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Here is my question:

Let $(R,\mathfrak m)$ be a Noetherian local ring with ideal $I\subset\mathfrak m$. Do we have $$\mathrm{depth}_IR+\dim R/I\geq\mathrm{depth}R?$$

If we have, then this is stronger than $\mathrm{depth}R\leq\dim R$ since we have $$\mathrm{depth}_IR+\dim R/I\leq\mathrm{height}(I)+\dim R/I\leq\dim R.$$ But I don not whether this is true or not! If this is not true, is there some counter-examples?

Thank you for your any help!

WakeUp-X.Liu
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  • Sorry, what is $\operatorname{depth}_I R$? Did you mean $\operatorname{depth}_R I$? You need an underlying ring to calculate the depth of a module! – PCeltide Mar 16 '24 at 10:42
  • @PCeltide The $\mathrm{depth}_IR$ is the maximal length of $R$-regular sequence in $I$. – WakeUp-X.Liu Mar 16 '24 at 12:01
  • This may not answer your question, but for a sufficiently nice ring (say, excellent and equidimensional, maybe not necessary) then $\operatorname{depth}(R)\le \operatorname{depth}(R_\mathfrak{p})+\dim(R/\mathfrak{p})$ for all $\mathfrak{p}\in \operatorname{Spec}(R)$ (in fact, $\operatorname{depth}(R)$ is equal to the infimum of the right hand side). I believe this to be an even more refined version of your inequality, but I don't see an immediate reduction of it which implies your inequality. – walkar Mar 16 '24 at 18:03
  • @walkar Yes, this is Tag 0FCC. But I don't know how to use this... May be we can take infimum in $\mathfrak p\in V(I)$? – WakeUp-X.Liu Mar 17 '24 at 04:58
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    Oh wow, I'm surprised that it has essentially no hypotheses. I'll keep thinking of how to use it. – walkar Mar 17 '24 at 15:39

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