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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. –  Miha Habič Nov 19 '12 at 23:19
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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. –  Keenan Kidwell Nov 19 '12 at 23:22
    
Keenan this is a good example. In particular it implies that the assumption of the ring being a PID would not be enough! –  Captain Darling Nov 19 '12 at 23:44
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3 Answers 3

Take your favorite infinite local ring $R$ with maximal ideal $\mathfrak{m}$. The ring $R \times R$ is infinite and has exactly two maximal ideals: $\mathfrak{m} \times R$ and $R \times \mathfrak{m}$.

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Good, nice counterexample which has shockingly eluded me. Is there a counterexample of an indecomposable ring ? –  Captain Darling Nov 19 '12 at 23:41
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Rings with finitely many maximal ideals are called semilocal. If your ring is commutative this is equivalent to $R/J(R)$ being Artinian, where $J(R)$ is the Jacobson radical.

Coming back to the example I gave in the comments, the product of fields, it seems that this is the only nice case that can arise, in the commutative case at least; if $R$ is semilocal the Chinese remainder theorem gives an isomorphism between $R/J(R)$ and a finite direct product of fields.

I'm not sure what can be said about the prime ideal version. I do know that there is a result, due to Bill Dubuque, who sometimes posts around here, which says that an infinite ring $R$, such that the set of invertible elements of $R$ has strictly smaller cardinality than $R$, will have infinitely many prime ideals.

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Thanks for your answer Miha. Sometimes people consider finite and countable infinite sets to have the same cardinality. Is this the case for the Dubuque result ? If so, this wouldn't apply to the case of the ring of integers. –  Captain Darling Nov 19 '12 at 23:55
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@CaptainDarling No, the cardinality here is what it should be; finite is very far from countably infinite. –  Miha Habič Nov 20 '12 at 0:00
    
Googling hasn't helped a lot in finding the reference you refer to :( I would be thankful if you could provide a link to the Dubuque result. –  Captain Darling Nov 20 '12 at 1:45
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@CaptainDarling He gives a proof in the answer to this question. –  Miha Habič Nov 20 '12 at 1:54
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There are examples of infinite indecomposable non-local commutative rings w/ 1 which have but finitely many maximal ideals (i.e. are semi-quasi-local). In fact, they are not hard to construct; some can be found lying inside of $\mathbb{Q}$. To see this, set $S:=(2\mathbb{Z}\cup3\mathbb{Z})^c$ (the complement in $\mathbb{Z}$ of the union of the principal ideals generated by $2$ and $3$). One checks easily that $S$ is a multiplicative subset of $\mathbb{Z}$. Now set $R:=\mathbb{Z}_{S}$ (this is simply the subring of $\mathbb{Q}$ containing fractions with numerator in $\mathbb{Z}$ and denominator in $S$). There is a one-to-one correspondence between the maximal ideals of $R$ and the ideals of $\mathbb{Z}$ which are maximal with respect to having empty intersection with $S$ (i.e. maximal with respect to being contained in $2\mathbb{Z}\cup 3\mathbb{Z}$). It is easy to see that these ideals (of $\mathbb{Z}$) are precisely $2\mathbb{Z}$ and $3\mathbb{Z}$. Hence $R$ has exactly $2$ maximal ideals. Every additive subgroup of $\mathbb{Q}$ is indecomposable (otherwise $\mathbb{Q}$ would contain two elements which are linearly independent over $\mathbb{Z}$). It follows immediately from this observation that the ring $R$ is indecomposable.

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