Non-Noetherian ring with a single prime ideal [closed]

What are the most simple examples of a commutative ring $R$ satisfying both of the following two properties:
1. $R$ is not Noetherian.
2. $R$ has exactly one prime ideal.

Solution 1:

Try $k[x_1, x_2, ...]/(x_i x_j)$ for all $i, j$.

Solution 2:

$$R:= k[x_1, x_2, x_3, \cdots ]/\langle x_1^2,\ x_2^2=x_1,\ x_3^2=x_2,\ x_4^2=x_3,\ \cdots \rangle$$

Remember that prime ideals are precisely the kernels of maps to fields. If $K$ is any field, and we have a map $\phi: R \to K$, then $\phi(x_1)=0$ and hence $\phi(x_2) =0$ and hence $\phi(x_3)=0$ and so forth, so the only prime ideal is $\langle x_1, x_2, x_3, \ldots \rangle$.

To see that this is not noetherian, note that $(0) \subsetneq (x_1) \subsetneq (x_2) \subsetneq (x_3) \subsetneq \cdots$ is an ascending chain of ideals that doesn't terminate.

Of course, the specific exponent $2$ isn't important. The key is to make a sequence of variables $x_1$, $x_2$, ..., all of which are nilpotent, but so that you can kill $x_1$, $x_2$, ..., $x_k$ while not killing the later $x$'s.

Solution 3:

If you're willing to ask for one nonzero prime ideal, then for any prime $p$ the ring of integers of ${\mathbf C}_p$ (the $p$-adic complex numbers) is an example. The only nonzero prime ideal in that ring is its maximal ideal, which is not finitely generated.

Solution 4:

I thought this observation might be useful to future viewers other than the OP. If a mod thinks this is old enough for a new topic, please do.

Since the nilradical is in every prime, the condition of having only one prime ideal (which will also be maximal) is equivalent to the condition that the nilradical is a maximal ideal.

Thus one immediately obtains a large class of examples as mentioned by the posters above as non noetherian rings of the form $k[N]$ where $k$ is a field and $N$ is a set of nilpotent elements. So in fact, in Qiaochu's example it is enough to take the simpler quotient $k[x_1,x_2,...]/(x_1^2,x_2^2,...)$.

Also note as in the examples of the form $k[N]$ given above, in any example the maximal ideal will need to be be infinitely generated since a ring is noetherian iff every prime is finitely generated.

I do wonder if there are examples not of the form $k[N]$.

Solution 5:

Take a localization $O:=R_M$ of the ring $R$ of all algebraic integers at a maximal ideal $M$. Then $O/xO$ has exactly one prime for every non-zero $x\in MO$.