Let $I$ and $J$ be two monomial ideals of $R=K[x_1,\dotsc,x_n]$ where $K$ is a field.
Could we say $\mid\operatorname{depth} (R/I)- \operatorname{depth}(R/(I:J))\mid\leq 1$, where $(I:J)$ is colon ideal and $I\nsubseteq J$?
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1$\begingroup$ Does $\operatorname{depth}(R/I)$ mean $\operatorname{depth}_{\mathfrak m}(M)$ for the ideal $\mathfrak m = (x_1,\ldots,x_n) \subseteq R$ and the $R$-module $M = R/I$, or do you mean some other notion of depth? (A local ring $R$ also has a depth $\operatorname{depth}(R)$, but $R/I$ is not local...) $\endgroup$– R. van Dobben de BruynCommented May 5, 2023 at 13:22
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$\begingroup$ Yes. $depth (R/I)$ mean $depth_m(M)$. $\endgroup$– Amir MafiCommented May 5, 2023 at 19:49
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1$\begingroup$ The edited version has an easy counterexample: let $I = (x_1,x_2,x_3)(x_4,x_5,x_6)$ be the ideal of two three-spaces in $\mathbf A^6$ meeting at the origin. Then $R/I$ is not (S2) by Hartshorne's connectedness theorem (see Eisenbud, Thm. 18.12). But if $J = (x_4,x_5,x_6)$, then $(I:J) = (x_1,x_2,x_3)$ and $R/(I:J)$ is regular, so $\operatorname{depth}_{\mathfrak m}(R/I) = 1$ and $\operatorname{depth}_{\mathfrak m}(R/(I:J)) = 3$. $\endgroup$– R. van Dobben de BruynCommented May 6, 2023 at 14:56
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$\begingroup$ For the original question whether $\operatorname{depth}_{\mathfrak m}(R/I) \leq \operatorname{depth}_{\mathfrak m}(R/(I:J))$, the only thing I was able to obtain is that a nonzerodivisor on $R/I$ is also a nonzerodivisor on $R/(I:J)$. But I cannot set up an induction because the inclusion $(I+(a):J) \supseteq (I:J) + (a)$ is not always an equality (and I was unable to choose a nonzerodivisor $a$ on $R/I$ cleverly to make this hold). $\endgroup$– R. van Dobben de BruynCommented May 6, 2023 at 15:02
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$\begingroup$ Out of curiosity, do you already know that this result is false for arbitrary ideals contained in $\mathfrak m$, in an arbitrary Noetherian domain $R$ with maximal ideal $\mathfrak m$? $\endgroup$– R. van Dobben de BruynCommented May 6, 2023 at 15:05
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