Maximal Ideal Must be Prime

Question

I am trying to prove that an ideal that is maximal with respect to not being finitely generated must be prime.

What does it mean to be an ideal that is maximal with respect to not being finitely generated?

I am pretty sure that I need to assume that we have a prime ideal $P \subseteq R$ such that $P=$ the intersection of some ideals. Then I can use that for $f \in R$, $(P:(f))$ equals R if $f \in P$ or $P$ if $f \notin P$.

Is this the right approach?

Answer 1

You can actually prove this from elementary principles:

Let $I$ be the ideal maximal with respect to not being finitely generated. Suppose $I$ is not prime. Then $\exists f,g\in R$ such that $f,g\notin I,\ fg\in I$. Then consider $J := I + (f)$. You can show that since $I$ not finitely generated, $J$ is not finitely generated. Moreover, $J \neq R$ (otherwise $g\in I$). Hence, $J$ is a proper ideal strictly greater than $I$ that is not finitely generated, which is a contradiction to our choice of $I$.