## 51.22 A bit of uniformity, III

In this section we fix a Noetherian ring $A$ and an ideal $I \subset A$. Our goal is to prove Lemma 51.22.7 which we will use in a later chapter to solve a lifting problem, see Algebraization of Formal Spaces, Lemma 88.5.3.

Throughout this section we denote

$p : X \to \mathop{\mathrm{Spec}}(A)$

the blowing up of $\mathop{\mathrm{Spec}}(A)$ in the ideal $I$. In other words, $X$ is the $\text{Proj}$ of the Rees algebra $\bigoplus _{n \geq 0} I^ n$. We also consider the fibre product

$\xymatrix{ Y \ar[r] \ar[d] & X \ar[d]^ p \\ \mathop{\mathrm{Spec}}(A/I) \ar[r] & \mathop{\mathrm{Spec}}(A) }$

Then $Y$ is the exceptional divisor of the blowup and hence an effective Cartier divisor on $X$ such that $\mathcal{O}_ X(-1) = \mathcal{O}_ X(Y)$. Since taking $\text{Proj}$ commutes with base change we have

$Y = \text{Proj}(\bigoplus \nolimits _{n \geq 0} I^ n/I^{n + 1}) = \text{Proj}(S)$

where $S = \text{Gr}_ I(A) = \bigoplus _{n \geq 0} I^ n/I^{n + 1}$.

We denote $d = d(S) = d(\text{Gr}_ I(A)) = d(\bigoplus _{n \geq 0} I^ n/I^{n + 1})$ the maximum of the dimensions of the fibres of $p$ (and we set it equal to $0$ if $X = \emptyset$). This is well defined. In fact, we have

1. $d \leq t - 1$ if $I = (a_1, \ldots , a_ t)$ since then $X \subset \mathbf{P}^{t - 1}_ A$, and

2. $d$ is also the maximal dimension of the fibres of $\text{Proj}(S) \to \mathop{\mathrm{Spec}}(S_0)$ provided that $Y$ is nonempty and $d = 0$ if $Y = \emptyset$ (equivalently $S = 0$, equivalently $I = A$).

Hence $d$ only depends on the isomorphism class of $S = \text{Gr}_ I(A)$. Observe that $H^ i(X, \mathcal{F}) = 0$ for every coherent $\mathcal{O}_ X$-module $\mathcal{F}$ and $i > d$ by Cohomology of Schemes, Lemmas 30.20.9 and 30.4.6. Of course the same is true for coherent modules on $Y$.

We denote $d = d(S) = d(\text{Gr}_ I(A)) = d(\bigoplus _{n \geq 0} I^ n/I^{n + 1})$ the integer defined as follows. Note that the algebra $S = \bigoplus _{n \geq 0} I^ n/I^{n + 1}$ is a Noetherian graded ring generated in degree $1$ over degree $0$. Hence by Cohomology of Schemes, Lemmas 30.14.2 and 30.14.3 we can define $q(S)$ as the smallest integer $q(S) \geq 0$ such that for all $q \geq q(S)$ we have $H^ i(Y, \mathcal{O}_ Y(q)) = 0$ for $1 \leq i \leq d$ and $H^0(Y, \mathcal{O}_ Y(q)) = I^ q/I^{q + 1}$. (If $S = 0$, then $q(S) = 0$.)

For $n \geq 1$ we may consider the effective Cartier divisor $nY$ which we will denote $Y_ n$.

Lemma 51.22.1. With $q_0 = q(S)$ and $d = d(S)$ as above, we have

1. for $n \geq 1$, $q \geq q_0$, and $i > 0$ we have $H^ i(X, \mathcal{O}_{Y_ n}(q)) = 0$,

2. for $n \geq 1$ and $q \geq q_0$ we have $H^0(X, \mathcal{O}_{Y_ n}(q)) = I^ q/I^{q + n}$,

3. for $q \geq q_0$ and $i > 0$ we have $H^ i(X, \mathcal{O}_ X(q)) = 0$,

4. for $q \geq q_0$ we have $H^0(X, \mathcal{O}_ X(q)) = I^ q$.

