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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?

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  • $\begingroup$ 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? $\endgroup$ – Robert Shore Mar 8 at 22:17
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    $\begingroup$ @RobertShore Finite length means “artinian and noetherian” or, which is the same, “having a composition series”. $\endgroup$ – egreg Mar 8 at 23:35
  • $\begingroup$ Please post the questions into the box question rather than in the title. $\endgroup$ – user26857 Mar 9 at 21:22
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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.

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  • $\begingroup$ 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? $\endgroup$ – davidh Mar 9 at 4:37
  • $\begingroup$ @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. $\endgroup$ – jgon Mar 9 at 17:41

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