# 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.

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?

• For those of us who'd like to follow along but are a couple of decades removed from our last abstract algebra course, would you mind providing (in a comment) a link to a definition of "length" or just provide a definition? – Robert Shore Mar 8 at 22:17
• @RobertShore Finite length means “artinian and noetherian” or, which is the same, “having a composition series”. – egreg Mar 8 at 23:35
• Please post the questions into the box question rather than in the title. – user26857 Mar 9 at 21:22

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.
• Just to clarify, is it that we need $J/I$ with finite length for a contradiction since that would imply $R/J \cong (R/I)/(J/I)$ is of infinite length, i.e. the quotient of infinite length module $R/I$ over finite length module $J/I$ is infinite length? – davidh Mar 9 at 4:37
• @davidh Sure, that's one way of seeing the contradiction. I was personally thinking that we would then have that $R/I$ fits into a short exact sequence as the middle term with both other terms of finite length, and therefore $R/I$ has finite length, contradiction. But that's basically the same as your argument, so either way. – jgon Mar 9 at 17:41