Proof. If $I = A$, then $X$ is affine and the statements are trivial. Hence we may and do assume $I \not= A$. Thus $Y$ and $X$ are nonempty schemes.

Let us prove (1) and (2) by induction on $n$. The base case $n = 1$ is our definition of $q_0$ as $Y_1 = Y$. Recall that $\mathcal{O}_ X(1) = \mathcal{O}_ X(-Y)$. Hence we have a short exact sequence

$0 \to \mathcal{O}_{Y_ n}(1) \to \mathcal{O}_{Y_{n + 1}} \to \mathcal{O}_ Y \to 0$

Hence for $i > 0$ we find

$H^ i(X, \mathcal{O}_{Y_ n}(q + 1)) \to H^ i(X, \mathcal{O}_{Y_{n + 1}}(q)) \to H^ i(X, \mathcal{O}_{Y}(q))$

and we obtain the desired vanishing of the middle term from the given vanishing of the outer terms. For $i = 0$ we obtain a commutative diagram

$\xymatrix{ 0 \ar[r] & I^{q + 1}/I^{q + 1 + n} \ar[d] \ar[r] & I^ q/I^{q + 1 + n} \ar[d] \ar[r] & I^ q/I^{q + 1} \ar[d] \ar[r] & 0 \\ 0 \ar[r] & H^0(X, \mathcal{O}_{Y_ n}(q + 1)) \ar[r] & H^0(X, \mathcal{O}_{Y_{n + 1}}(q)) \ar[r] & H^0(Y, \mathcal{O}_ Y(q)) \ar[r] & 0 }$

with exact rows for $q \geq q_0$ (for the bottom row observe that the next term in the long exact cohomology sequence vanishes for $q \geq q_0$). Since $q \geq q_0$ the left and right vertical arrows are isomorphisms and we conclude the middle one is too.

We omit the proofs of (3) and (4) which are similar. In fact, one can deduce (3) and (4) from (1) and (2) using the theorem on formal functors (but this would be overkill). $\square$

Let us introduce a notation: given $n \geq c \geq 0$ an $(A, n, c)$-module is a finite $A$-module $M$ which is annihilated by $I^ n$ and which as an $A/I^ n$-module is $I^ c/I^ n$-projective, see More on Algebra, Section 15.70.

We will use the following abuse of notation: given an $A$-module $M$ we denote $p^*M$ the quasi-coherent module gotten by pulling back by $p$ the quasi-coherent module $\widetilde{M}$ on $\mathop{\mathrm{Spec}}(A)$ associated to $M$. For example we have $\mathcal{O}_{Y_ n} = p^*(A/I^ n)$. For a short exact sequence $0 \to K \to L \to M \to 0$ of $A$-modules we obtain an exact sequence

$p^*K \to p^*L \to p^*M \to 0$

as $\widetilde{\ }$ is an exact functor and $p^*$ is a right exact functor.

Lemma 51.22.2. Let $0 \to K \to L \to M \to 0$ be a short exact sequence of $A$-modules such that $K$ and $L$ are annihilated by $I^ n$ and $M$ is an $(A, n, c)$-module. Then the kernel of $p^*K \to p^*L$ is scheme theoretically supported on $Y_ c$.

