Finding a ring isomorphism

Question

Let $\phi : R \to R'$ be a ring epimorphism and $J\lhd R'$ an ideal of $R'$.
Indicate a ring isomorphism $\psi: R/\phi^{-1}(J) \to R'/J$

The only thing i know about this problem is that $\phi^{-1}(J)\lhd R$ But how can i find a ring isomorphism? Quotient rings are somehow strange to me.

My idea: $\psi(a+\phi^{-1}(J))=\psi(a)+\psi(\phi^{-1}(J))$ and i guess that $\psi(\phi^{-1}(J))$ should be $J$. Unfortunatley that is all i got.

Answer 1

Define $$\psi : R/\phi^{-1}(J) \to R'/J$$ $$\psi(r + \phi^{-1}(J) )= \phi(r) + J$$ Then this map is:

i) well-defined: infact if $r + \phi^{-1}(J) = r' + \phi^{-1}(J) $ then $r-r' \in \phi^{-1}(J)$ and so $\phi(r) - \phi(r') \in J$

ii) a ring homomorphism: because $\phi$ is a ring homomorphism

iii) surjective: because $\phi$ is surjective

iv) injective: if $\psi(r + \phi^{-1}(J) )= \phi(r) + J= 0$ then $\phi(r) \in J \Rightarrow r \in \phi^{-1}(J) \Rightarrow r + \phi^{-1}(J) = 0$

Answer 2

your idea is correct, it is better to write it as $$ \psi(a+\phi^{-1}(J))=\psi(a)+J $$ then it is easy to show that it is a homomorphis, surjective and it's kernel is $(\phi^{-1}(J))$

then apply http://en.wikipedia.org/wiki/Isomorphism_theorem#First_isomorphism_theorem_2