I am working on the following exercise. Let $A$ be a local ring with maximal ideal $m$. Let $A\to B$ be an injective finite map of rings (commutative, with 1), so $A$ is a subring of $B$ and $B$ is a finitely generated $A$-module. Show that there is a prime $q$ of $B$ lying over $m$ ($q\cap A = m$).
I became worried that there is something wrong with my attempt because I only used the injectivity hypothesis to say that $B$ is not the zero ring. Here is what I tried:
Since $B$ is a f.g. $A$-module and $A$ is local with maximal ideal $m$, Nakayama’s lemma implies that $mB\neq B$, since otherwise we would have $B=0$, which is impossible since $A\to B$ is injective and $A$ is not the zero ring. Hence there is some maximal ideal $q$ of $B$ containing the ideal $mB$. Now $q\cap A$ is a prime ideal of $A$ containing $m$, and since $m$ is maximal, this means $q\cap A = m$.
Do we really only need the injectivity hypothesis to argue that $B\neq 0$? (And if so, why are we given the stronger hypothesis? Are they somehow actually equivalent in light of the other assumptions?) Or is there something wrong in the argument?