# Preimage of a maximal ideal.

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?

• 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. – Tobias Kildetoft Jun 3 '13 at 16:00
• Yes, that proof is actually also incorrect (not sure how those who answered it missed that). But in that case it can be salvaged by using that the quotient is isomorphic to a subring of $S/P$ and a subring of an integral domain is itself an integral domain. – Tobias Kildetoft Jun 3 '13 at 16:25
• @TobiasKildetoft does it work if it is surjective? my concern is that kernel(h) might not be $f^{-1} (P)$ but only $h^{-1}(P)$. Or is it true that these two are the same? – user10024395 Apr 14 '15 at 16:12