Semiprime ring

1. Aug 15, 2008

peteryellow

Let A be semiprime ring and e a non-zero idempotent.

If Ae is a minimal left ideal then eAe is a division ring.

Proof:
Suppose that Ae is a minimal left ideal and that exe is different from 0 for x in A.
Then $Aexe \subset Ae$ since Ae is an ideal and since Ae is minimal hence Aexe = Ae.

Thus there exists a in A such that e = aexe and we get that
(eae)(exe)=eae^2xe=eaexe = e^2 =e.

The only thing I dont understand is that why is e = aexe?

We have that Aexe = Ae so I will say that aexe =ae? but then the rest of proof will not hold?

Any suggestions? Thanks.

2. Aug 15, 2008

morphism

e is in A, so e=ee is in Ae. Thus e is in Aexe as well, so there exists an a such that e=aexe.

A word of caution: Aexe=Ae in general does not imply that aexe=ae for all a in A. Rather, it implies that if a is in A, then there exists an a' in A such that aexe=a'e.

3. Aug 15, 2008

peteryellow

Thank you very much for your help

4. Aug 15, 2008

morphism

You're welcome!