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I know that in a commutative ring with identity element, every maximal ideal is prime. I thought that maybe the converse is true too. Namely, that every prime ideal is a maximal in a commutative ring with identity. But I can't find a way to prove it. Any hints or suggestions? Thanks in advance!

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  • $\begingroup$ Proposition from Dummit and Foote. Every nonzero prime ideal in a principal ideal domain is a maximal ideal. I can write the proof if you want. $\endgroup$ – richitesenpai Aug 13 '15 at 6:11
  • $\begingroup$ Have you tried to think on this: if the converse were true then why people bothered to call them by two names??? $\endgroup$ – user26857 Aug 13 '15 at 6:57
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$\{0\}$ is prime in $\mathbb Z$ but not a maximal ideal.

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It is not true in general. For instance take an integral domain which is not a field and you see that $(0)$ is prime, but not maximal.

Any Principal Ideal Domain does have the property that non-zero prime ideals are maximal (try to prove it). In general, the property you are talking about is what is called 0-dimensional (Krull dimension) when prime implies maximal.

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If you want there is something weaker, in the sense that you can converse your statement, by that I mean: if $A$ is a unital finite commutative ring then $I$ is a prime ideal if and only if $I$ is a maximal ideal. I can sketch a proof, only for the needed part which is this implication $(\Rightarrow)$ then you can make all the details: Suppose $I$ is a prime ideal $\Rightarrow$ $\frac{A}{I}$ is a domain. Now because $\frac{A}{I}$ is a finite domain and every finite domain is a field then $\frac{A}{I}$ is a field $\Rightarrow$ $I$ is maximal.

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