Can you prove that if f is a surjective ring homomorphism, ?
Hi:
I have the Jacobson radical of a ring R defined as J(R) = {a 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.
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.