Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 1 down vote accepted

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 $\mathrm{depth}(I_1, A_1)>\mathrm{depth}(I_2, A_2)$.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.