Ring isomorphism maps a Jacobson radical into a Jacobson radical?

**ENRIQUESTEFANINI** · June 20th 2010, 03:02 AM

Hi:
I have the Jacobson radical of a ring R defined as J(R) = {a $\in$ 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.

**maddas** · June 20th 2010, 05:39 PM

Can you prove that if f is a surjective ring homomorphism, $J(R)f \subseteq J(S)$ ?

**ENRIQUESTEFANINI** · June 20th 2010, 08:43 PM

Originally Posted by maddas

Can you prove that if f is a surjective ring homomorphism, $J(R)f \subseteq J(S)$ ?

Thanks for your reply.

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

Let now $s\in S$ . Hence $(\exists u\in R) uf=s$ because f is surjective. Then by (1), $(\exists r'\in R) p(uf)+r'f-p(uf)(r'f)=0$ . Hence $(\exists v\in S) p(uf)\circ v=0$ . But $uf=s$ . Hence $(\exists v\in S) ps\circ v=0$ . But $s$ was an arbitrary element of S. Therefor, $pS$ is r.q.r. But then $p\in J(S)$ and $J(R)f\subseteq J(S)$ , q.e.d.

**maddas** · June 21st 2010, 09:25 AM

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 $J(R)f\subseteq J(S)$ , $J(S)g \subseteq J(R)$ , and $J(S)gf \subseteq J(R)f$ . But $gf=1$ , so $J(S)\subseteq J(R)f \subseteq J(S)$ . (I'm not sure why I typed this, I guess just confirming you got your answer?)

**ENRIQUESTEFANINI** · June 21st 2010, 10:45 AM

Originally Posted by maddas

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 $J(R)f\subseteq J(S)$ , $J(S)g \subseteq J(R)$ , and $J(S)gf \subseteq J(R)f$ . But $gf=1$ , so $J(S)\subseteq J(R)f \subseteq J(S)$ . (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.

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