I have to solve the following exercise.
Let $R$ be an integral domain that is not a field. Suppose that for all $x\in R\setminus \{0\}$ we have the quotient $R/(x)$ is a finite set. Show that $\dim R = 1$.
I understand that in order to prove $\dim R = 1$ we have to prove that all prime ideal different from $(0)$ is maximal, but I don't know what else to do.
Can someone give me a hint of how I do that?
Suppose that there is a chain $0 \subsetneq \mathfrak{p} \subsetneq \mathfrak{m}$ in $R$.
That implies that $P$ contains a nonunit $x$. Since $R / (x)$ is finite, argue that $R / \mathfrak{p}$ is finite too, and in particular is a finite integral domain.
From there you should be able to conclude that $\mathfrak{p} = \mathfrak{m}$.