Lemma 51.9.1. Let $A$ be a Noetherian ring. Let $I \subset A$ be an ideal. Let $M$ be a finite $A$-module. Let $\mathfrak p \in V(I)$ be a prime ideal. Assume $e = \text{depth}_{IA_\mathfrak p}(M_\mathfrak p) < \infty $. Then there exists a nonempty open $U \subset V(\mathfrak p)$ such that $\text{depth}_{IA_\mathfrak q}(M_\mathfrak q) \geq e$ for all $\mathfrak q \in U$.

## 51.9 Depth and dimension

Some helper lemmas.

**Proof.**
By definition of depth we have $IM_\mathfrak p \not= M_\mathfrak p$ and there exists an $M_\mathfrak p$-regular sequence $f_1, \ldots , f_ e \in IA_\mathfrak p$. After replacing $A$ by a principal localization we may assume $f_1, \ldots , f_ e \in I$ form an $M$-regular sequence, see Algebra, Lemma 10.68.6. Consider the module $M' = M/IM$. Since $\mathfrak p \in \text{Supp}(M')$ and since the support of a finite module is closed, we find $V(\mathfrak p) \subset \text{Supp}(M')$. Thus for $\mathfrak q \in V(\mathfrak p)$ we get $IM_\mathfrak q \not= M_\mathfrak q$. Hence, using that localization is exact, we see that $\text{depth}_{IA_\mathfrak q}(M_\mathfrak q) \geq e$ for any $\mathfrak q \in V(I)$ by definition of depth.
$\square$

Lemma 51.9.2. Let $A$ be a Noetherian ring. Let $M$ be a finite $A$-module. Let $\mathfrak p$ be a prime ideal. Assume $e = \text{depth}_{A_\mathfrak p}(M_\mathfrak p) < \infty $. Then there exists a nonempty open $U \subset V(\mathfrak p)$ such that $\text{depth}_{A_\mathfrak q}(M_\mathfrak q) \geq e$ for all $\mathfrak q \in U$ and for all but finitely many $\mathfrak q \in U$ we have $\text{depth}_{A_\mathfrak q}(M_\mathfrak q) > e$.

**Proof.**
By definition of depth we have $\mathfrak p M_\mathfrak p \not= M_\mathfrak p$ and there exists an $M_\mathfrak p$-regular sequence $f_1, \ldots , f_ e \in \mathfrak p A_\mathfrak p$. After replacing $A$ by a principal localization we may assume $f_1, \ldots , f_ e \in \mathfrak p$ form an $M$-regular sequence, see Algebra, Lemma 10.68.6. Consider the module $M' = M/(f_1, \ldots , f_ e)M$. Since $\mathfrak p \in \text{Supp}(M')$ and since the support of a finite module is closed, we find $V(\mathfrak p) \subset \text{Supp}(M')$. Thus for $\mathfrak q \in V(\mathfrak p)$ we get $\mathfrak q M_\mathfrak q \not= M_\mathfrak q$. Hence, using that localization is exact, we see that $\text{depth}_{A_\mathfrak q}(M_\mathfrak q) \geq e$ for any $\mathfrak q \in V(I)$ by definition of depth. Moreover, as soon as $\mathfrak q$ is not an associated prime of the module $M'$, then the depth goes up. Thus we see that the final statement holds by Algebra, Lemma 10.63.5.
$\square$

Lemma 51.9.3. Let $X$ be a Noetherian scheme with dualizing complex $\omega _ X^\bullet $. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. Let $k \geq 0$ be an integer. Assume $\mathcal{F}$ is $(S_ k)$. Then there is a finite number of points $x \in X$ such that

**Proof.**
We will prove this lemma by induction on $k$. The base case $k = 0$ says that $\mathcal{F}$ has a finite number of embedded associated points, which follows from Divisors, Lemma 31.2.5.

