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In THIS PROBLEM it is proved that for any non-zero ring $R$ with identity, $R[x]$ has an infinite number of maximal left ideals. Is it possible for an uncountable non-zero ring $R$ with identity, $R[x]$ has only a countable number of maximal left ideals ??

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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.

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@Martin: Right, thanks. Lets continue: How about reduced rings ? Do you have any reduced example ? – user30230 Jan 13 '13 at 12:47
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. – Martin Brandenburg Jan 13 '13 at 15:36
@Martin: I'm afraid there is no such Boolean algebra. Since maximal ideals are in two-sided 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. – 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.

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

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