The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

15.86 Some evaluation maps

In this section we prove that certain canonical maps of $R\mathop{\mathrm{Hom}}\nolimits $'s are isomorphisms for suitable types of complexes.

Lemma 15.86.1. Let $R$ be a ring. Let $K, L, M$ be objects of $D(R)$. the map

\[ R\mathop{\mathrm{Hom}}\nolimits _ R(L, M) \otimes _ R^\mathbf {L} K \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _ R(R\mathop{\mathrm{Hom}}\nolimits _ R(K, L), M) \]

of Lemma 15.68.3 is an isomorphism in the following two cases

  1. $K$ perfect, or

  2. $K$ is pseudo-coherent, $L \in D^+(R)$, and $M$ finite injective dimension.

Proof. Choose a K-injective complex $I^\bullet $ representing $M$, a K-injective complex $J^\bullet $ representing $L$, and a bounded above complex of finite projective modules $K^\bullet $ representing $K$. Consider the map of complexes

\[ \text{Tot}(\mathop{\mathrm{Hom}}\nolimits ^\bullet (J^\bullet , I^\bullet ) \otimes _ R K^\bullet ) \longrightarrow \mathop{\mathrm{Hom}}\nolimits ^\bullet (\mathop{\mathrm{Hom}}\nolimits ^\bullet (K^\bullet , J^\bullet ), I^\bullet ) \]

of Lemma 15.67.3. Note that

\[ \left(\prod \nolimits _{p + r = t} \mathop{\mathrm{Hom}}\nolimits _ R(J^{-r}, I^ p)\right) \otimes _ R K^ s = \prod \nolimits _{p + r = t} \mathop{\mathrm{Hom}}\nolimits _ R(J^{-r}, I^ p) \otimes _ R K^ s \]

because $K^ s$ is finite projective. The map is given by the maps

\[ c_{p, r, s} : \mathop{\mathrm{Hom}}\nolimits _ R(J^{-r}, I^ p) \otimes _ R K^ s \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ R(\mathop{\mathrm{Hom}}\nolimits _ R(K^ s, J^{-r}), I^ p) \]

which are isomorphisms as $K^ s$ is finite projective. For every element $\alpha = (\alpha ^{p, r, s})$ of degree $n$ of the left hand side, there are only finitely many values of $s$ such that $\alpha ^{p, r, s}$ is nonzero (for some $p, r$ with $n = p + r + s$). Hence our map is an isomorphism if the same vanishing condition is forced on the elements $\beta = (\beta ^{p, r, s})$ of the right hand side. If $K^\bullet $ is a bounded complex of finite projective modules, this is clear. On the other hand, if we can choose $I^\bullet $ bounded and $J^\bullet $ bounded below, then $\beta ^{p, r, s}$ is zero for $p$ outside a fixed range, for $s \gg 0$, and for $r \gg 0$. Hence among solutions of $n = p + r + s$ with $\beta ^{p, r, s}$ nonzero only a finite number of $s$ values occur. $\square$

Lemma 15.86.2. Let $R$ be a ring. Let $K, L, M$ be objects of $D(R)$. the map

\[ R\mathop{\mathrm{Hom}}\nolimits _ R(L, M) \otimes _ R^\mathbf {L} K \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _ R(R\mathop{\mathrm{Hom}}\nolimits _ R(K, L), M) \]

of Lemma 15.68.3 is an isomorphism if the following three conditions are satisfied

  1. $L, M$ have finite injective dimension,

  2. $R\mathop{\mathrm{Hom}}\nolimits _ R(L, M)$ has finite tor dimension,

  3. for every $n \in \mathbf{Z}$ the truncation $\tau _{\leq n}K$ is pseudo-coherent

Proof. Pick an integer $n$ and consider the distinguished triangle

\[ \tau _{\leq n}K \to K \to \tau _{\geq n + 1}K \to \tau _{\leq n}K[1] \]

see Derived Categories, Remark 13.12.4. By assumption (3) and Lemma 15.86.1 the map is an isomorphism for $\tau _{\leq n}K$. Hence it suffices to show that both

\[ R\mathop{\mathrm{Hom}}\nolimits _ R(L, M) \otimes _ R^\mathbf {L} \tau _{\geq n + 1}K \quad \text{and}\quad R\mathop{\mathrm{Hom}}\nolimits _ R(R\mathop{\mathrm{Hom}}\nolimits _ R(\tau _{\geq n + 1}K, L), M) \]

have vanishing cohomology in degrees $\leq n - c$ for some $c$. This follows immediately from assumptions (2) and (1). $\square$

