Counterexample: Local Artinian ring is not a PID

Question

Specifically, I am looking for a commutative ring with unity, which is local and Artinian, but not a PID. I'm not too sure how to figure it out. I know the power series $\mathbb{F}\lbrack\lbrack x \rbrack\rbrack$ over a field is local and Artinian, but it is also a PID. I also know a local Artinian ring has only one maximal ideal and no other prime ideals, and furthermore every element of the ring is either a unit or nilpotent. The text I am reading seems to hint that this information is useful in finding a counterexample, but I'm just not seeing it.

I just tried something by following an example on another problem (found here), which takes $R = \mathbb{C}[x,y]$ together with an ideal $I = \langle x, y \rangle$. Then we construct a quotient ring $R/I^2$, which can be proven to be local (its maximal ideal is $\langle \overline{x}, \overline{y} \rangle$), and as such is obviously not a PID. However the answer in question claims there are no other prime ideals than $\langle \overline{x}, \overline{y} \rangle$, which I do not seem to follow. Are the ideals $\langle \overline{x} \rangle, \langle \overline{y} \rangle$ not prime themselves?

Answer 1

Some DaRT searches:

Commutative, local, Artinian, not a PID

Commutative, local, Artinian, not a principal ideal ring

In particular $F_2[x,y]/(x,y)^2$ and $\mathbb R[x,y,z]/(x^2,y^2, xz,yz,z^2-xy)$.

The first one, for example, has $8$ elements, and the the unique maximal ideal is generated by $x$ and $y$, and is nonprinciple.

The easiest one in the first list is probably just $\mathbb Z/p^k\mathbb Z$ where $p$ is a prime and $k>1$. It simply fails to be a domain, but it is still a PIR. So $\mathbb Z/4\mathbb Z$ would be a small example like that.

However the answer in question claims there are no other prime ideals than $\langle \overline{x}, \overline{y} \rangle$, which I do not seem to follow.

In an Artinian ring, the prime ideals are maximal, so it suffices to show there is a unique maximal ideal. That is easy to show since $x$ and $y$ are both nilpotent, and therefore the ideal they generate has only nilpotent elements. It's obviously also maximal since the quotient is isomorphic to $\mathbb C$. A maximal ideal of nilpotent elements is always unique. (easy exercise.)

Are the ideals $\langle \overline{x} \rangle, \langle \overline{y} \rangle$ not prime themselves?

Not in this ring: Note that $(x+I^2)=(x, y^2)+I^2$ and then

$(\mathbb C[x,y]/I^2)/(x+I^2)\cong \mathbb C[x,y]/(x,y^2)\cong \mathbb C[y]/(y^2)$ which has a nilpotent element. A similar thing can be said for the other quotient.