Hence we have a bimodule.

Conversely, a bimodule M is an R module, and if f is an element of S, it rearranges the elements of M; a function from M into itself. Since f commutes with R, and with addition in M, it is a valid M endomorphism. Since M is a right S module, functions in S add and multiply as described above. Thus S is a subring of the endomorphism ring of M.

R Equals the Endomorphisms of R

When R is a ring, it acts as a left R module, which admits a right R module S, namely the R endomorphisms of R.

We can build a ring homomorphism from R into S. Let f be an element of R, and let the corresponding function be R*f, i.e. scaling by f on the right. Verify that this is a module homomorphism. Showing (xr)f = x(rf) is merely restating associativity within R.

Next, show the map is a ring homomorphism from R into S. The function induced by f+g is indeed rf+rg. This is a restatement of the distributive property of R. Also, the function induced by fg is the composition of the function of f with the function of g, namely rfg.

Different elements map 1 to different elements in R, and represent distinct endomorphisms in S. The map is injective, and R embeds in S.

The image of 1 completely determines the endomorphism, and when e(1) = x, right multiplication by x induces that very endomorphism. The map is onto, and R = S.

Don't confuse module endomorphisms with ring endomorphisms. Complex conjugation fixes 1, and swaps i and -i. This is a ring endomorphism that fixes 1, but there is no left R module endomorphism that fixes 1, other than the identity map. It's a different world.