Let $(A,\mathfrak{m})$ is a Noetherian local ring and $P\neq0$ is a finitely generated projective $A$-module. Then by Auslander-Buchbaum formula, $\operatorname{depth}P=\operatorname{depth} A$. But is this true even if the ring is not local? That is, if $A$ is a Noetherian ring, $I\subset A$ is an ideal, $P$ is a f.g. $A$-module such that $IP\neq P$, then is $$\operatorname{depth}(I,P)=\operatorname{depth}(I,A)$$ true? I can see that $$\operatorname{depth}(I,P)\geq\operatorname{depth}(I,A),$$ but I can't seem to derive the equality.


This is true if $P$ is faithfully flat: let $a\in I$ be $P$-regular. Let $\phi: A\to A$ be the multiplication-by-$a$ map. Then $$\phi\otimes \mathrm{Id}_P: P\to P$$ is the multiplication by $a$ on $P$. Hence $0=\ker(\phi\otimes\mathrm{Id}_P)=\ker(\phi)\otimes_A P$ by flatness. This implies $\ker\phi=0$ by faithful flatness. Hence $a$ is regular in $A$.

Counterexample if $P$ is not faithfully flat. Let $A=k[x,y]$, where $k$ is a field, with the relation $x(x-1)=0$. Let $I=yA$, $P=A/xA$ and $a=(x-1)y\in I$. As $A=A/xA \oplus A/(x-1)A$, $P$ is projective. The multiplication by $a$ on $P$ is the same as the multiplication by $-y$, hence is injective. But $a$ is not a regular element because $xa=0$.

Update according to ashpool's comments. Consider two noetherian rings $A_1, A_2$, and two proper ideals $I_1\subset A_1, I_2\subset A_2$. Let $A=A_1\oplus A_2$, $I=I_1\oplus I_2$ and $P=A_1$. Then $P$ is projective and finitely generated over $A$, $IP\ne P$, and $\mathrm{depth}(I, P)=\mathrm{depth}(I_1, A_1)$. But $$\mathrm{depth}(I, A)=\min\{\mathrm{depth}(I_1, A_1), \mathrm{depth}(I_2, A_2) \}.$$ So to have a real counterexample, it is enough to choose the $A_i$ and $I_i$ with $\mathrm{depth}(I_1, A_1)>\mathrm{depth}(I_2, A_2)$.

  • $\begingroup$ That was a nice example! But does the fact that $I$ has a $P$-regular element that is not $A$-regular automatically imply that $\operatorname{depth}P>\operatorname{depth}A$? $\endgroup$
    – ashpool
    Nov 17 '11 at 19:57
  • 1
    $\begingroup$ You are right @ashpool, I completed the counterexample. $\endgroup$
    – user18119
    Nov 17 '11 at 20:17

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