ring of endomorphisms
Let be a ring and let be a right -module.
and multiply them
Instead of writing endomorphisms as functions, it is often convenient to write them multiplicatively: we simply write the application of the endomorphism as . Then the fact that each is an -module homomorphism can be expressed as:
for all and and . With this notation, it is clear that becomes an --bimodule.
Now, let be a left -module. We can construct the ring 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 into a bimodule, we need to define an action of on the right of : 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 when it acts on the right.
There are essentially two different ways to go from here. One is to define the multiplication in 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 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 is a --bimodule. We will adopt this convention for the lemma below.
Let be a ring with an identity element. Then and .
Define by .
A calculation shows that (functions written on the left) from which it is easily seen that the map is a ring homomorphism from to .
We must show that this is an isomorphism.
If , then . So is injective.
Let be an arbitrary element of , and let . Then for any , , so .
The proof of the other isomorphism is similar. ∎
|Title||ring of endomorphisms|
|Date of creation||2013-03-22 14:04:26|
|Last modified on||2013-03-22 14:04:26|
|Last modified by||mclase (549)|