A question on ring homomorphisms and maximal ideals.

Question

Let $A,B$ be commutative rings, and let $\phi: A \to B$ be a ring homomorphism where $B$ has finitely many elements. Prove that if $I \subset B$ is a maximal ideal then $\phi^{-1}(I)$ is also a maximal ideal in $A$

I've tried to construct a proof by contradiction but with no success, im not looking for answers but rather hints.

Answer 1

Hint: every subring of a finite field is also a finite field.

Same hint, more algebraic: if $x\notin \phi^{-1} (I)$, then multiplication by $x$ gives an injective map $A/\phi^{-1}(I) \to A/\phi^{-1}(I)$.

Answer 2

This sort of thing screams "isomorphism theorems!"

We know $\phi(A)$ is a subring of $B$, and we'd like to talk about "$\phi(A)/I$" but unfortunately there's no guarantee $I\subseteq \phi(A)$. To overcome this, you can instead note that $S:=\phi(A)+I$ is a subring of $B$, and that $I\lhd S$, so $S/I$ is a ring, actually a subring of $B/I$.

$B/I$ is of course a field, and by assumption $B$ and $B/I$ are finite, so $B/I$ is a finite field. $S/I$, being a subring of a field, is a domain, and being a finite domain, it's also a field.

The second isomorphism theorem says that $S/I\cong \phi(A)/(\phi(A)\cap I)$. There is a very natural candidate for a surjective map $\psi:A\to \phi(A)/(\phi(A)\cap I)$. If you discover this map, you'll find its kernel is $\phi^{-1}(I)$.

Applying the first isomorphism theorem to $\psi$, this would prove $A/\phi^{-1}(I)$ is isomorphic to the field $S/I$, hence $\phi^{-1}(I)$ would be maximal in $A$. So there is only one thing left for you to do: can you see what $\psi$ should be?