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Semiprime ring

  1. Aug 15, 2008 #1
    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. jcsd
  3. Aug 15, 2008 #2

    morphism

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    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.
     
  4. Aug 15, 2008 #3
    Thank you very much for your help:smile:
     
  5. Aug 15, 2008 #4

    morphism

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    You're welcome!
     
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