Lemma 15.86.3. Let $R$ be a ring. Let $K, L, M$ be objects of $D(R)$. The map

\[ K \otimes _ R^\mathbf {L} R\mathop{\mathrm{Hom}}\nolimits _ R(M, L) \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _ R(M, K \otimes _ R^\mathbf {L} L) \]

of Lemma 15.68.5 is an isomorphism in the following cases

  1. $M$ perfect, or

  2. $K$ is perfect, or

  3. $M$ is pseudo-coherent, $L \in D^+(R)$, and $K$ has tor amplitude in $[a, \infty ]$.

Proof. Proof in case $M$ is perfect. Note that both sides of the arrow transform distinguished triangles in $M$ into distinguished triangles and commute with direct sums. Hence it suffices to check it holds when $M = R[n]$, see Derived Categories, Remark 13.33.5 and Lemma 15.72.1. In this case the result is obvious.

Proof in case $K$ is perfect. Same argument as in the previous case.

Proof in case (3). We may represent $K$ and $L$ by bounded below complexes of $R$-modules $K^\bullet $ and $L^\bullet $. We may assume that $K^\bullet $ is a K-flat complex consisting of flat $R$-modules, see Lemma 15.63.4. We may represent $M$ by a bounded above complex $M^\bullet $ of finite free $R$-modules, see Definition 15.62.1. Then the object on the LHS is represented by

\[ \text{Tot}(K^\bullet \otimes _ R \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , L^\bullet )) \]

and the object on the RHS by

\[ \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , \text{Tot}(K^\bullet \otimes _ R L^\bullet )) \]

This uses Lemma 15.68.2. Both complexes have in degree $n$ the module

\[ \bigoplus \nolimits _{p + q + r = n} K^ p \otimes \mathop{\mathrm{Hom}}\nolimits _ R(M^{-r}, L^ q) = \bigoplus \nolimits _{p + q + r = n} \mathop{\mathrm{Hom}}\nolimits _ R(M^{-r}, K^ p \otimes _ R L^ q) \]

because $M^{-r}$ is finite free (as well these are finite direct sums). The map defined in Lemma 15.68.5 comes from the map of complexes defined in Lemma 15.67.5 which uses the canonical isomorphisms between these modules. $\square$

Lemma 15.86.4. Let $R$ be a ring. Let $P^\bullet $ be a bounded above complex of projective $R$-modules. Let $K^\bullet $ be a K-flat complex of $R$-modules. If $P^\bullet $ is a perfect object of $D(R)$, then $\mathop{\mathrm{Hom}}\nolimits ^\bullet (P^\bullet , K^\bullet )$ is K-flat and represents $R\mathop{\mathrm{Hom}}\nolimits _ R(P^\bullet , K^\bullet )$.

Proof. The last statement is Lemma 15.68.2. Since $P^\bullet $ represents a perfect object, there exists a finite complex of finite projective $R$-modules $F^\bullet $ such that $P^\bullet $ and $F^\bullet $ are isomorphic in $D(R)$, see Definition 15.69.1. Then $P^\bullet $ and $F^\bullet $ are homotopy equivalent, see Derived Categories, Lemma 13.19.8. Then $\mathop{\mathrm{Hom}}\nolimits ^\bullet (P^\bullet , K^\bullet )$ and $\mathop{\mathrm{Hom}}\nolimits ^\bullet (F^\bullet , K^\bullet )$ are homotopy equivalent. Hence the first is K-flat if and only if the second is (follows from Definition 15.57.3 and Lemma 15.57.1). It is clear that

\[ \mathop{\mathrm{Hom}}\nolimits ^\bullet (F^\bullet , K^\bullet ) = \text{Tot}(E^\bullet \otimes _ R K^\bullet ) \]

where $E^\bullet $ is the dual complex to $F^\bullet $ with terms $E^ n = \mathop{\mathrm{Hom}}\nolimits _ R(F^{-n}, R)$, see Lemma 15.69.14 and its proof. Since $E^\bullet $ is a bounded complex of projectives we find that it is K-flat by Lemma 15.57.9. Then we conclude by Lemma 15.57.6. $\square$


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