The Stacks project

Lemma 10.72.3. Let $(R, \mathfrak m)$ be a Noetherian local ring. Let $M$ be a nonzero finite $R$-module. Then $\dim (\text{Supp}(M)) \geq \text{depth}(M)$.

Proof. The proof is by induction on $\dim (\text{Supp}(M))$. If $\dim (\text{Supp}(M)) = 0$, then $\text{Supp}(M) = \{ \mathfrak m\} $, whence $\text{Ass}(M) = \{ \mathfrak m\} $ (by Lemmas 10.63.2 and 10.63.7), and hence the depth of $M$ is zero for example by Lemma 10.63.18. For the induction step we assume $\dim (\text{Supp}(M)) > 0$. Let $f_1, \ldots , f_ d$ be a sequence of elements of $\mathfrak m$ such that $f_ i$ is a nonzerodivisor on $M/(f_1, \ldots , f_{i - 1})M$. According to Lemma 10.72.2 it suffices to prove $\dim (\text{Supp}(M)) \geq d$. We may assume $d > 0$ otherwise the lemma holds. By Lemma 10.63.10 we have $\dim (\text{Supp}(M/f_1M)) = \dim (\text{Supp}(M)) - 1$. By induction we conclude $\dim (\text{Supp}(M/f_1M)) \geq d - 1$ as desired. $\square$

Comments (8)

Comment #1665 by on

Suggested slogan: Lemma 10.71.3 states that if R is a Noetherian local ring and M is a nonzero finite R-module, then the dimension of the support of M is greater than or equal to the depth of M.

Comment #1670 by on

Slogans are not just supposed to restate the lemma...

Comment #2210 by David Savitt on

I take the proof to be an induction on dim(Supp(M)), in which case I think the base case needs an argument ([Tag 00LD], say). [It can't be an induction on depth because depth isn't "known" to be finite until this item.]

Comment #2220 by on

OK, yes, this proof was a bit too terse. Thanks! Fixed here.

Comment #2827 by Dario Weißmann on

The last inequality in proof should read: .

Comment #2828 by Dario Weißmann on

And I think we should cite lemma 10.62.10 as . I found the inequality somewhat misleading.

Comment #2829 by Dario Weißmann on

And I think we should cite lemma 10.62.10 as . I found the inequality somewhat misleading.

Edit: There is also a typo, should read .

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