ring of endomorphisms

Let R be a ring and let M be a right R-module.

An endomorphismPlanetmathPlanetmathPlanetmath of M is a R-module homomorphismMathworldPlanetmath from M to itself. We shall write endomorphisms on the left, so that f:MM maps xf(x). If f,g:MM are two endomorphisms, we can add them:


and multiply them


With these operationsMathworldPlanetmath, the set of endomorphisms of M becomes a ring, which we call the ring of endomorphisms of M, written EndR(M).

Instead of writing endomorphisms as functionsMathworldPlanetmath, it is often convenient to write them multiplicatively: we simply write the application of the endomorphism f as xfx. Then the fact that each f is an R-module homomorphism can be expressed as:


for all xM and rR and fEndR(M). With this notation, it is clear that M becomes an EndR(M)-R-bimodule.

Now, let N be a left R-module. We can construct the ring EndR(N) in the same way. There is a complication, however, if we still think of endomorphism as functions written on the left. In order to make M into a bimodule, we need to define an action of EndR(N) on the right of N: say


But then we have a problem with the multiplication:




In order to make this work, we need to reverse the order of composition when we define multiplication in the ring EndR(N) when it acts on the right.

There are essentially two different ways to go from here. One is to define the multiplication in EndR(N) the other way, which is most natural if we write the endomorphisms as functions on the right. This is the approach taken in many older books.

The other is to leave the multiplication in EndR(N) the way it is, but to use the opposite ring to define the bimodule. This is the approach that is generally taken in more recent works. Using this approach, we conclude that N is a R-EndR(N)op-bimodule. We will adopt this convention for the lemma below.

Considering R as a right and a left module over itself, we can construct the two endomorphism rings EndR(RR) and EndR(RR).


Let R be a ring with an identity elementMathworldPlanetmath. Then REndR(RR) and REndR(RR)𝑜𝑝.


Define ρrEndR(RR) by xxr.

A calculation shows that ρrs=ρsρr (functions written on the left) from which it is easily seen that the map θ:rρr is a ring homomorphismMathworldPlanetmath from R to EndR(RR)op.

We must show that this is an isomorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmath.

If ρr=0, then r=1r=ρr(1)=0. So θ is injectivePlanetmathPlanetmath.

Let f be an arbitrary element of EndR(RR), and let r=f(1). Then for any xR, f(x)=f(x1)=xf(1)=xr=ρr(x), so f=ρr=θ(r).

The proof of the other isomorphism is similar. ∎

Title ring of endomorphisms
Canonical name RingOfEndomorphisms
Date of creation 2013-03-22 14:04:26
Last modified on 2013-03-22 14:04:26
Owner mclase (549)
Last modified by mclase (549)
Numerical id 8
Author mclase (549)
Entry type Definition
Classification msc 16S20
Classification msc 16W20
Synonym endomorphism ring