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An application of Zorn's Lemma shows that each non-trivial ring has a maximal ideal. When it comes to the ring $ \mathbb{Z},$ this shows that there exists at least one rational prime $p,$ as maximal ideals in this particular ring are ideals of the form $m\mathbb{Z},$ where $m$ is a prime.

Therefore, if one could show that any infinite ring has infinitely many maximal ideals, this would yield another proof of the existence of infinitely many primes.

However, the example of the local ring $\mathbb{Z}_p$ shows that this does not hold in general, as this ring has a unique maximal ideal, namely $p\mathbb{Z}_p.$

One therefore poses the following question:

Does an infinite non-local ring always possess infinitely many maximal ideals ?

If the answer is no, what would be a counterexample ? And in that case, can one provide a further assumption to ensure that infinitely many maximal ideals can be found ? What happens if one replaces the word maximal by prime ?

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    Direct products of (infinite) fields would give a trivial counterexample to the title question.2012-11-19
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    Take $R$ to be the ring of integers in a number field (or any infinite Dedekind domain). Let $P_1,\ldots,P_n$ be maximal ideals of $R$. Put $S=\bigcap_{i=1}^n(R\setminus P_i)$, a multiplicative subset of $R$. Then the localization $S^{-1}R$ is a principal ideal domain containing $R$ (so infinite) whose maximal ideals are $P_1,\ldots,P_n$. Together with $(0)$, these are all the prime ideals.2012-11-19
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    Keenan this is a good example. In particular it implies that the assumption of the ring being a PID would not be enough!2012-11-19

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