Ring theory

1. Aug 16, 2008

peteryellow

Let A be a ring.

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

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. Aug 16, 2008

morphism

Your reasoning about why e=xa is fine. And as to why we have a=ae, this is because A is unital.

3. Aug 16, 2008

peteryellow

What does unital means?

4. Aug 16, 2008

NoMoreExams

That you have an element "1" such that 1x = x1 = x for all x in your A.

5. Aug 16, 2008

peteryellow

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?

6. Aug 16, 2008

morphism

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?

7. Aug 16, 2008

peteryellow

No, I cant see that why s =ae and where we use that e is idempotent. Can you please explain. Thanks.

8. Aug 17, 2008

morphism

Hint: a=a'e=(a'e)e.

9. Aug 17, 2008

peteryellow

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.

10. Aug 17, 2008

morphism

I don't understand what it is you're asking exactly. Could you clarify?

11. Aug 17, 2008

peteryellow

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.

12. Aug 18, 2008

morphism

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