I've just started a commutative algebra course and I'm stuck on the very first homework problem:
Let $A \not= \{0\}$ be a commutative ring. Let $\Phi : A \longrightarrow B$ be a ring homomorphism where $B$ has finitely many elements. Show that if $I$ is a maximal ideal of $B$, then $\Phi^{-1}(I)$ is a maximal ideal of $A$.
It's easy enough to prove that if $I$ is an ideal then $\Phi^{-1}(I)$ is also an ideal, but I'm stuck on the maximality. My idea was to assume, for a contradiction, that there exists some proper ideal $J \subset A$ that strictly contains $\Phi^{-1}(I)$; then its image would perhaps be a proper ideal in $B$ that strictly contains $I$, which would contradict the maximality of $I$. However, this doesn't seem to work, since I can't prove that the image of $J$ is a proper ideal. It is not given that $\Phi$ is surjective. I'm also not sure how to exploit the fact that $B$ has finitely many elements.
Any hints on this problem (no complete solutions please)?
Hints. $A/\Phi^{-1}(I)$ is isomorphic to a subring of $B/I$. The last one is a finite field. Finite integral domains are fields. An ideal is maximal iff its quotient ring is a field.