Let R be noetherian. If ideal $I$ is maximal with respect to the property that $R/I$ is not of finite length, prove $I$ is prime.

Question

I'm quite lost as to how to go about proving this. My instinct is to try find some sort of contradiction, but I couldn't formulate any concrete arguments. Any ideas to prove this?

Answer 1

Suppose $I$ were not prime. Then there are $f,g\in R\setminus I$ with $fg\in I$. Then consider the ideal $J=I+(f)$.

Then consider the exact sequence $$0\to J/I \to R/I \to R/J\to 0$$ Since $I$ was maximal among ideals not having finite length quotients, $R/J$ has finite length. Thus to find a contradiction, it suffices to show that $J/I$ has finite length. However, $J/I$ is annihilated by $I+(g)$, so it is in fact an $R/I+(g)$ module. Again, because $I$ was maximal, $R/I+(g)$ has finite length, so it is Artinian. Thus $J/I$ is a finitely generated (by $f$) module over an Artinian ring, and thus has finite length.

Note: We don't actually need that $R$ is Noetherian for this proof to work, but I assume it is being used to justify the existence of an ideal maximal with respect to the given property.