Lemma 47.13.1. Let $A \to B$ be a ring homomorphism. The functor $R\mathop{\mathrm{Hom}}\nolimits (B, -)$ constructed above is right adjoint to the restriction functor $D(B) \to D(A)$.

## 47.13 Trivial duality for a ring map

Let $A \to B$ be a ring homomorphism. Consider the functor

This functor is left exact and has a derived extension $R\mathop{\mathrm{Hom}}\nolimits (B, -) : D(A) \to D(B)$.

**Proof.**
This is a consequence of the fact that restriction and $\mathop{\mathrm{Hom}}\nolimits _ A(B, -)$ are adjoint functors by Algebra, Lemma 10.14.4. See Derived Categories, Lemma 13.30.3.
$\square$

Lemma 47.13.2. Let $A \to B \to C$ be ring maps. Then $R\mathop{\mathrm{Hom}}\nolimits (C, -) \circ R\mathop{\mathrm{Hom}}\nolimits (B, -) : D(A) \to D(C)$ is the functor $R\mathop{\mathrm{Hom}}\nolimits (C, -) : D(A) \to D(C)$.

**Proof.**
Follows from uniqueness of right adjoints and Lemma 47.13.1.
$\square$

Lemma 47.13.3. Let $\varphi : A \to B$ be a ring homomorphism. For $K$ in $D(A)$ we have

where $\varphi _* : D(B) \to D(A)$ is restriction. In particular $R^ q\mathop{\mathrm{Hom}}\nolimits (B, K) = \mathop{\mathrm{Ext}}\nolimits _ A^ q(B, K)$.

**Proof.**
Choose a K-injective complex $I^\bullet $ representing $K$. Then $R\mathop{\mathrm{Hom}}\nolimits (B, K)$ is represented by the complex $\mathop{\mathrm{Hom}}\nolimits _ A(B, I^\bullet )$ of $B$-modules. Since this complex, as a complex of $A$-modules, represents $R\mathop{\mathrm{Hom}}\nolimits _ A(B, K)$ we see that the lemma is true.
$\square$

Let $A$ be a Noetherian ring. We will denote

the full subcategory consisting of those objects $K$ of $D(A)$ whose cohomology modules are all finite $A$-modules. This makes sense by Derived Categories, Section 13.17 because as $A$ is Noetherian, the subcategory of finite $A$-modules is a Serre subcategory of $\text{Mod}_ A$.

Lemma 47.13.4. With notation as above, assume $A \to B$ is a finite ring map of Noetherian rings. Then $R\mathop{\mathrm{Hom}}\nolimits (B, -)$ maps $D^+_{\textit{Coh}}(A)$ into $D^+_{\textit{Coh}}(B)$.

**Proof.**
We have to show: if $K \in D^+(A)$ has finite cohomology modules, then the complex $R\mathop{\mathrm{Hom}}\nolimits (B, K)$ has finite cohomology modules too. This follows for example from Lemma 47.13.3 if we can show the ext modules $\mathop{\mathrm{Ext}}\nolimits ^ i_ A(B, K)$ are finite $A$-modules. Since $K$ is bounded below there is a convergent spectral sequence

This finishes the proof as the modules $\mathop{\mathrm{Ext}}\nolimits ^ p_ A(B, H^ q(K))$ are finite by Algebra, Lemma 10.71.9. $\square$

Remark 47.13.5. Let $A$ be a ring and let $I \subset A$ be an ideal. Set $B = A/I$. In this case the functor $\mathop{\mathrm{Hom}}\nolimits _ A(B, -)$ is equal to the functor

which sends $M$ to the submodule of $I$-torsion.

Situation 47.13.6. Let $R \to A$ be a ring map. We will give an alternative construction of $R\mathop{\mathrm{Hom}}\nolimits (A, -)$ which will stand us in good stead later in this chapter. Namely, suppose we have a differential graded algebra $(E, d)$ over $R$ and a quasi-isomorphism $E \to A$ where we view $A$ as a differential graded algebra over $R$ with zero differential. Then we have commutative diagrams

where the horizontal arrows are equivalences of categories (Differential Graded Algebra, Lemma 22.37.1). It is clear that the first diagram commutes. The second diagram commutes because the first one does and our functors are their left adjoints (Differential Graded Algebra, Example 22.33.6) or because we have $E \otimes ^\mathbf {L}_ E A = E \otimes _ E A$ and we can use Differential Graded Algebra, Lemma 22.34.1.

