Proof: Why a = axa When A is a Ring & Jac(A)=0

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Discussion Overview

The discussion revolves around a proof concerning the Jacobson radical of a ring and the properties of idempotent elements within that context. Participants explore the implications of the statement that if every finitely generated right ideal is generated by an idempotent, then the Jacobson radical is zero. The focus is on understanding specific steps in the proof and the definitions involved, particularly the concept of a unital ring.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Post 1 introduces the proof and questions the reasoning behind the equalities a = ae and e = xa.
  • Post 2 affirms the reasoning about e = xa and explains that a = ae follows from A being unital.
  • Post 3 and Post 4 seek clarification on the term "unital," leading to a definition that a unital ring has an identity element.
  • Post 5 suggests that since Ae = Aa, one can conclude a = ae using the identity element.
  • Post 6 challenges this conclusion, indicating that while a = a'e for some a' in A, the idempotent nature of e needs to be considered.
  • Post 7 expresses confusion about the role of idempotency in the proof.
  • Post 8 provides a hint to help clarify the relationship between a, a', and e.
  • Post 9 attempts to apply the hint but questions the assumption that a = a'e.
  • Post 11 elaborates on the structure of Aa and its relationship to Ae, questioning the choice of x as the neutral element.
  • Post 12 confirms that the choice of x corresponds to the neutral element in A.

Areas of Agreement / Disagreement

Participants generally agree on the definition of a unital ring and the implications of idempotent elements, but there is ongoing uncertainty regarding the specific steps in the proof and the reasoning behind certain equalities. The discussion remains unresolved as participants seek clarification on these points.

Contextual Notes

There are limitations in the understanding of how idempotency affects the proof, and the discussion highlights dependencies on definitions and assumptions regarding the elements of the ring.

peteryellow
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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.
 
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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.
 
What does unital means?
 
peteryellow said:
What does unital means?

That you have an element "1" such that 1x = x1 = x for all x in your A.
 
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?
 
peteryellow said:
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?
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?
 
No, I can't see that why s =ae and where we use that e is idempotent. Can you please explain. Thanks.
 
Hint: a=a'e=(a'e)e.
 
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.
 
  • #10
I don't understand what it is you're asking exactly. Could you clarify?
 
  • #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.
 
  • #12
peteryellow said:
but it seems that we are choosing x to be the neutral element in A? Is it so?
Yes, that's exactly what we're doing.
 

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