• June 20th 2010, 04:02 AM
ENRIQUESTEFANINI
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.
• June 20th 2010, 06:39 PM
Can you prove that if f is a surjective ring homomorphism, $J(R)f \subseteq J(S)$?
• June 20th 2010, 09:43 PM
ENRIQUESTEFANINI
Quote:

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

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.
• June 21st 2010, 10:25 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?)
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?)