Proof. Let $\mathop{\mathrm{Spec}}(B) \subset X$ be an affine open. The restriction of the exact sequence over $\mathop{\mathrm{Spec}}(B)$ corresponds to the sequence of $B$-modules

$K \otimes _ A B \to L \otimes _ A B \to M \otimes _ A B \to 0$

which is isomorphismic to the sequence

$K \otimes _{A/I^ n} B/I^ nB \to L \otimes _{A/I^ n} B/I^ nB \to M \otimes _{A/I^ n} B/I^ nB \to 0$

Hence the kernel of the first map is the image of the module $\text{Tor}_1^{A/I^ n}(M, B/I^ nB)$. Recall that the exceptional divisor $Y$ is cut out by $I\mathcal{O}_ X$. Hence it suffices to show that $\text{Tor}_1^{A/I^ n}(M, B/I^ nB)$ is annihilated by $I^ c$. Since multiplication by $a \in I^ c$ on $M$ factors through a finite free $A/I^ n$-module, this is clear. $\square$

We have the canonical map $\mathcal{O}_ X \to \mathcal{O}_ X(1)$ which vanishes exactly along $Y$. Hence for every coherent $\mathcal{O}_ X$-module $\mathcal{F}$ we always have canonical maps $\mathcal{F}(q) \to \mathcal{F}(q + n)$ for any $q \in \mathbf{Z}$ and $n \geq 0$.

Lemma 51.22.3. Let $\mathcal{F}$ be a coherent $\mathcal{O}_ X$-module. Then $\mathcal{F}$ is scheme theoretically supported on $Y_ c$ if and only if the canonical map $\mathcal{F} \to \mathcal{F}(c)$ is zero.

Proof. This is true because $\mathcal{O}_ X \to \mathcal{O}_ X(1)$ vanishes exactly along $Y$. $\square$

Lemma 51.22.4. With $q_0 = q(S)$ and $d = d(S)$ as above, suppose we have integers $n \geq c \geq 0$, an $(A, n, c)$-module $M$, an index $i \in \{ 0, 1, \ldots , d\}$, and an integer $q$. Then we distinguish the following cases

1. In the case $i = d \geq 1$ and $q \geq q_0$ we have $H^ d(X, p^*M(q)) = 0$.

2. In the case $i = d - 1 \geq 1$ and $q \geq q_0$ we have $H^{d - 1}(X, p^*M(q)) = 0$.

3. In the case $d - 1 > i > 0$ and $q \geq q_0 + (d - 1 - i)c$ the map $H^ i(X, p^*M(q)) \to H^ i(X, p^*M(q - (d - 1 - i)c))$ is zero.

4. In the case $i = 0$, $d \in \{ 0, 1\}$, and $q \geq q_0$, there is a surjection

$I^ qM \longrightarrow H^0(X, p^*M(q))$
5. In the case $i = 0$, $d > 1$, and $q \geq q_0 + (d - 1)c$ the map

$H^0(X, p^*M(q)) \to H^0(X, p^*M(q - (d - 1)c))$

has image contained in the image of the canonical map $I^{q - (d - 1)c}M \to H^0(X, p^*M(q - (d - 1)c))$.

Proof. Let $M$ be an $(A, n, c)$-module. Choose a short exact sequence

$0 \to K \to (A/I^ n)^{\oplus r} \to M \to 0$

We will use below that $K$ is an $(A, n, c)$-module, see More on Algebra, Lemma 15.70.6. Consider the corresponding exact sequence

$p^*K \to (\mathcal{O}_{Y_ n})^{\oplus r} \to p^*M \to 0$

We split this into short exact sequences

$0 \to \mathcal{F} \to p^*K \to \mathcal{G} \to 0 \quad \text{and}\quad 0 \to \mathcal{G} \to (\mathcal{O}_{Y_ n})^{\oplus r} \to p^*M \to 0$

By Lemma 51.22.2 the coherent module $\mathcal{F}$ is scheme theoretically supported on $Y_ c$.

Proof of (1). Assume $d > 0$. We have to prove $H^ d(X, p^*M(q)) = 0$ for $q \geq q_0$. By the vanishing of the cohomology of twists of $\mathcal{G}$ in degrees $> d$ and the long exact cohomology sequence associated to the second short exact sequence above, it suffices to prove that $H^ d(X, \mathcal{O}_{Y_ n}(q)) = 0$. This is true by Lemma 51.22.1.

