Let $R:=k[x]/x^p$ with a field $k$ of characteristic $p$.
Then this local ring $R$ is regular or not?
commutative-algebra
asked 2012-10-18
user id:39198
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No, because regular local rings are domains. It doesn't matter that $k$ has characteristic $p$, only that $p>1$. Alternatively, and more easily, this ring has dimension zero. If a regular local ring has dimension zero, then its maximal ideal is equal to its square, hence is zero by Nakayama. So regular local rings of dimension zero are fields. Your ring is not a field. – 2012-10-18
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Thak you very much. I understood. – 2012-10-18
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@Keenan: In order to apply Nakayama, you need that the maximal is finitely generated. So probably add "noetherian" somewhere. – 2012-10-19
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