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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.

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    $\begingroup$ 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... $\endgroup$
    – Alex Wertheim
    Commented Apr 27, 2022 at 5:28
  • $\begingroup$ 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. $\endgroup$
    – nolatos
    Commented Apr 27, 2022 at 5:39

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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.)

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