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If $R$ is regular local ring then $R$ is also Cohen-Macaulay ring and integral domain. I want to show that the converse is false.

I come up with this example.

Let $K$ be a field and $R=K[X,Y,Z]/(XY-Z^2)$, $M=(X,Y,Z)/(XY-Z^2)$ is the maximal ideal of $R$, $P=(X,Z)/(XY-Z^2)\subset M$ is the prime ideal in $R$. Because $K[X,Y,Z]$ is Cohen-Macaulay then so is $R$ (for $XY-Z^2$ is not zero-divisor) . Also, $(XY-Z^2)$ is irreducible element in UFD ring $K[X,Y,Z]$ so $R$ is integral domain.

Note that $htP=1$ hence $htP_M=1$. Now, $R_M$ is Cohen-Macaulay local domain (localization preserves Cohen-Macaulay property). If $R_M$ is regular local ring then by Auslander-Buchsbaum theorem it is UFD.

If I can show that $P_M$ is not principal ideal in $R_M$ then I get a needed contradiction and $R_M$ would be the perfect counterexample. But I stuck at this point so I'm looking for some helps.

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    $\begingroup$ In every textbook in CA I know it's proved (or it's an exercise) the following result: if $(R,m)$ is a local regular ring, and $x\in m$ then $R/(x)$ is regular iff $x\notin m^2$. $\endgroup$
    – user26857
    May 14, 2017 at 14:26
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    $\begingroup$ A simpler example: $K[X,Y]_{(X,Y)}/(X^2-Y^3)$. $\endgroup$
    – user26857
    May 14, 2017 at 14:32
  • $\begingroup$ Hello @user26857, I think the reason I stuck is that I dont see $R_M$ clearly.In general, let $\phi: R\to R^*$ is surjective homomorphism let $S$ be multiplicatively closed set of $R$ can you show me what is the relation between $S^{-1}R $ and $\phi (S)^{-1}R^*$ ? Thank you $\endgroup$
    – Anh_Rose 1210
    May 14, 2017 at 14:49
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    $\begingroup$ In fact, the relation is as follows: set $R^*=R/I$; then $S^{-1}R/S^{-1}I\simeq\phi(S)^{-1}(R/I)$. $\endgroup$
    – user26857
    May 14, 2017 at 14:51
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    $\begingroup$ @rschwieb The completion with respect to a maximal ideal gives raise to a local ring. $\endgroup$
    – user26857
    May 15, 2017 at 20:53

1 Answer 1

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The easier path is to realize that if $R_M$ is a regular local ring, it has dimension 2 and thus the maximal ideal must be two generated. This is not the case (it is three generated and not two) follows easily since $M/M^2$ is a three dimensional vector space over $R/M$.

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