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Thread: Ring isomorphism maps a Jacobson radical into a Jacobson radical?

  1. #1
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    Ring isomorphism maps a Jacobson radical into a Jacobson radical?

    Hi:
    I have the Jacobson radical of a ring R defined as J(R) = {a $\displaystyle \in $R: aR is r.q.r}, where r.q.r stands for right quasi-regular. If S is another ring and f:R-->S is an isomorphism of R onto S, I want to prove (I know, there's no need) that J(R)f = J(S).

    Could somebody give me a hint? Thanks in advance.
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  2. #2
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    Can you prove that if f is a surjective ring homomorphism, $\displaystyle J(R)f \subseteq J(S)$?
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  3. #3
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    Quote Originally Posted by maddas View Post
    Can you prove that if f is a surjective ring homomorphism, $\displaystyle J(R)f \subseteq J(S)$?
    Thanks for your reply.

    Let $\displaystyle p\in J(R)f$. Then $\displaystyle p=af$ where $\displaystyle a\in J(R)$. But $\displaystyle a\in J(R) \Rightarrow aR$ is r.q.r $\displaystyle \Rightarrow (\forall r\in R)(\exists r'\in R) ar\circ r'=0\Rightarrow (\forall r\in R)(\exists r'\in R)ar+r'-arr'=0 \Rightarrow (\forall r\in R)(\exists r'\in R)p(rf)+r'f-p(rf)(r'f) =0 $. Therefore
    $\displaystyle (\forall r\in R)(\exists r'\in R)p(rf)+r'f-p(rf)(r'f) =0$ ************* (1)

    Let now $\displaystyle s\in S$. Hence $\displaystyle (\exists u\in R) uf=s$ because f is surjective. Then by (1), $\displaystyle (\exists r'\in R) p(uf)+r'f-p(uf)(r'f)=0 $. Hence $\displaystyle (\exists v\in S) p(uf)\circ v=0$. But $\displaystyle uf=s$. Hence $\displaystyle (\exists v\in S) ps\circ v=0$. But $\displaystyle s$ was an arbitrary element of S. Therefor, $\displaystyle pS$ is r.q.r. But then $\displaystyle p\in J(S)$ and $\displaystyle J(R)f\subseteq J(S)$, q.e.d.
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  4. #4
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    Er, I'll take your word for it. Anyway, if f is a ring isomorphism, it has an inverse g. Then both f and g are surjective ring homomorphisms, so $\displaystyle J(R)f\subseteq J(S)$, $\displaystyle J(S)g \subseteq J(R)$, and $\displaystyle J(S)gf \subseteq J(R)f$. But $\displaystyle gf=1$, so $\displaystyle J(S)\subseteq J(R)f \subseteq J(S)$. (I'm not sure why I typed this, I guess just confirming you got your answer?)
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  5. #5
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    Quote Originally Posted by maddas View Post
    Er, I'll take your word for it. Anyway, if f is a ring isomorphism, it has an inverse g. Then both f and g are surjective ring homomorphisms, so $\displaystyle J(R)f\subseteq J(S)$, $\displaystyle J(S)g \subseteq J(R)$, and $\displaystyle J(S)gf \subseteq J(R)f$. But $\displaystyle gf=1$, so $\displaystyle J(S)\subseteq J(R)f \subseteq J(S)$. (I'm not sure why I typed this, I guess just confirming you got your answer?)
    How simple! Well, thanks a lot.

    P.S.: I was reading a proof the author begins with "We may as well assume that R is itsef an irreducible ring of endomorphisms of an abelian group V (not just isomorphic to such a ring)". That is why I issued the first post. I want to prove the theorem without making that assumption. I don't get used to the idea that isomorphic structures can be treated as if they were exactly the same structure. Forgive my lengthyness.
    Last edited by ENRIQUESTEFANINI; Jun 21st 2010 at 11:09 AM.
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