# Reduced noetherian local ring of depth zero is artinian?

It is well-known that a local ring $$A$$ with maximal ideal $$\mathfrak{m}$$ of depth zero is not necessarily artinian (e.g. $$k[x,y]/(xy, x^2)$$ localised at the origin), but what if we further require that $$A$$ is reduced? I know (or at least am very sure) the answer is yes, a reduced noetherian local ring of depth zero is artinian, but I'm having trouble showing it. Another way to phrase the problem is that if every element of $$\mathfrak{m}$$ is a zero divisor, then $$\operatorname{Ann} \mathfrak{m} \neq 0$$, and the key difficulty for me is going from "everything is annihilated by something" to "something annihilates everything". Any help would be much appreciated!

EDIT: and of course, such an artinian ring would be a field.

• Isn't the maximal ideal of an Artinian ring nilpotent? It sounds like you want to show such a ring is a field, in that case... Commented Apr 27, 2022 at 5:28
• Yeah basically. The original formulation is that in a positive dimensional reduced local ring there is some element in the maximal ideal which is not a zero divisor; it eventually became the question you see. Commented Apr 27, 2022 at 5:39

In a reduced ring the set of zerodivisors equals the union of minimal primes; see here. Since the depth is zero the maximal ideal is contained in this union, so it is minimal. This shows that $$\dim A=0$$. (In fact, we get $$\mathfrak m=(0)$$, so $$A$$ is a field.)