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$1. 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] The argument given for $d > 0$ works but gives a slightly weaker result.

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