Assume $k > 0$ and the result holds for all smaller $k$. We can cover $X$ by finitely many affine opens, hence we may assume $X = \mathop{\mathrm{Spec}}(A)$ is affine. Then $\mathcal{F}$ is the coherent $\mathcal{O}_ X$-module associated to a finite $A$-module $M$ which satisfies $(S_ k)$. We will use Algebra, Lemmas 10.63.10 and 10.72.7 without further mention.

Let $f \in A$ be a nonzerodivisor on $M$. Then $M/fM$ has $(S_{k - 1})$. By induction we see that there are finitely many primes $\mathfrak p \in V(f)$ with $\text{depth}((M/fM)_\mathfrak p) = k - 1$ and $\dim (\text{Supp}((M/fM)_\mathfrak p)) > k - 1$. These are exactly the primes $\mathfrak p \in V(f)$ with $\text{depth}(M_\mathfrak p) = k$ and $\dim (\text{Supp}(M_\mathfrak p)) > k$. Thus we may replace $A$ by $A_ f$ and $M$ by $M_ f$ in trying to prove the finiteness statement.

Since $M$ satisfies $(S_ k)$ and $k > 0$ we see that $M$ has no embedded associated primes (Algebra, Lemma 10.157.2). Thus $\text{Ass}(M)$ is the set of generic points of the support of $M$. Thus Dualizing Complexes, Lemma 47.20.4 shows the set $U = \{ \mathfrak q \mid M_\mathfrak q\text{ is Cohen-Macaulay}\} $ is an open containing $\text{Ass}(M)$. By prime avoidance (Algebra, Lemma 10.15.2) we can pick $f \in A$ with $f \not\in \mathfrak p$ for $\mathfrak p \in \text{Ass}(M)$ such that $D(f) \subset U$. Then $f$ is a nonzerodivisor on $M$ (Algebra, Lemma 10.63.9). After replacing $A$ by $A_ f$ and $M$ by $M_ f$ (see above) we find that $M$ is Cohen-Macaulay. Thus for all $\mathfrak q \subset A$ we have $\dim (M_\mathfrak q) = \text{depth}(M_\mathfrak q)$ and hence the set described in the lemma is empty and a fortiori finite. $\square$

Lemma 51.9.4. Let $(A, \mathfrak m)$ be a Noetherian local ring with normalized dualizing complex $\omega _ A^\bullet $. Let $M$ be a finite $A$-module. Set $E^ i = \text{Ext}_ A^{-i}(M, \omega _ A^\bullet )$. Then

$E^ i$ is a finite $A$-module nonzero only for $0 \leq i \leq \dim (\text{Supp}(M))$,

$\dim (\text{Supp}(E^ i)) \leq i$,

$\text{depth}(M)$ is the smallest integer $\delta \geq 0$ such that $E^\delta \not= 0$,

$\mathfrak p \in \text{Supp}(E^0 \oplus \ldots \oplus E^ i) \Leftrightarrow \text{depth}_{A_\mathfrak p}(M_\mathfrak p) + \dim (A/\mathfrak p) \leq i$,

the annihilator of $E^ i$ is equal to the annihilator of $H^ i_\mathfrak m(M)$.

**Proof.**
Parts (1), (2), and (3) are copies of the statements in Dualizing Complexes, Lemma 47.16.5. For a prime $\mathfrak p$ of $A$ we have that $(\omega _ A^\bullet )_\mathfrak p[-\dim (A/\mathfrak p)]$ is a normalized dualzing complex for $A_\mathfrak p$. See Dualizing Complexes, Lemma 47.17.3. Thus

is zero for $i - \dim (A/\mathfrak p) < \text{depth}_{A_\mathfrak p}(M_\mathfrak p)$ and nonzero for $i = \dim (A/\mathfrak p) + \text{depth}_{A_\mathfrak p}(M_\mathfrak p)$ by part (3) over $A_\mathfrak p$. This proves part (4). If $E$ is an injective hull of the residue field of $A$, then we have

by the local duality theorem (in the form of Dualizing Complexes, Lemma 47.18.4). Since $A \to A^\wedge $ is faithfully flat, we find (5) is true by Matlis duality (Dualizing Complexes, Proposition 47.7.8). $\square$

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