Preimage of a maximal ideal.

Question

My textbook says that if $f: R \rightarrow S$ is a ring homomorphism, where $R$ and S are commutative; then if $P$ is a maximal ideal of $S$, it might not necessarily be a maximal ideal of $R$. A counterexample is if we choose $R = \Bbb{Z}$, $S = \Bbb{Q}$, and let $f$ be the inclusion map.

However, when I try to "prove" that the preimage must also be maximal, my proof makes sense to me. I know that I must be doing something wrong in the proof, but I couldn't really see it.

"Proof"

Define $g: S \rightarrow S/P$ with kernel $P$. Let $h = g \circ f: R \rightarrow S/P$. Since $h$ is a ring homomorphism, the kernel is an ideal of $R$. Also, from the first isomorphism theorem, we know that $R/\ker(h) \cong S/P$. Since $P$ is a maximal ideal of $S$, we know that $S/P$ is a field. Since $R/\ker(h)$ is isomorphic to $S/P$, it must also be a field, which implies that the kernel of $R$ (which is the preimage of $P$) is a maximal ideal of $R$.

Can anybody tell me where I went wrong in this proof?

Thank you in advance

Answer 1

To remove this from the unanswered queue, I am copying the comment of Tobias Kildetoft above.

You need the map to be surjective to use the isomorphism theorem like that. You only get that the quotient is isomorphic to the image of h in general.