Lemma 15.91.1. Let $A$ be a ring. Let $f \in A$. Let $K \in D(A)$. The following are equivalent

1. $\mathop{\mathrm{Ext}}\nolimits ^ n_ A(A_ f, K) = 0$ for all $n$,

2. $\mathop{\mathrm{Hom}}\nolimits _{D(A)}(E, K) = 0$ for all $E$ in $D(A_ f)$,

3. $T(K, f) = 0$,

4. for every $p \in \mathbf{Z}$ we have $T(H^ p(K), f) = 0$,

5. for every $p \in \mathbf{Z}$ we have $\mathop{\mathrm{Hom}}\nolimits _ A(A_ f, H^ p(K)) = 0$ and $\mathop{\mathrm{Ext}}\nolimits ^1_ A(A_ f, H^ p(K)) = 0$,

6. $R\mathop{\mathrm{Hom}}\nolimits _ A(A_ f, K) = 0$,

7. the map $\prod _{n \geq 0} K \to \prod _{n \geq 0} K$, $(x_0, x_1, \ldots ) \mapsto (x_0 - fx_1, x_1 - fx_2, \ldots )$ is an isomorphism in $D(A)$, and

Proof. It is clear that (2) implies (1) and that (1) is equivalent to (6). Assume (1). Let $I^\bullet$ be a K-injective complex of $A$-modules representing $K$. Condition (1) signifies that $\mathop{\mathrm{Hom}}\nolimits _ A(A_ f, I^\bullet )$ is acyclic. Let $M^\bullet$ be a complex of $A_ f$-modules representing $E$. Then

$\mathop{\mathrm{Hom}}\nolimits _{D(A)}(E, K) = \mathop{\mathrm{Hom}}\nolimits _{K(A)}(M^\bullet , I^\bullet ) = \mathop{\mathrm{Hom}}\nolimits _{K(A_ f)}(M^\bullet , \mathop{\mathrm{Hom}}\nolimits _ A(A_ f, I^\bullet ))$

by Algebra, Lemma 10.14.4. As $\mathop{\mathrm{Hom}}\nolimits _ A(A_ f, I^\bullet )$ is a K-injective complex of $A_ f$-modules by Lemma 15.56.3 the fact that it is acyclic implies that it is homotopy equivalent to zero (Derived Categories, Lemma 13.31.2). Thus we get (2).

A free resolution of the $A$-module $A_ f$ is given by

$0 \to \bigoplus \nolimits _{n \in \mathbf{N}} A \to \bigoplus \nolimits _{n \in \mathbf{N}} A \to A_ f \to 0$

where the first map sends the $(a_0, a_1, a_2, \ldots )$ to $(a_0, a_1 - fa_0, a_2 - fa_1, \ldots )$ and the second map sends $(a_0, a_1, a_2, \ldots )$ to $a_0 + a_1/f + a_2/f^2 + \ldots$. Applying $\mathop{\mathrm{Hom}}\nolimits _ A(-, I^\bullet )$ we get

$0 \to \mathop{\mathrm{Hom}}\nolimits _ A(A_ f, I^\bullet ) \to \prod I^\bullet \to \prod I^\bullet \to 0$

Since $\prod I^\bullet$ represents $\prod _{n \geq 0} K$ this proves the equivalence of (1) and (7). On the other hand, by construction of derived limits in Derived Categories, Section 13.34 the displayed exact sequence shows the object $T(K, f)$ is a representative of $R\mathop{\mathrm{Hom}}\nolimits _ A(A_ f, K)$ in $D(A)$. Thus the equivalence of (1) and (3).

There is a spectral sequence

$E_2^{p, q} = \mathop{\mathrm{Ext}}\nolimits ^ p_ A(A_ f, H^ q(K)) \Rightarrow \mathop{\mathrm{Ext}}\nolimits ^{p + q}_ A(A_ f, K)$

See Equation (15.67.0.1). This spectral sequence degenerates at $E_2$ because $A_ f$ has a length $1$ resolution by projective $A$-modules (see above) hence the $E_2$-page has only 2 nonzero columns. Thus we obtain short exact sequences

$0 \to \mathop{\mathrm{Ext}}\nolimits ^1_ A(A_ f, H^{p - 1}(K)) \to \mathop{\mathrm{Ext}}\nolimits ^ p_ A(A_ f, K) \to \mathop{\mathrm{Hom}}\nolimits _ A(A_ f, H^ p(K)) \to 0$

This proves (4) and (5) are equivalent to (1). $\square$

Comment #3060 by Noah Olander on

I think the resolution of A_f is not quite right. I think the first map should be changed to (a_1, a_2,a_3,...) maps to (-a_1, fa_1 - a_2, fa_2 - a_3, ...).

Also a possibility for (7) "Add more here" is that the map from the infinite product of K to itself given by (x_0,x_1, ...) maps to (x_0 - fx_1, x_1 - fx_2, ...) be a quasi-isomorphism.

Comment #6496 by on

In #6491 Daichi Takeuchi suggested swapping the indices in the spectral sequence. We have changed them here and added a reference to Section 15.67.

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