# Prove that an infinite ring with finite quotient rings is an integral domain

How can we show that if $R$ is an infinite commutative ring and $R/I$ is finite for every nonzero $I \unlhd R$, then $R$ is an integral domain?

I tried proceeding by contradiction: assume $a$,$b$ $\in R \backslash \{0\}$ and $ab=0$; then $R/(a)$ and $R/(b)$ must be finite, say $R/(a)=\{k_i + (a) : 1 \leq i \leq m\}$ and $R/(b)=\{l_j + (b) : 1 \leq j \leq n\}$. Does this mean $R$ must be finite? Or what about using the fact that $R/(a,b)$ finite?

Thanks for any help with this!

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If $ab=0$, and if $k_i$, $1\leq i\leq m$, are representatives mod $a$, then $(b)=\{bk_1,\dots,bk_m\}$, i.e. $(b)$ is finite and hence $R/(b)$ infinite.

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Dear user, I don't see why $(b)=\{bk_1,\dots,bk_m\}$. Could you please explain that? – Georges Elencwajg Mar 30 '12 at 11:09
+1 Very nice proof, if a bit succinct. – Marc van Leeuwen Mar 30 '12 at 11:43
@GeorgesElencwajg: Maybe you understand better in abstract terms: whenver $ab=0$ in a commutative ring $R$, the surjective map $r\mapsto br:R\to bR$ induces an surjective $R$-module morphism $R/aR\to bR$, so if $R/aR$ and $R/bR$ are both finite, then $R$ must be finite as well (and in fact its order divides $\#(R/aR)\times\#(R/bR)$). – Marc van Leeuwen Mar 30 '12 at 11:44
@user8268 Thanks! – Harry Macpherson Mar 30 '12 at 11:59
@Georges Elencwajg: any $c\in R$ is of the form $k_i+ad$ for some $d\in R$. And $bc=bk_i$, as $ba=0$. (the comment by Marc van Leeuwen makes it much nicer) – user8268 Mar 30 '12 at 12:00

I give a solution when $R$ is commutative and noetherian.

Suppose absurdly that $R$ is not a domain, hence $0$ is not a prime ideal. Therefore, for every prime ideal $\mathfrak{p}$ of $R$, the domain $R / \mathfrak{p}$ is finite, hence a field. So every prime ideal of $R$ is maximal, i.e. $R$ is artinian. Therefore $R$ is a direct product of some of its quotients, hence $R$ is finite.

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A few comments: the condition on $R$ implies ACC, so $R$ is always Noetherian. Second, this reasoning only reduces to the case where $R$ is local, i.e. the direct product has only 1 factor. To finish from here, one can argue as in the other answer: if $(R,m)$ is Artinian local, and $m^k = 0, m^{k-1} \ne 0$, then pick $0 \ne x \in m^{k-1}$, and note that $|(x)| = |R/m| < \infty$ (since $mx = 0$), so $|R| \le |(x)||R/(x)| < \infty$ – zcn Oct 20 '14 at 22:35
Another way to phrase the argument once you have deduced that $R$ is artinian is that it has finite length as a module over itself, and hence is finite since each simple $R$-module is finite by hypothesis. – Eric Wofsey Jan 29 at 7:40

Here's another way to look at it: suppose that $R$ is as you say, and assume by way of contradiction that $R$ is not a domain. Then $ab=0$ for some nonzero $a,b\in R$. Let $\operatorname{Ann}_R(b):=\{x\in R\colon xb=0\}$. Then both $R/Rb$ and $R/\operatorname{Ann}_R(b)$ are finite. Since $R$ is infinite, both $Rb$ and $\operatorname{Ann}_R(b)$ are infinite. But $Rb\cong R/\operatorname{Ann}_R(b)$, thus $Rb$ is finite, a contradiction.

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This idea also appears in the accepted answer: if $Rb$ and $R/Rb$ are finite, then $R$ is finite. – user26857 Jan 30 at 19:00