Finitely generated $k$-modules have finite spectrum 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.
 A: 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.

Let $a\in D\setminus\{0\}$; we wish to show that $a$ is invertible. Note that every ideal of $D$ is also a $k$-subspace of $D$, and so – since the dimension of $D$ is finite – any descending chain of ideals must stabilize. In particular, considering the chain $$\langle a\rangle\supseteq\langle a^2\rangle\supseteq\dots,$$ there exists $n\in\mathbb{N}$ with $a^{n}\in\langle a^{n+1}\rangle$; say $a^n=\lambda a^{n+1}$. Since $a\neq 0$ and $D$ is a domain, hence $1=\lambda a$, so indeed $a$ is a unit and we are done.

Step 2: Using Step 1, deduce that every prime ideal of $R$ is maximal.

Let $P$ be a prime ideal of $R$. Then the composition of $\phi$ with the projection map $R\to R\big/P$ is injective (why?) and realizes $k$ as a subring of the integral domain $R\big/P$. Since the projection $R\to R\big/P$ is surjective, $R\big/P$ has $k$-dimension at most that of $R$, and in particular has finite $k$-dimension; by Step 1 $R\big/P$ is thus a field, ie $P$ 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.

Suppose $P_1\dots P_k=P_1\dots P_{k+1}$; then in particular $P_1\dots P_k\subseteq P_{k+1}$, and so since $P_{k+1}$ is prime $P_i\subseteq P_{k+1}$ for some $i$. But $P_i$ is maximal by Step 2, so $P_i=P_{k+1}$, contradicting the hypothesis that the $P_i$ are distinct.

A: Your $R$ is a finite $k$-algebra, so it is an Artinian ring: in fact the submodules of a finite $k$-module satisfy the descending chain condition, and the ideals of $R$ are also $k$-submodules of $R$. Then $R$ is Artinian,  and it is known that an Artinian ring has only finitely many prime ideal, which are all  maximal.
