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!

Solution 1:

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.

Solution 2:

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.

Solution 3:

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.