Proof of (2). Assume $d > 1$. We have to prove $H^{d - 1}(X, p^*M(q)) = 0$ for $q \geq q_0$. Arguing as in the previous paragraph, we see that it suffices to show that $H^ d(X, \mathcal{G}(q)) = 0$. Using the first short exact sequence and the vanishing of the cohomology of twists of $\mathcal{F}$ in degrees $> d$ we see that it suffices to show $H^ d(X, p^*K(q))$ is zero which is true by (1) and the fact that $K$ is an $(A, n, c)$-module (see above).

Proof of (3). Let $0 < i < d - 1$ and assume the statement holds for $i + 1$ except in the case $i = d - 2$ we have statement (2). Using the long exact sequence of cohomology associated to the second short exact sequence above we find an injection

$H^ i(X, p^*M(q - (d - 1 - i)c)) \subset H^{i + 1}(X, \mathcal{G}(q - (d - 1 - i)c))$

as $q - (d - 1 - i)c \geq q_0$ gives the vanishing of $H^ i(X, \mathcal{O}_{Y_ n}(q - (d - 1 - i)c))$ (see above). Thus it suffices to show that the map $H^{i + 1}(X, \mathcal{G}(q)) \to H^{i + 1}(X, \mathcal{G}(q - (d - 1 - i)c))$ is zero. To study this, we consider the maps of exact sequences

$\xymatrix{ H^{i + 1}(X, p^*K(q)) \ar[r] \ar[d] & H^{i + 1}(X, \mathcal{G}(q)) \ar[r] \ar[d] \ar@{..>}[ld] & H^{i + 2}(X, \mathcal{F}(q)) \ar[d] \\ H^{i + 1}(X, p^*K(q - c)) \ar[r] \ar[d] & H^{i + 1}(X, \mathcal{G}(q - c)) \ar[r] \ar[d] & H^{i + 2}(X, \mathcal{F}(q - c)) \\ H^{i + 1}(X, p^*K(q - (d - 1 - i)c)) \ar[r] & H^{i + 1}(X, \mathcal{G}(q - (d - 1 - i)c)) }$

Since $\mathcal{F}$ is scheme theoretically supported on $Y_ c$ we see that the canonical map $\mathcal{G}(q) \to \mathcal{G}(q - c)$ factors through $p^*K(q - c)$ by Lemma 51.22.3. This gives the dotted arrow in the diagram. (In fact, for the proof it suffices to observe that the vertical arrow on the extreme right is zero in order to get the dotted arrow as a map of sets.) Thus it suffices to show that $H^{i + 1}(X, p^*K(q - c)) \to H^{i + 1}(X, p^*K(q - (d - 1 - i)c))$ is zero. If $i = d - 2$, then the source of this arrow is zero by (2) as $q - c \geq q_0$ and $K$ is an $(A, n, c)$-module. If $i < d - 2$, then as $K$ is an $(A, n, c)$-module, we get from the induction hypothesis that the map is indeed zero since $q - c - (q - (d - 1 - i)c) = (d - 2 - i)c = (d - 1 - (i + 1))c$ and since $q - c \geq q_0 + (d - 1 - (i + 1))c$. In this way we conclude the proof of (3).

Proof of (4). Assume $d \in \{ 0, 1\}$ and $q \geq q_0$. Then the first short exact sequence gives a surjection $H^1(X, p^*K(q)) \to H^1(X, \mathcal{G}(q))$ and the source of this arrow is zero by case (1). Hence for all $q \in \mathbf{Z}$ we see that the map

$H^0(X, (\mathcal{O}_{Y_ n})^{\oplus r}(q)) \longrightarrow H^0(X, p^*M(q))$

is surjective. For $q \geq q_0$ the source is equal to $(I^ q/I^{q + n})^{\oplus r}$ by Lemma 51.22.1 and this easily proves the statement.

