# $\text{depth }A\leq \text{depth }_{\frak{p}}A + \dim(A/{\frak{p}})$, where $A$ is a Noetherian local ring.

This is Exercise 16.5 in Commutative Algebra, Matsumura or Exercise 17.5 in Commutative Ring Theory by the same author.

Let $$(A,\frak{m})$$ be a Noetherian local ring and $$\frak{p}$$ be a prime ideal. Show that $$\text{depth }A\leq \text{depth }_{\frak{p}}A + \dim(A/\frak{p})$$

If $$\textbf{x}$$ is a maximal $$A$$-regular sequence in $$\frak{p}$$, then it suffices to show that $$\text{depth }A/\textbf{x} \leq \dim A/\frak{p}$$ and according to a theorem in Matsumura's book, we may consider showing that $$\mathfrak{p} \in \text{Ass } (A/\textbf{x})$$. I can see that $$\frak{p}$$ is in the support of $$A/\textbf{x}$$ so if $$\frak{p}$$ is minimal prime ideal above $$\textbf{x}A$$, then it is a minimal prime in the support, hence necessarily an associated prime of $$A/\textbf{x}$$. If this is a right approach, can some please give me a hint about showing that $$\frak{p}$$ is a minimal prime above the ideal $$\textbf{x}A$$?

Thank you very much.

• Why don't follow Matsumura's hint to extend the regular sequence to $\mathfrak m$? Commented May 10 at 6:58
• @user26857 Thank you for the hint! With this, I can show that any associated prime of $A/\textbf{x}A$ contains $\textbf{x}A$, but I don't see why this implies that $\frak{p}$ is contained in an associated prime, maybe we still need $\frak{p}$ be minimal above $\textbf{x}A$?
– Ubik
Commented May 10 at 13:29
• Not every prime containing $x_1,\cdots,x_t$ as a maximal length regular sequence need be minimal over the ideal generated by $x_1,\cdots,x_t$. For example, if $\operatorname{depth} A = 0$ but $\dim A > 0$, then the maximal ideal of $A$ is associated to the zero ideal (the ideal generated by the empty sequence) but is not minimal. Commented May 10 at 15:00
• Of course, the conclusion of the exercise still holds in this setting, since $\operatorname{depth} A = 0 \le 0+\dim A/\mathfrak{p} = \dim A/\mathfrak{p}$. Commented May 10 at 15:04

So I find a proof thanks to the hint given by @user26857. By extending x to a maximal regular sequence in $$\frak{m}$$, it is clear that $$\text{depth }{A} = \text{depth }A/\textbf{x}A + \text{depth }_{\frak{p}} A$$ so it suffices to show that $$\text{depth }A/I \leq \dim A/\frak{p}$$. To see this, notice that $$\frak{p}$$ is contained in the set of zero divisors of $$A/I$$, where the latter one is equal to $$\displaystyle{\bigcup_{\mathfrak{q} \in \text{Ass}(A/I)}}\frak{q}$$, and it can be shown that in this case $$\frak{p} \subset q$$ for some associated prime $$\frak{q}$$ necessarily, which means $$\dim A/\mathfrak{p} \geq \dim A/\mathfrak{q} \geq \text{depth }A/I$$, as desired.