I've been doing an exercise, and reduced the problem to the following claim. I don't know how to prove it yet, hints are appreciated.
Let $R$ be a ring and $k$ be a field such that $R$ is finitely generated as a $k$-module, meaning there exists an injective ring homomorphism $\phi:k\to R$ and $x_1,\dots,x_n\in R$ so that for each $r\in R$, there exist $a_1,\dots,a_n\in k$ such that
$$ \sum_{i=1}^n \phi(a_i) x_i = r.$$
Prove that $R$ has finitely many prime ideals.
What I tried: I wanted to prove the stronger claim that $R$ has finitely many submodules, but this is false (consider $R = \Bbb{R}^2$ which has infinitely many 1-dimensional submodules). I no longer know how to proceed.
Best Answer
I've outlined an argument in 3 steps; since you only asked for a hint, I've put the proofs of the steps under spoiler tags. Let me know if anything is unclear!
Step 1: Show that, for an integral domain $D$, if $k$ is a field contained in $D$ and $\operatorname{dim}_kD$ is finite, then $D$ is a field.
Step 2: Using Step 1, deduce that every prime ideal of $R$ is maximal.
Step 3: Suppose for contradiction that $P_1,P_2,\dots$ are infinitely many distinct prime ideals in $R$, and use Step 2 to show that the chain $P_1\supseteq P_1P_2\supseteq P_1P_2P_3\supseteq\dots$ is strictly descending. Since any ideal of $R$ is in particular a $k$-subspace of $R$, this will contradict that $\dim_kR$ is finite.