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Let $(R,m,k)$ be a local Noetherian ring such that $\operatorname{depth}R=0$.

Question: Is it true that $R$ is Artinian?

PS: If it is true then please only say so, as i am still attempting to prove it. If it is not true then please provide a counterexample, preferably using the polynomial ring over a field.

Manos
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1 Answers1

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It is not true. Let $R = k[x,y]_{(x,y)}/(x^2, xy)$. Then the maximal ideal $m = (x,y)$ annihilates the nonunit $x$, so $m = \text{ann}_R(x)$ is an associated prime, i.e. $\text{depth}(R) = 0$, but $R$ is not Artinian.

(By the way, it is well worth keeping this example in mind, because it occurs so frequently.)

zcn
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  • @Manos By the same token, you can construct a depth zero Noetherian local ring, essentially of finite type over a field, with arbitrary Krull dimension $d$. Just let $x$, $y_1, \ldots, y_d$ be indeterminates over the field $k$, and let $R = k[x, y_1, \ldots, y_d]_{(x, y_1, \ldots, y_d)} / x \cdot \mathfrak m$, where $\mathfrak m$ is the unique maximal ideal. – neilme Feb 28 '14 at 02:53