Proof of (5). Assume $d > 1$. Arguing as in the proof of (4) we see that it suffices to show that the image of

$H^0(X, p^*M(q)) \longrightarrow H^0(X, p^*M(q - (d - 1)c))$

is contained in the image of

$H^0(X, (\mathcal{O}_{Y_ n})^{\oplus r}(q - (d - 1)c)) \longrightarrow H^0(X, p^*M(q - (d - 1)c))$

To show the inclusion above, it suffices to show that for $\sigma \in H^0(X, p^*M(q))$ with boundary $\xi \in H^1(X, \mathcal{G}(q))$ the image of $\xi$ in $H^1(X, \mathcal{G}(q - (d - 1)c))$ is zero. This follows by the exact same arguments as in the proof of (3). $\square$

Remark 51.22.5. Given a pair $(M, n)$ consisting of an integer $n \geq 0$ and a finite $A/I^ n$-module $M$ we set $M^\vee = \mathop{\mathrm{Hom}}\nolimits _{A/I^ n}(M, A/I^ n)$. Given a pair $(\mathcal{F}, n)$ consisting of an integer $n$ and a coherent $\mathcal{O}_{Y_ n}$-module $\mathcal{F}$ we set

$\mathcal{F}^\vee = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_{Y_ n}}(\mathcal{F}, \mathcal{O}_{Y_ n})$

Given $(M, n)$ as above, there is a canonical map

$can : p^*(M^\vee ) \longrightarrow (p^*M)^\vee$

Namely, if we choose a presentation $(A/I^ n)^{\oplus s} \to (A/I^ n)^{\oplus r} \to M \to 0$ then we obtain a presentation $\mathcal{O}_{Y_ n}^{\oplus s} \to \mathcal{O}_{Y_ n}^{\oplus r} \to p^*M \to 0$. Taking duals we obtain exact sequences

$0 \to M^\vee \to (A/I^ n)^{\oplus r} \to (A/I^ n)^{\oplus s}$

and

$0 \to (p^*M)^\vee \to \mathcal{O}_{Y_ n}^{\oplus r} \to \mathcal{O}_{Y_ n}^{\oplus s}$

Pulling back the first sequence by $p$ we find the desired map $can$. The construction of this map is functorial in the finite $A/I^ n$-module $M$. The kernel and cokernel of $can$ are scheme theoretically supported on $Y_ c$ if $M$ is an $(A, n, c)$-module. Namely, in that case for $a \in I^ c$ the map $a : M \to M$ factors through a finite free $A/I^ n$-module for which $can$ is an isomorphism. Hence $a$ annihilates the kernel and cokernel of $can$.

Lemma 51.22.6. With $q_0 = q(S)$ and $d = d(S)$ as above, let $M$ be an $(A, n, c)$-module and let $\varphi : M \to I^ n/I^{2n}$ be an $A$-linear map. Assume $n \geq \max (q_0 + (1 + d)c, (2 + d)c)$ and if $d = 0$ assume $n \geq q_0 + 2c$. Then the composition

$M \xrightarrow {\varphi } I^ n/I^{2n} \to I^{n - (1 + d)c}/I^{2n - (1 + d)c}$

is of the form $\sum a_ i \psi _ i$ with $a_ i \in I^ c$ and $\psi _ i : M \to I^{n - (2 + d)c}/I^{2n - (2 + d)c}$.

Proof. The case $d > 1$. Since we have a compatible system of maps $p^*(I^ q) \to \mathcal{O}_ X(q)$ for $q \geq 0$ there are canonical maps $p^*(I^ q/I^{q + \nu }) \to \mathcal{O}_{Y_\nu }(q)$ for $\nu \geq 0$. Using this and pulling back $\varphi$ we obtain a map

$\chi : p^*M \longrightarrow \mathcal{O}_{Y_ n}(n)$

such that the composition $M \to H^0(X, p^*M) \to H^0(X, \mathcal{O}_{Y_ n}(n))$ is the given homomorphism $\varphi$ combined with the map $I^ n/I^{2n} \to H^0(X, \mathcal{O}_{Y_ n}(n))$. Since $\mathcal{O}_{Y_ n}(n)$ is invertible on $Y_ n$ the linear map $\chi$ determines a section

