$R,S$ are commutative rings with $\phi : R \rightarrow S$ an isomorphism between them.
$I \subset R$ is an ideal and $\phi(I) \subset S$ is an ideal as well.
Show that $R/I$ and $S/\phi(I)$ are isomorphic.
This is how I approached this problem:
The natural homomorphism $\pi : R \rightarrow R/I$ is a surjective ring homomorphism with $ker(\phi) = I$.
Given that $\phi : R \rightarrow S$ an isomorphism, then it should follow that $\psi : R \rightarrow S/\phi(I)$ is also a ring isomorphism, meaning it is both surjective and injective.
Thus, by the isomorphism theorem, because there exists a surjective homomorphism $\pi : R \rightarrow R/I$ and a surjective homomorphism $\psi : R \rightarrow S/\phi(I)$, then it follows that there exists an isomorphism $\sigma : R/I \rightarrow S/\phi(I)$.
The reason I am not sure about this proof is because I have made an assumption that I am not sure is correct and also because I haven't really done anything with the elements within these rings.
Any help would be greatly appreciated.
There is a natural surjective homomorphism $\pi \colon S \to S/\phi(I)$. Also $\phi$ is surjective since it is an isomorphism. Then $\psi = \pi \circ \phi$ is a surjective homomorphism $R \to S/\phi(I)$. Check that $\ker \psi = I$, so then by the first isomorphism theorem $R/I \cong S/\phi(I)$.