The Stacks project

51.13 Annihilators of local chomology, II

We extend the discussion of annihilators of local cohomology in Section 51.10 to bounded below complexes with finite cohomology modules.

Definition 51.13.1. Let $I$ be an ideal of a Noetherian ring $A$. Let $K \in D^+_{\textit{Coh}}(A)$. We define the $I$-depth of $K$, denoted $\text{depth}_ I(K)$, to be the maximal $m \in \mathbf{Z} \cup \{ \infty \} $ such that $H^ i_ I(K) = 0$ for all $i < m$. If $A$ is local with maximal ideal $\mathfrak m$ then we call $\text{depth}_\mathfrak m(K)$ simply the depth of $K$.

This definition does not conflict with Algebra, Definition 10.71.1 by Dualizing Complexes, Lemma 47.11.1.

Proposition 51.13.2. Let $A$ be a Noetherian ring which has a dualizing complex. Let $T \subset T' \subset \mathop{\mathrm{Spec}}(A)$ be subsets stable under specialization. Let $s \in \mathbf{Z}$. Let $K$ be an object of $D_{\textit{Coh}}^+(A)$. The following are equivalent

  1. there exists an ideal $J \subset A$ with $V(J) \subset T'$ such that $J$ annihilates $H^ i_ T(K)$ for $i \leq s$, and

  2. for all $\mathfrak p \not\in T'$, $\mathfrak q \in T$ with $\mathfrak p \subset \mathfrak q$ we have

    \[ \text{depth}_{A_\mathfrak p}(K_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) > s \]

Proof. This lemma is the natural generalization of Proposition 51.10.1 whose proof the reader should read first. Let $\omega _ A^\bullet $ be a dualizing complex. Let $\delta $ be its dimension function, see Dualizing Complexes, Section 47.17. An important role will be played by the finite $A$-modules

\[ E^ i = \mathop{\mathrm{Ext}}\nolimits _ A^ i(K, \omega _ A^\bullet ) \]

For $\mathfrak p \subset A$ we will write $H^ i_\mathfrak p$ to denote the local cohomology of an object of $D(A_\mathfrak p)$ with respect to $\mathfrak pA_\mathfrak p$. Then we see that the $\mathfrak pA_\mathfrak p$-adic completion of

\[ (E^ i)_\mathfrak p = \mathop{\mathrm{Ext}}\nolimits ^{\delta (\mathfrak p) + i}_{A_\mathfrak p}(K_\mathfrak p, (\omega _ A^\bullet )_\mathfrak p[-\delta (\mathfrak p)]) \]

is Matlis dual to

\[ H^{-\delta (\mathfrak p) - i}_{\mathfrak p}(K_\mathfrak p) \]

by Dualizing Complexes, Lemma 47.18.4. In particular we deduce from this the following fact: an ideal $J \subset A$ annihilates $(E^ i)_\mathfrak p$ if and only if $J$ annihilates $H^{-\delta (\mathfrak p) - i}_{\mathfrak p}(K_\mathfrak p)$.

Set $T_ n = \{ \mathfrak p \in T \mid \delta (\mathfrak p) \leq n\} $. As $\delta $ is a bounded function, we see that $T_ a = \emptyset $ for $a \ll 0$ and $T_ b = T$ for $b \gg 0$.

Assume (2). Let us prove the existence of $J$ as in (1). We will use a double induction to do this. For $i \leq s$ consider the induction hypothesis $IH_ i$: $H^ a_ T(K)$ is annihilated by some $J \subset A$ with $V(J) \subset T'$ for $a \leq i$. The case $IH_ i$ is trivial for $i$ small enough because $K$ is bounded below.

Induction step. Assume $IH_{i - 1}$ holds for some $i \leq s$. Pick $J'$ with $V(J') \subset T'$ annihilating $H^ a_ T(K)$ for $a \leq i - 1$ (the induction hypothesis guarantees we can do this). We will show by descending induction on $n$ that there exists an ideal $J$ with $V(J) \subset T'$ such that the associated primes of $J H^ i_ T(K)$ are in $T_ n$. For $n \ll 0$ this implies $JH^ i_ T(K) = 0$ (Algebra, Lemma 10.62.7) and hence $IH_ i$ will hold. The base case $n \gg 0$ is trivial because $T = T_ n$ in this case and all associated primes of $H^ i_ T(K)$ are in $T$.