$\sigma \in \Gamma (X, (p^*M)^\vee (n))$

with notation as in Remark 51.22.5. The discussion in Remark 51.22.5 shows the cokernel and kernel of $can : p^*(M^\vee ) \to (p^*M)^\vee$ are scheme theoretically supported on $Y_ c$. By Lemma 51.22.3 the map $(p^*M)^\vee (n) \to (p^*M)^\vee (n - 2c)$ factors through $p^*(M^\vee )(n - 2c)$; small detail omitted. Hence the image of $\sigma$ in $\Gamma (X, (p^*M)^\vee (n - 2c))$ comes from an element

$\sigma ' \in \Gamma (X, p^*(M^\vee )(n - 2c))$

By Lemma 51.22.4 part (5), the fact that $M^\vee$ is an $(A, n, c)$-module by More on Algebra, Lemma 15.70.7, and the fact that $n \geq q_0 + (1 + d)c$ so $n - 2c \geq q_0 + (d - 1)c$ we see that the image of $\sigma '$ in $H^0(X, p^*M^\vee (n - (1 + d)c))$ is the image of an element $\tau$ in $I^{n - (1 + d)c}M^\vee$. Write $\tau = \sum a_ i \tau _ i$ with $\tau _ i \in I^{n - (2 + d)c}M^\vee$; this makes sense as $n - (2 + d)c \geq 0$. Then $\tau _ i$ determines a homomorphism of modules $\psi _ i : M \to I^{n - (2 + d)c}/I^{2n - (2 + d)c}$ using the evaluation map $M \otimes M^\vee \to A/I^ n$.

Let us prove that this works1. Pick $z \in M$ and let us show that $\varphi (z)$ and $\sum a_ i \psi _ i(z)$ have the same image in $I^{n - (1 + d)c}/I^{2n - (1 + d)c}$. First, the element $z$ determines a map $p^*z : \mathcal{O}_{Y_ n} \to p^*M$ whose composition with $\chi$ is equal to the map $\mathcal{O}_{Y_ n} \to \mathcal{O}_{Y_ n}(n)$ corresponding to $\varphi (z)$ via the map $I^ n/I^{2n} \to \Gamma (\mathcal{O}_{Y_ n}(n))$. Next $z$ and $p^*z$ determine evaluation maps $e_ z : M^\vee \to A/I^ n$ and $e_{p^*z} : (p^*M)^\vee \to \mathcal{O}_{Y_ n}$. Since $\chi (p^*z)$ is the section corresponding to $\varphi (z)$ we see that $e_{p^*z}(\sigma )$ is the section corresponding to $\varphi (z)$. Here and below we abuse notation: for a map $a : \mathcal{F} \to \mathcal{G}$ of modules on $X$ we also denote $a : \mathcal{F}(t) \to \mathcal{F}(t)$ the corresponding map of twisted modules. The diagram

$\xymatrix{ p^*(M^\vee ) \ar[d]_{can} \ar[r]_{p^*e_ z} & \mathcal{O}_{Y_ n} \ar@{=}[d] \\ (p^*M)^\vee \ar[r]^{e_{p^*z}} & \mathcal{O}_{Y_ n} }$

commutes by functoriality of the construction $can$. Hence $(p^*e_ z)(\sigma ')$ in $\Gamma (Y_ n, \mathcal{O}_{Y_ n}(n - 2c))$ is the section corresponding to the image of $\varphi (z)$ in $I^{n - 2c}/I^{2n - 2c}$. The next step is that $\sigma '$ maps to the image of $\sum a_ i \tau _ i$ in $H^0(X, p^*M^\vee (n - (1 + d)c))$. This implies that $(p^*e_ z)(\sum a_ i \tau _ i) = \sum a_ i p^*e_ z(\tau _ i)$ in $\Gamma (Y_ n, \mathcal{O}_{Y_ n}(n - (1 + d)c))$ is the section corresponding to the image of $\varphi (z)$ in $I^{n - (1 + d)c}/I^{2n - (1 + d)c}$. Recall that $\psi _ i$ is defined from $\tau _ i$ using an evaluation map. Hence if we denote

