In THIS PROBLEM it is proved that for any nonzero ring $R$ with identity, $R[x]$ has an infinite number of maximal left ideals. Is it possible for an uncountable nonzero ring $R$ with identity, $R[x]$ has only a countable number of maximal left ideals ??
Generalizing the answer by wccanard: Let $R$ be a commutative ring. Then the kernel of $R[x] \twoheadrightarrow R_{red}[x]$ consists of nilpotent elements, hence this map induces a homeomorphism $\mathrm{Spec}(R_{red}[x]) \cong \mathrm{Spec}(R[x])$. It restricts to a homeomorphism between the subspaces of closed points $\mathrm{Spm}(R_{red}[x]) \cong \mathrm{Spm}(R[x])$. Now take any uncountable $R$ such that $R_{red}$ is a finite field.

$\begingroup$ @Martin: Right, thanks. Lets continue: How about reduced rings ? Do you have any reduced example ? $\endgroup$ – user30230 Jan 13 '13 at 12:47

$\begingroup$ Is there an uncountable boolean ring $R$ such that $\mathrm{Spec}(R)$ is countable? This would answer your question, because $\mathrm{Spec}(R[x]) = \coprod_{m \in \mathrm{Spec}(R)} \mathrm{Spec}(R/m[x]) = \coprod_{m \in \mathrm{Spec}(R)} \mathrm{Spec}(\mathbb{F}_2[x])$ is countable. $\endgroup$ – Martin Brandenburg Jan 13 '13 at 15:36

6$\begingroup$ @Martin: I'm afraid there is no such Boolean algebra. Since maximal ideals are in twosided correspondence with ultrafilters and if $R$ is a superatomic Boolean algebra then $Ult(R) = R$ and if $R$ is not superatomic then $Ult(R) > 2^{\aleph_0}$. So it seems that there is no uncountable Boolean algebra with only a countable number of ultrafilters. $\endgroup$ – user30230 Jan 13 '13 at 19:59
Yes. Take your favorite finite field, now form $R$ by adjoining uncountably many new variables each with square equal to zero and product of any two equal to zero (more precisely, let's say any element of $R$ is only allowed to mention finitely many of these variables) and now any maximal ideal will have to contain all of them, so done.

$\begingroup$ Which is to say, the ideals in $R[x]$ correspond to the ideals in $F[x]$, where $F$ was the finite field. $\endgroup$ – Allen Knutson Jan 13 '13 at 12:13