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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