**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 10.63.7) 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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