Lemma 15.83.6. Let $R \to A$ be a flat ring map. Let $K, L \in D(A)$ with $K$ pseudo-coherent and $L$ finite tor dimension over $R$. We may choose

1. a bounded above complex $P^\bullet$ of finite free $A$-modules representing $K$, and

2. a bounded complex of $R$-flat $A$-modules $F^\bullet$ representing $L$.

Given these choices we have

1. $E^\bullet = \mathop{\mathrm{Hom}}\nolimits ^\bullet (P^\bullet , F^\bullet )$ is a bounded below complex of $R$-flat $A$-modules representing $R\mathop{\mathrm{Hom}}\nolimits _ A(K, L)$,

2. for any ring map $R \to R'$ with $A' = A \otimes _ R R'$ the complex $E^\bullet \otimes _ R R'$ represents $R\mathop{\mathrm{Hom}}\nolimits _{A'}(K \otimes _ A^\mathbf {L} A', L \otimes _ A^\mathbf {L} A')$.

If in addition $R \to A$ is of finite presentation and $L$ is $R$-perfect, then we may choose $F^ p$ to be finitely presented $A$-modules and consequently $E^ n$ will be finitely presented $A$-modules as well.

Proof. The existence of $P^\bullet$ is the definition of a pseudo-coherent complex. We first represent $L$ by a bounded above complex $F^\bullet$ of free $A$-modules (this is possible because bounded tor dimension in particular implies bounded). Next, say $L$ viewed as an object of $D(R)$ has tor amplitude in $[a, b]$. Then, after replacing $F^\bullet$ by $\tau _{\geq a}F^\bullet$, we get a complex as in (2). This follows from Lemma 15.66.2.

Proof of (a). Since $F^\bullet$ is bounded an since $P^\bullet$ is bounded above, we see that $E^ n = 0$ for $n \ll 0$ and that $E^ n$ is a finite (!) direct sum

$E^ n = \bigoplus \nolimits _{p + q = n} \mathop{\mathrm{Hom}}\nolimits _ A(P^{-q}, F^ p)$

and since $P^{-q}$ is finite free, this is indeed an $R$-flat $A$-module. The fact that $E^\bullet$ represents $R\mathop{\mathrm{Hom}}\nolimits _ A(K, L)$ follows from Lemma 15.73.2.

Proof of (b). Let $R \to R'$ be a ring map and $A' = A \otimes _ R R'$. By Lemma 15.61.2 the object $L \otimes _ A^\mathbf {L} A'$ is represented by $F^\bullet \otimes _ R R'$ viewed as a complex of $A'$-modules (by flatness of $F^ p$ over $R$). Similarly for $P^\bullet \otimes _ R R'$. As above $R\mathop{\mathrm{Hom}}\nolimits _{A'}(K \otimes _ A^\mathbf {L} A', L \otimes _ A^\mathbf {L} A')$ is represented by

$\mathop{\mathrm{Hom}}\nolimits ^\bullet (P^\bullet \otimes _ R R', F^\bullet \otimes _ R R') = E^\bullet \otimes _ R R'$

The equality holds by looking at the terms of the complex individually and using that $\mathop{\mathrm{Hom}}\nolimits _{A'}(P^{-q} \otimes _ R R', F^ p \otimes _ R R') = \mathop{\mathrm{Hom}}\nolimits _ A(P^{-q}, F^ p) \otimes _ R R'$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).