Lemma 15.60.1. The construction above is independent of choices and defines an exact functor of triangulated categories $- \otimes _ R^\mathbf {L} N^\bullet : D(R) \to D(A)$. There is a functorial isomorphism

for $E^\bullet $ in $D(R)$.

Let $R \to A$ be a ring map. Let $N^\bullet $ be a complex of $A$-modules. We can also use K-flat resolutions to define a functor

\[ - \otimes _ R^{\mathbf{L}} N^\bullet : D(R) \to D(A) \]

as the left derived functor of the functor $K(R) \to K(A)$, $M^\bullet \mapsto \text{Tot}(M^\bullet \otimes _ R N^\bullet )$. In particular, taking $N^\bullet = A[0]$ we obtain a derived base change functor

\[ - \otimes _ R^{\mathbf{L}} A : D(R) \to D(A) \]

extending the functor (15.57.0.2). Namely, for every complex of $R$-modules $M^\bullet $ we can choose a K-flat resolution $K^\bullet \to M^\bullet $ and set

\[ M^\bullet \otimes _ R^{\mathbf{L}} N^\bullet = \text{Tot}(K^\bullet \otimes _ R N^\bullet ). \]

You can use Lemmas 15.59.10 and 15.59.12 to see that this is well defined. However, to cross all the t's and dot all the i's it is perhaps more convenient to use some general theory.

Lemma 15.60.1. The construction above is independent of choices and defines an exact functor of triangulated categories $- \otimes _ R^\mathbf {L} N^\bullet : D(R) \to D(A)$. There is a functorial isomorphism

\[ E^\bullet \otimes _ R^\mathbf {L} N^\bullet = (E^\bullet \otimes _ R^\mathbf {L} A) \otimes _ A^\mathbf {L} N^\bullet \]

for $E^\bullet $ in $D(R)$.

**Proof.**
To prove the existence of the derived functor $- \otimes _ R^\mathbf {L} N^\bullet $ we use the general theory developed in Derived Categories, Section 13.14. Set $\mathcal{D} = K(R)$ and $\mathcal{D}' = D(A)$. Let us write $F : \mathcal{D} \to \mathcal{D}'$ the exact functor of triangulated categories defined by the rule $F(M^\bullet ) = \text{Tot}(M^\bullet \otimes _ R N^\bullet )$. To prove the stated properties of $F$ use Lemmas 15.58.2 and 15.58.4. We let $S$ be the set of quasi-isomorphisms in $\mathcal{D} = K(R)$. This gives a situation as in Derived Categories, Situation 13.14.1 so that Derived Categories, Definition 13.14.2 applies. We claim that $LF$ is everywhere defined. This follows from Derived Categories, Lemma 13.14.15 with $\mathcal{P} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ the collection of K-flat complexes: (1) follows from Lemma 15.59.10 and (2) follows from Lemma 15.59.12. Thus we obtain a derived functor

\[ LF : D(R) = S^{-1}\mathcal{D} \longrightarrow \mathcal{D}' = D(A) \]

see Derived Categories, Equation (13.14.9.1). Finally, Derived Categories, Lemma 13.14.15 guarantees that $LF(K^\bullet ) = F(K^\bullet ) = \text{Tot}(K^\bullet \otimes _ R N^\bullet )$ when $K^\bullet $ is K-flat, i.e., $LF$ is indeed computed in the way described above. Moreover, by Lemma 15.59.3 the complex $K^\bullet \otimes _ R A$ is a K-flat complex of $A$-modules. Hence

\[ (K^\bullet \otimes _ R^\mathbf {L} A) \otimes _ A^\mathbf {L} N^\bullet = \text{Tot}((K^\bullet \otimes _ R A) \otimes _ A N^\bullet ) = \text{Tot}(K^\bullet \otimes _ A N^\bullet ) = K^\bullet \otimes _ A^\mathbf {L} N^\bullet \]

which proves the final statement of the lemma. $\square$

Lemma 15.60.2. Let $R \to A$ be a ring map. Let $f : L^\bullet \to N^\bullet $ be a map of complexes of $A$-modules. Then $f$ induces a transformation of functors

\[ 1 \otimes f : - \otimes _ A^\mathbf {L} L^\bullet \longrightarrow - \otimes _ A^\mathbf {L} N^\bullet \]

If $f$ is a quasi-isomorphism, then $1 \otimes f$ is an isomorphism of functors.

