If $R/(x)$ is finite, every non-zero prime ideal of $R$ is maximal

Question

Let $R$ a commutative ring (with unity) satisfying that for all $x\in R-\{0\},$ $R/(x)$ is finite (where $(x)=xR$. I need to prove that every prime ideal of $R$, $I\neq0$, is maximal, and every ideal of $R$ is finitely generated.

I don't know how to start. Any hints?

Answer 1

The second fact results from the fact that if $I$ is a nonzero ideal, and $0\ne x\in I$, as $R/(x)$ is finite, $I/(x)$ is also finite, hence $a_1+(x),\dots, a_n+(x)$ are its elements, we have $$I=(x, a_1,\dots, a_n).$$

Answer 2

The first fact follow from the fact that a finite integral domain with a unit is a field.

Answer 3

Let $P$ be a nonzero prime ideal of $R$. Thus, there is a non zero element $x\in P$. By assumption $R/(x)$ is finite. Hence $\frac{R/(x)}{P/(x)}\cong R/P$ is finite. Thus, $R/P$ if a finite integral domain, and so $R/P$ is a field. Thus $P$ is a maximal ideal of $R$.

Note that the above result is not true for nonzero prime ideal, for example $(\mathbb{Z}, +, .)$ has the above condition, but the zero ideal is not maximal.