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Ring theory

  1. Aug 16, 2008 #1
    Let A be a ring.

    If every finitely generated right ideal is genreated by an idempotent then Jac(A) =0.

    Here Jac(A) means jacobson radical.

    Proof: Let a be in Jac(A) and pick an idempotent element e such that Ae = Aa, thus a = ae and e=xa for x in A. Hence a =axa so a(1-xa)=0. Since a is in Jac(A) also xa is in Jac(A)
    and 1-xa is a unit , hence a=0.


    My question is that why is a = ae and e=xa , please help me with this.

    Because Ae = Aa, so I will say that ae = a'a. and we have that e=xa, is it because we can pick e in A so ee=e =xa, is it so, please help.
     
  2. jcsd
  3. Aug 16, 2008 #2

    morphism

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    Your reasoning about why e=xa is fine. And as to why we have a=ae, this is because A is unital.
     
  4. Aug 16, 2008 #3
    What does unital means?
     
  5. Aug 16, 2008 #4
    That you have an element "1" such that 1x = x1 = x for all x in your A.
     
  6. Aug 16, 2008 #5
    Oh i see unital means that the ring has an identity element. Since Ae=Aa, we choose identity element in A say 1, so we have 1a=ae so a = ae. Is it correct?
     
  7. Aug 16, 2008 #6

    morphism

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    Not quite. We're not guaranteed immediately that the element we get from A is a, i.e. all we can say at this point is that a=a'e for some a' in A. But there is one more ingredient we haven't used: e is idempotent. Can you see how this factors in?
     
  8. Aug 16, 2008 #7
    No, I cant see that why s =ae and where we use that e is idempotent. Can you please explain. Thanks.
     
  9. Aug 17, 2008 #8

    morphism

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    Hint: a=a'e=(a'e)e.
     
  10. Aug 17, 2008 #9
    Thanks for your hint.

    Let a = a'e =a'ee. by multiplying with e we have that ae =(a'e)e =a'e =a. Is it so but my question is we do we have that a =a'e, since Aa =Ae so it is natural for me to say that a'a =xe for x in A.
     
  11. Aug 17, 2008 #10

    morphism

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    I don't understand what it is you're asking exactly. Could you clarify?
     
  12. Aug 17, 2008 #11
    What I am asking is :

    I think that Aa = {a_0a,a_1a,a_2a,......}
    so every element in Aa has a form xa for x in A. Since Aa =Ae so we should have xa=be but it seems that we are choosing x to be the neutral element in A? Is it so? I hope you understand what I mean.
     
  13. Aug 18, 2008 #12

    morphism

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    Yes, that's exactly what we're doing.
     
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