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$R,S$ 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.

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flutterbug98
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  • "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. " How are you defining $\psi$? Why should it follow? – Ben Grossmann Nov 11 '19 at 07:30
  • This is the assumption I am not too sure about. I am simply assuming that since $R \rightarrow S$ is an isomorphism, then $R \rightarrow S/\phi(I)$ is also an isomorphism – flutterbug98 Nov 11 '19 at 07:44
  • We would have an easier time examining your assumption if you explained how exactly the map from $R$ to $S/\phi(I)$ is supposed to be defined. That being said, here's a clear indication is false: if we have isomorphisms from $R$ to $S$ and from $R$ to $S/\phi(I)$, then we would be forced to conclude that the rings $R,S,S/\phi(I)$ are all isomorphic. Certainly, we shouldn't expect to have $S$ be isomorphic to $S/\phi(I)$. – Ben Grossmann Nov 11 '19 at 08:20
  • I remember proving this lemma on this post – rschwieb Nov 11 '19 at 17:16

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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 $R/I \cong S/\phi(I)$.

Ethan Alwaise
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