$\chi _ i : p^*M \longrightarrow \mathcal{O}_{Y_ n}(n - (2 + d)c)$

the map we get from $\psi _ i$, then we see by the same reasoning as above that the section corresponding to $\psi _ i(z)$ is $\chi _ i(p^*z) = e_{p^*z}(\chi _ i) = p^*e_ z(\tau _ i)$. Hence we conclude that the image of $\varphi (z)$ in $\Gamma (Y_ n, \mathcal{O}_{Y_ n}(n - (1 + d)c))$ is equal to the image of $\sum a_ i\psi _ i(z)$. Since $n - (1 + d)c \geq q_0$ we have $\Gamma (Y_ n, \mathcal{O}_{Y_ n}(n - (1 + d)c)) = I^{n - (1 + d)c}/I^{2n - (1 + d)c}$ by Lemma 51.22.1 and we conclude the desired compatibility is true.

The case $d = 1$. Here we argue as above that we get

$\chi : p^*M \longrightarrow \mathcal{O}_{Y_ n}(n),\quad \sigma \in \Gamma (X, (p^*M)^\vee (n)),\quad \sigma ' \in \Gamma (X, p^*(M^\vee )(n - 2c)),$

and then we use Lemma 51.22.4 part (4) to see that $\sigma '$ is the image of some element $\tau \in I^{n - 2c}M^\vee$. The rest of the argument is the same.

The case $d = 0$. Argument is exactly the same as in the case $d = 1$. $\square$

Lemma 51.22.7. With $d = d(S)$ and $q_0 = q(S)$ as above. Then

1. for integers $n \geq c \geq 0$ with $n \geq \max (q_0 + (1 + d)c, (2 + d)c)$,

2. for $K$ of $D(A/I^ n)$ with $H^ i(K) = 0$ for $i \not= -1, 0$ and $H^ i(K)$ finite for $i = -1, 0$ such that $\mathop{\mathrm{Ext}}\nolimits ^1_{A/I^ c}(K, N)$ is annihilated by $I^ c$ for all finite $A/I^ n$-modules $N$

the map

$\mathop{\mathrm{Ext}}\nolimits ^1_{A/I^ n}(K, I^ n/I^{2n}) \longrightarrow \mathop{\mathrm{Ext}}\nolimits ^1_{A/I^ n}(K, I^{n - (1 + d)c}/I^{2n - 2(1 + d)c})$

is zero.

Proof. The case $d > 0$. Let $K^{-1} \to K^0$ be a complex representing $K$ as in More on Algebra, Lemma 15.84.5 part (5) with respect to the ideal $I^ c/I^ n$ in the ring $A/I^ n$. In particular $K^{-1}$ is $I^ c/I^ n$-projective as multiplication by elements of $I^ c/I^ n$ even factor through $K^0$. By More on Algebra, Lemma 15.84.4 part (1) we have

$\mathop{\mathrm{Ext}}\nolimits ^1_{A/I^ n}(K, I^ n/I^{2n}) = \mathop{\mathrm{Coker}}(\mathop{\mathrm{Hom}}\nolimits _{A/I^ n}(K^0, I^ n/I^{2n}) \to \mathop{\mathrm{Hom}}\nolimits _{A/I^ n}(K^{-1}, I^ n/I^{2n}))$