**Proof.**
Since the functors are computing by evaluating on K-flat complexes $K^\bullet $ we can simply use the functoriality

\[ \text{Tot}(K^\bullet \otimes _ R L^\bullet ) \to \text{Tot}(K^\bullet \otimes _ R N^\bullet ) \]

to define the transformation. The last statement follows from Lemma 15.59.2. $\square$

Lemma 15.60.3. Let $R \to A$ be a ring map. The functor $D(R) \to D(A)$, $E \mapsto E \otimes _ R^\mathbf {L} A$ of Lemma 15.60.1 is left adjoint to the restriction functor $D(A) \to D(R)$.

**Proof.**
This follows from Derived Categories, Lemma 13.30.1 and the fact that $- \otimes _ R A$ and restriction are adjoint by Algebra, Lemma 10.14.3.
$\square$

Remark 15.60.4 (Warning). Let $R \to A$ be a ring map, and let $N$ and $N'$ be $A$-modules. Denote $N_ R$ and $N'_ R$ the restriction of $N$ and $N'$ to $R$-modules, see Algebra, Section 10.14. In this situation, the objects $N_ R \otimes _ R^\mathbf {L} N'$ and $N \otimes _ R^\mathbf {L} N'_ R$ of $D(A)$ are in general not isomorphic! In other words, one has to pay careful attention as to which of the two sides is being used to provide the $A$-module structure.

For a specific example, set $R = k[x, y]$, $A = R/(xy)$, $N = R/(x)$ and $N' = A = R/(xy)$. The resolution $0 \to R \xrightarrow {xy} R \to N'_ R \to 0$ shows that $N \otimes _ R^\mathbf {L} N'_ R = N[1] \oplus N$ in $D(A)$. The resolution $0 \to R \xrightarrow {x} R \to N_ R \to 0$ shows that $N_ R \otimes _ R^\mathbf {L} N'$ is represented by the complex $A \xrightarrow {x} A$. To see these two complexes are not isomorphic, one can show that the second complex is not isomorphic in $D(A)$ to the direct sum of its cohomology groups, or one can show that the first complex is not a perfect object of $D(A)$ whereas the second one is. Some details omitted.

Lemma 15.60.5. Let $A \to B \to C$ be ring maps. Let $N^\bullet $ be a complex of $B$-modules and $K^\bullet $ a complex of $C$-modules. The compositions of the functors

\[ D(A) \xrightarrow {- \otimes _ A^\mathbf {L} N^\bullet } D(B) \xrightarrow {- \otimes _ B^\mathbf {L} K^\bullet } D(C) \]

is the functor $- \otimes _ A^\mathbf {L} (N^\bullet \otimes _ B^\mathbf {L} K^\bullet ) : D(A) \to D(C)$. If $M$, $N$, $K$ are modules over $A$, $B$, $C$, then we have

\[ (M \otimes _ A^\mathbf {L} N) \otimes _ B^\mathbf {L} K = M \otimes _ A^\mathbf {L} (N \otimes _ B^\mathbf {L} K) = (M \otimes _ A^\mathbf {L} C) \otimes _ C^\mathbf {L} (N \otimes _ B^\mathbf {L} K) \]

in $D(C)$. We also have a canonical isomorphism

\[ (M \otimes _ A^\mathbf {L} N) \otimes _ B^\mathbf {L} K \longrightarrow (M \otimes _ A^\mathbf {L} K) \otimes _ C^\mathbf {L} (N \otimes _ B^\mathbf {L} C) \]

using signs. Similar results holds for complexes.