Thus we assume given $J$ with the property for $n$. Let $\mathfrak q \in T_ n$. Let $T_\mathfrak q \subset \mathop{\mathrm{Spec}}(A_\mathfrak q)$ be the inverse image of $T$. We have $H^ j_ T(K)_\mathfrak q = H^ j_{T_\mathfrak q}(K_\mathfrak q)$ by Lemma 51.5.7. Consider the spectral sequence

\[ H_\mathfrak q^ p(H^ q_{T_\mathfrak q}(K_\mathfrak q)) \Rightarrow H^{p + q}_\mathfrak q(K_\mathfrak q) \]

of Lemma 51.5.8. Below we will find an ideal $J'' \subset A$ with $V(J'') \subset T'$ such that $H^ i_\mathfrak q(K_\mathfrak q)$ is annihilated by $J''$ for all $\mathfrak q \in T_ n \setminus T_{n - 1}$. Claim: $J (J')^ i J''$ will work for $n - 1$. Namely, let $\mathfrak q \in T_ n \setminus T_{n - 1}$. The spectral sequence above defines a filtration

\[ E_\infty ^{0, i} = E_{i + 2}^{0, i} \subset \ldots \subset E_3^{0, i} \subset E_2^{0, i} = H^0_\mathfrak q(H^ i_{T_\mathfrak q}(K_\mathfrak q)) \]

The module $E_\infty ^{0, i}$ is annihilated by $J''$. The subquotients $E_ j^{0, i}/E_{j + 1}^{0, i}$ for $i + 1 \geq j \geq 2$ are annihilated by $J'$ because the target of $d_ j^{0, i}$ is a subquotient of

\[ H^ j_\mathfrak q(H^{i - j + 1}_{T_\mathfrak q}(K_\mathfrak q)) = H^ j_\mathfrak q(H^{i - j + 1}_ T(K)_\mathfrak q) \]

and $H^{i - j + 1}_ T(K)_\mathfrak q$ is annihilated by $J'$ by choice of $J'$. Finally, by our choice of $J$ we have $J H^ i_ T(K)_\mathfrak q \subset H^0_\mathfrak q(H^ i_ T(K)_\mathfrak q)$ since the non-closed points of $\mathop{\mathrm{Spec}}(A_\mathfrak q)$ have higher $\delta $ values. Thus $\mathfrak q$ cannot be an associated prime of $J(J')^ iJ'' H^ i_ T(K)$ as desired.

By our initial remarks we see that $J''$ should annihilate

\[ (E^{-\delta (\mathfrak q) - i})_\mathfrak q = (E^{-n - i})_\mathfrak q \]

for all $\mathfrak q \in T_ n \setminus T_{n - 1}$. But if $J''$ works for one $\mathfrak q$, then it works for all $\mathfrak q$ in an open neighbourhood of $\mathfrak q$ as the modules $E^{-n - i}$ are finite. Since every subset of $\mathop{\mathrm{Spec}}(A)$ is Noetherian with the induced topology (Topology, Lemma 5.9.2), we conclude that it suffices to prove the existence of $J''$ for one $\mathfrak q$.

Since the ext modules are finite the existence of $J''$ is equivalent to

\[ \text{Supp}(E^{-n - i}) \cap \mathop{\mathrm{Spec}}(A_\mathfrak q) \subset T'. \]

This is equivalent to showing the localization of $E^{-n - i}$ at every $\mathfrak p \subset \mathfrak q$, $\mathfrak p \not\in T'$ is zero. Using local duality over $A_\mathfrak p$ we find that we need to prove that

\[ H^{i + n - \delta (\mathfrak p)}_\mathfrak p(K_\mathfrak p) = H^{i - \dim ((A/\mathfrak p)_\mathfrak q)}_\mathfrak p(K_\mathfrak p) \]

is zero (this uses that $\delta $ is a dimension function). This vanishes by the assumption in the lemma and $i \leq s$ and our definition of depth in Definition 51.13.1.

To prove the converse implication we assume (2) does not hold and we work backwards through the arguments above. First, we pick a $\mathfrak q \in T$, $\mathfrak p \subset \mathfrak q$ with $\mathfrak p \not\in T'$ such that

\[ i = \text{depth}_{A_\mathfrak p}(K_\mathfrak p) + \dim ((A/\mathfrak p)_\mathfrak q) \leq s \]

is minimal. Then $H^{i - \dim ((A/\mathfrak p)_\mathfrak q)}_\mathfrak p(K_\mathfrak p)$ is nonzero by the our definition of depth in Definition 51.13.1. Set $n = \delta (\mathfrak q)$. Then there does not exist an ideal $J \subset A$ with $V(J) \subset T'$ such that $J(E^{-n - i})_\mathfrak q = 0$. Thus $H^ i_\mathfrak q(K_\mathfrak q)$ is not annihilated by an ideal $J \subset A$ with $V(J) \subset T'$. By minimality of $i$ it follows from the spectral sequence displayed above that the module $H^ i_ T(K)_\mathfrak q$ is not annihilated by an ideal $J \subset A$ with $V(J) \subset T'$. Thus $H^ i_ T(K)$ is not annihilated by an ideal $J \subset A$ with $V(J) \subset T'$. This finishes the proof of the proposition. $\square$


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