Prove that $ k[x_1,\ldots,x_4]/ \langle x_1x_2,x_2x_3,x_3x_4,x_4x_1 \rangle$ is not Cohen-Macaulay.

We have $\langle x_1x_2,x_2x_3,x_3x_4,x_4x_1 \rangle=\langle x_1,x_3 \rangle \cap \langle x_2,x_4\rangle$. Therefore $\dim k[x_1,\ldots,x_4]/ \langle x_1x_2,x_2x_3,x_3x_4,x_4x_1 \rangle=2$.

How do I prove $ k[x_1,\ldots,x_4]/ \langle x_1x_2,x_2x_3,x_3x_4,x_4x_1 \rangle$ is not Cohen-Macaulay?


I believe this is the idea: Look in the local ring at the origin. Mod out by $x_2-x_1$. The quotient ring is now the localization of $k[x_2,x_3,x_4]/(x_2^2,x_2x_3,x_3x_4,x_2x_4)$, and everybody in the maximal ideal is now a zero-divisor. So the depth at the origin is only $1$.

  • $\begingroup$ In order to conclude that the depth is 1 it remains to show that $x_2-x_1$ is a non-zero divisor on the original ring. $\endgroup$ – user26857 Aug 31 '15 at 6:34
  • 1
    $\begingroup$ Fair enough; I'll "leave that as an exercise for the interested reader." In any event the depth is $\le 1$, so we get non-CM ;) $\endgroup$ – John Brevik Aug 31 '15 at 17:08

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