and similarly for other Ext groups. Hence any class $\xi$ in $\mathop{\mathrm{Ext}}\nolimits ^1_{A/I^ n}(K, I^ n/I^{2n})$ comes from an element $\varphi \in \mathop{\mathrm{Hom}}\nolimits _{A/I^ n}(K^{-1}, I^ n/I^{2n})$. Denote $\varphi '$ the image of $\varphi$ in $\mathop{\mathrm{Hom}}\nolimits _{A/I^ n}(K^{-1}, I^{n - (1 + d)c}/I^{2n - (1 + d)c})$. By Lemma 51.22.6 we can write $\varphi ' = \sum a_ i \psi _ i$ with $a_ i \in I^ c$ and $\psi _ i \in \mathop{\mathrm{Hom}}\nolimits _{A/I^ n}(M, I^{n - (2 + d)c}/I^{2n - (2 + d)c})$. Choose $h_ i : K^0 \to K^{-1}$ such that $a_ i \text{id}_{K^{-1}} = h_ i \circ d_ K^{-1}$. Set $\psi = \sum \psi _ i \circ h_ i : K^0 \to I^{n - (2 + d)c}/I^{2n - (2 + d)c}$. Then $\varphi ' = \psi \circ \text{d}_ K^{-1}$ and we conclude that $\xi$ already maps to zero in $\mathop{\mathrm{Ext}}\nolimits ^1_{A/I^ n}(K, I^{n - (1 + d)c}/I^{2n - (1 + d)c})$ and a fortiori in $\mathop{\mathrm{Ext}}\nolimits ^1_{A/I^ n}(K, I^{n - (1 + d)c}/I^{2n - 2(1 + d)c})$.

The case $d = 0$2. Let $\xi$ and $\varphi$ be as above. We consider the diagram

$\xymatrix{ K^0 \\ K^{-1} \ar[u] \ar[r]^\varphi & I^ n/I^{2n} \ar[r] & I^{n - c}/I^{2n - c} }$

Pulling back to $X$ and using the map $p^*(I^ n/I^{2n}) \to \mathcal{O}_{Y_ n}(n)$ we find a solid diagram

$\xymatrix{ p^*K^0 \ar@{..>}[rrd] \\ p^*K^{-1} \ar[u] \ar[r] & \mathcal{O}_{Y_ n}(n) \ar[r] & \mathcal{O}_{Y_ n}(n - c) }$

We can cover $X$ by affine opens $U = \mathop{\mathrm{Spec}}(B)$ such that there exists an $a \in I$ with the following property: $IB = aB$ and $a$ is a nonzerodivisor on $B$. Namely, we can cover $X$ by spectra of affine blowup algebras, see Divisors, Lemma 31.32.2. The restriction of $\mathcal{O}_{Y_ n}(n) \to \mathcal{O}_{Y_ n}(n - c)$ to $U$ is isomorphic to the map of quasi-coherent $\mathcal{O}_ U$-modules corresponding to the $B$-module map $a^ c : B/a^ nB \to B/a^ nB$. Since $a^ c : K^{-1} \to K^{-1}$ factors through $K^0$ we see that the dotted arrow exists over $U$. In other words, locally on $X$ we can find the dotted arrow! Now the sheaf of dotted arrows fitting into the diagram is principal homogeneous under

$\mathcal{F} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}( \mathop{\mathrm{Coker}}(p^*K^{-1} \to p^*K^0), \mathcal{O}_{Y_ n}(n - c))$

which is a coherent $\mathcal{O}_ X$-module. Hence the obstruction for finding the dotted arrow is an element of $H^1(X, \mathcal{F})$. This cohomology group is zero as $1 > d = 0$, see discussion following the definition of $d = d(S)$. This proves that we can find a dotted arrow $\psi : p^*K^0 \to \mathcal{O}_{Y_ n}(n - c)$ fitting into the diagram. Since $n - c \geq q_0$ we find that $\psi$ induces a map $K^0 \to I^{n - c}/I^{2n - c}$. Chasing the diagram we conclude that $\varphi ' = \psi \circ \text{d}_ K^{-1}$ and the proof is finished as before. $\square$

[1] We hope some reader will suggest a less dirty proof of this fact.
[2] The argument given for $d > 0$ works but gives a slightly weaker result.

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