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Let $A$ be a commutative ring, such that $A^*$ is finite, and that for any maximal ideal $\frak m$, $A/\frak m$ is finite. Is it true that $A$ is denumerable? I've been trying to disprove this by looking at $\mathbf{K}[X]$, where $X=(X_i)_{i \in I}$ and $I$ is non-denumerable but I can't prove that the maximal ideals of this ring are trivial, so that the residue fields would be isomorphic to $\mathbf{K}$ (then choose $\mathbf{K}$ finite). Another idea was looking at maximal ideals of some non-denumerable product $\prod_i K$ but again there may be some strange maximal ideals (like ultrafilters).

What do you think? Maybe the answer to the question is in fact yes?

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up vote 4 down vote accepted

Any Boolean ring $B$, such as a product $\prod_i \mathbb{F}_2$ over an arbitrary index set, has this property. If $u$ is a unit, then $u^2 = u$ implies $u = 1$, so $1$ is the only unit. If $P$ is a prime ideal, then $B/P$ is a Boolean ring which is also an integral domain, and considering the equation $u(1 - u) = 0$ it follows that the only such Boolean ring is $\mathbb{F}_2$. Hence strange maximal ideals are not an issue.

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@QiaochuYuan What about the case when $A$ is required to be an integral domain? – user26857 Dec 22 '12 at 11:13
Indeed! You made my day, thanks. – timofei Dec 22 '12 at 11:33

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