Let $I$ be a Cohen-Macaulay monomial ideal of $R=K[x_1,...,x_n]$, where $K$ is a field. Can we say the ideal $(x_1)+I$ is Cohen Macaulay?
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Not even when $I$ is square-free. Let $I$ be the ideal $(x_2x_4,x_2x_5,x_3x_4,x_3x_5, x_1x_2,x_1x_5)$. It is Cohen-Macaulay because it is the Stanley-Reisner ideal of a path of $5$ vertices (with $x_1$ in the middle). Killing $x_1$ will disconnect the graph, and indeed the ideal $(x_2x_4,x_2x_5,x_3x_4,x_3x_5)$ is not Cohen-Macaulay. You can check both statements by Macaulay 2.
answered Jan 18, 2020 at 1:31
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$\begingroup$ In fact, you can get rid of $x_5$ to make a smaller example. $\endgroup$ Commented Jan 18, 2020 at 1:38
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$\begingroup$ Thank you so much for your nice example $\endgroup$ Commented Jan 18, 2020 at 4:59