**Proof.**
Choose a K-flat complex $P^\bullet $ of $B$-modules and a quasi-isomorphism $P^\bullet \to N^\bullet $ (Lemma 15.59.10). Let $M^\bullet $ be a K-flat complex of $A$-modules representing an arbitrary object of $D(A)$. Then we see that

\[ (M^\bullet \otimes _ A^\mathbf {L} P^\bullet ) \otimes _ B^\mathbf {L} K^\bullet \longrightarrow (M^\bullet \otimes _ A^\mathbf {L} N^\bullet ) \otimes _ B^\mathbf {L} K^\bullet \]

is an isomorphism by Lemma 15.60.2 applied to the material inside the brackets. By Lemmas 15.59.3 and 15.59.4 the complex

\[ \text{Tot}(M^\bullet \otimes _ A P^\bullet ) = \text{Tot}((M^\bullet \otimes _ R A) \otimes _ A P^\bullet \]

is K-flat as a complex of $B$-modules and it represents the derived tensor product in $D(B)$ by construction. Hence we see that $(M^\bullet \otimes _ A^\mathbf {L} P^\bullet ) \otimes _ B^\mathbf {L} K^\bullet $ is represented by the complex

\[ \text{Tot}(\text{Tot}(M^\bullet \otimes _ A P^\bullet )\otimes _ B K^\bullet ) = \text{Tot}(M^\bullet \otimes _ A \text{Tot}(P^\bullet \otimes _ B K^\bullet )) \]

of $C$-modules. Equality by Homology, Remark 12.18.4. Going back the way we came we see that this is equal to

\[ M^\bullet \otimes _ A^\mathbf {L} (P^\bullet \otimes _ B^\mathbf {L} K^\bullet ) \longleftarrow M^\bullet \otimes _ A^\mathbf {L} (N^\bullet \otimes _ B^\mathbf {L} K^\bullet ) \]

The arrow is an isomorphism by definition of the functor $-\otimes _ B^\mathbf {L} K^\bullet $. All of these constructions are functorial in the complex $M^\bullet $ and hence we obtain our isomorphism of functors.

By the above we have the first equality in

\[ (M \otimes _ A^\mathbf {L} N) \otimes _ B^\mathbf {L} K = M \otimes _ A^\mathbf {L} (N \otimes _ B^\mathbf {L} K) = (M \otimes _ A^\mathbf {L} C) \otimes _ C^\mathbf {L} (N \otimes _ B^\mathbf {L} K) \]

The second equality follows from the final statement of Lemma 15.60.1. The same thing allows us to write $N \otimes _ B^\mathbf {L} K = (N \otimes _ B^\mathbf {L} C) \otimes _ C^\mathbf {L} K$ and substituting we get

\begin{align*} (M \otimes _ A^\mathbf {L} N) \otimes _ B^\mathbf {L} K & = (M \otimes _ A^\mathbf {L} C) \otimes _ C^\mathbf {L} ((N \otimes _ B^\mathbf {L} C) \otimes _ C^\mathbf {L} K) \\ & = (M \otimes _ A^\mathbf {L} C) \otimes _ C^\mathbf {L} (K \otimes _ C^\mathbf {L} (N \otimes _ B^\mathbf {L} C)) \\ & = ((M \otimes _ A^\mathbf {L} C) \otimes _ C^\mathbf {L} K) \otimes _ C^\mathbf {L} (N \otimes _ B^\mathbf {L} C)) \\ & = (M \otimes _ C^\mathbf {L} K) \otimes _ C^\mathbf {L} (N \otimes _ B^\mathbf {L} C) \end{align*}

by Lemmas 15.59.14 and 15.59.15 as well as the previously mentioned lemma. $\square$

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