Lemma 47.13.7. In Situation 47.13.6 the functor $R\mathop{\mathrm{Hom}}\nolimits (A, -)$ is equal to the composition of $R\mathop{\mathrm{Hom}}\nolimits (E, -) : D(R) \to D(E, \text{d})$ and the equivalence $- \otimes ^\mathbf {L}_ E A : D(E, \text{d}) \to D(A)$.

**Proof.**
This is true because $R\mathop{\mathrm{Hom}}\nolimits (E, -)$ is the right adjoint to $- \otimes ^\mathbf {L}_ R E$, see Differential Graded Algebra, Lemma 22.33.5. Hence this functor plays the same role as the functor $R\mathop{\mathrm{Hom}}\nolimits (A, -)$ for the map $R \to A$ (Lemma 47.13.1), whence these functors must correspond via the equivalence $- \otimes ^\mathbf {L}_ E A : D(E, \text{d}) \to D(A)$.
$\square$

Lemma 47.13.8. In Situation 47.13.6 assume that

$E$ viewed as an object of $D(R)$ is compact, and

$N = \mathop{\mathrm{Hom}}\nolimits ^\bullet _ R(E^\bullet , R)$ computes $R\mathop{\mathrm{Hom}}\nolimits (E, R)$.

Then $R\mathop{\mathrm{Hom}}\nolimits (E, -) : D(R) \to D(E)$ is isomorphic to $K \mapsto K \otimes _ R^\mathbf {L} N$.

**Proof.**
Special case of Differential Graded Algebra, Lemma 22.33.9.
$\square$

Lemma 47.13.9. In Situation 47.13.6 assume $A$ is a perfect $R$-module. Then

is given by $K \mapsto K \otimes _ R^\mathbf {L} M$ where $M = R\mathop{\mathrm{Hom}}\nolimits (A, R) \in D(A)$.

**Proof.**
We apply Divided Power Algebra, Lemma 23.6.10 to choose a Tate resolution $(E, \text{d})$ of $A$ over $R$. Note that $E^ i = 0$ for $i > 0$, $E^0 = R[x_1, \ldots , x_ n]$ is a polynomial algebra, and $E^ i$ is a finite free $E^0$-module for $i < 0$. It follows that $E$ viewed as a complex of $R$-modules is a bounded above complex of free $R$-modules. We check the assumptions of Lemma 47.13.8. The first holds because $A$ is perfect (hence compact by More on Algebra, Proposition 15.78.3) and the second by More on Algebra, Lemma 15.73.2. From the lemma conclude that $K \mapsto R\mathop{\mathrm{Hom}}\nolimits (E, K)$ is isomorphic to $K \mapsto K \otimes _ R^\mathbf {L} N$ for some differential graded $E$-module $N$. Observe that

in $D(A)$. Hence by Differential Graded Algebra, Lemma 22.34.2 we conclude that the composition of $- \otimes _ R^\mathbf {L} N$ and $- \otimes _ R^\mathbf {L} A$ is of the form $- \otimes _ R M$ for some $M \in D(A)$. To finish the proof we apply Lemma 47.13.7. $\square$

Lemma 47.13.10. Let $R \to A$ be a surjective ring map whose kernel $I$ is an invertible $R$-module. The functor $R\mathop{\mathrm{Hom}}\nolimits (A, -) : D(R) \to D(A)$ is isomorphic to $K \mapsto K \otimes _ R^\mathbf {L} N[-1]$ where $N$ is inverse of the invertible $A$-module $I \otimes _ R A$.

**Proof.**
Since $A$ has the finite projective resolution

we see that $A$ is a perfect $R$-module. By Lemma 47.13.9 it suffices to prove that $R\mathop{\mathrm{Hom}}\nolimits (A, R)$ is represented by $N[-1]$ in $D(A)$. This means $R\mathop{\mathrm{Hom}}\nolimits (A, R)$ has a unique nonzero cohomology module, namely $N$ in degree $1$. As $\text{Mod}_ A \to \text{Mod}_ R$ is fully faithful it suffice to prove this after applying the restriction functor $i_* : D(A) \to D(R)$. By Lemma 47.13.3 we have

Using the finite projective resolution above we find that the latter is represented by the complex $R \to I^{\otimes -1}$ with $R$ in degree $0$. The map $R \to I^{\otimes -1}$ is injective and the cokernel is $N$. $\square$

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