Let $A$ be a Cohen-Macaulay ring, and $I$ an ideal of $A$.
What can we say about $\operatorname{depth}(A/I)$?
I know that $\operatorname{depth}(A/I)\le \dim(A/I)$.
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1$\begingroup$ What would you like to say? You can choose ideals so that the depth can be any number in the allowed range of $[0,\dim (A/I)]$. $\endgroup$– MohanCommented Jul 18, 2016 at 15:27
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$\begingroup$ For example, depth(A) -depth(I) less than or equal depth(A/I)??? $\endgroup$– Paulo RossiCommented Jul 18, 2016 at 15:29
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1$\begingroup$ No. For example, consider the case of $I$ a maximal ideal. Then depth of $A/I=0$ and if $\dim A>0$ (locally near $I$), the depth of $I=1$. But, depth of $A$ can be as large as you want. $\endgroup$– MohanCommented Jul 18, 2016 at 15:35
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1$\begingroup$ A is cohen-macaulay, so, dim(A) = depth(A) = depth(I) (because I is a maximal ideal) $\endgroup$– Paulo RossiCommented Jul 18, 2016 at 15:43
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$\begingroup$ @PauloRossi. You are wrong to write that $\text{depth}(I)$ equals $\text{depth}(A)$. If $R$ is a local integral domain, and if $(x,y)$ is any ordered pair of nonzero elements of the maximal ideal $I$, then $\overline{x} \in I/xI$ is nonzero, but $y\cdot \overline{x}$ equals $0$ in $I/xI$. Thus, the depth of $I$ is at most $1$. $\endgroup$– Jason StarrCommented Jul 18, 2016 at 16:05
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