Are there infinitely many maximal ideals in an infinite non-local ring? 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 ?
 A: 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 $R \times \mathfrak{m}$. 
A: 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.
A: 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.
