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Let $R:=k[x_1,\ldots, x_n]$ be the standard polynomial ring. Let $I\subseteq R$ be a monomial ideal of height $\ge 2,$ and $\{\ell_1, \ldots, \ell_{t}\}\subseteq R_1$ an $R$-regular sequence. Assume $depth \frac{R}{I+(\ell_1, \ldots, \ell_{t})}=0.$ Then $depth (R/I)\le t?$

Thanks so much for any suggestions.

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  • $\begingroup$ This is an interesting question. Do you have some motivations? $\endgroup$ – Hailong Dao Jul 22 '16 at 18:21
  • $\begingroup$ @HailongDao Thanks for the question. The main motivation, currently, came from Proposition 1.2 in arxiv.org/pdf/0812.2080.pdf This implies that if $depth \frac{R}{I+(x_1, \ldots, x_t)}=0$ then $depth \frac{R}{I}\le t.$ On the other hand, if $\{\ell_1,\ldots, \ell_t \}$ is an $R/I$-sequence the statement is trivially true. $\endgroup$ – user327174 Jul 24 '16 at 19:16
  • $\begingroup$ Ah, yes, that makes sense. I think example 1.4 in that paper gave a counter example to your question. $\endgroup$ – Hailong Dao Jul 25 '16 at 5:32
  • $\begingroup$ Thanks for your help @HailongDao, it was under my eyes... With this settings the answer is NO! as Example 1.4 shows. Let be $R:=k[x_1,x_2,x_3,x_4$ and $I=(x_1x_2, x_2x_3, x_3x_4)\subseteq R.$ $I$ is an aCM ideal of height 2 so $depth R/I=2,$ but one can check that $depth \frac{R}{I+(x_2+x_3)}=0.$ $\endgroup$ – user327174 Jul 26 '16 at 13:59

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