Lemma 91.28.1. In the situation above, if $X$ and $Y$ are Tor independent over $B$, then the object $E$ in (91.28.0.1) is zero. In this case we have

## 91.28 Fibre products of algebraic spaces and the cotangent complex

Let $S$ be a scheme. Let $X \to B$ and $Y \to B$ be morphisms of algebraic spaces over $S$. Consider the fibre product $X \times _ B Y$ with projection morphisms $p : X \times _ B Y \to X$ and $q : X \times _ B Y \to Y$. In this section we discuss $L_{X \times _ B Y/B}$. Most of the information we want is contained in the following diagram

Explanation: The middle row is the fundamental triangle of Lemma 91.22.3 for the morphisms $X \times _ B Y \to X \to B$. The middle column is the fundamental triangle for the morphisms $X \times _ B Y \to Y \to B$. Next, $E$ is an object of $D(\mathcal{O}_{X \times _ B Y})$ which “fits” into the upper right corner, i.e., which turns both the top row and the right column into distinguished triangles. Such an $E$ exists by Derived Categories, Proposition 13.4.23 applied to the lower left square (with $0$ placed in the missing spot). To be more explicit, we could for example define $E$ as the cone (Derived Categories, Definition 13.9.1) of the map of complexes

and get the two maps with target $E$ by an application of TR3. In the Tor independent case the object $E$ is zero.

**Proof.**
Choose a scheme $W$ and a surjective étale morphism $W \to B$. Choose a scheme $U$ and a surjective étale morphism $U \to X \times _ B W$. Choose a scheme $V$ and a surjective étale morphism $V \to Y \times _ B W$. Then $U \times _ W V \to X \times _ B Y$ is surjective étale too. Hence it suffices to prove that the restriction of $E$ to $U \times _ W V$ is zero. By Lemma 91.26.3 and Derived Categories of Spaces, Lemma 74.20.3 this reduces us to the case of schemes. Taking suitable affine opens we reduce to the case of affine schemes. Using Lemma 91.24.2 we reduce to the case of a tensor product of rings, i.e., to Lemma 91.15.1.
$\square$

In general we can say the following about the object $E$.

Lemma 91.28.2. Let $S$ be a scheme. Let $X \to B$ and $Y \to B$ be morphisms of algebraic spaces over $S$. The object $E$ in (91.28.0.1) satisfies $H^ i(E) = 0$ for $i = 0, -1$ and for a geometric point $(\overline{x}, \overline{y}) : \mathop{\mathrm{Spec}}(k) \to X \times _ B Y$ we have

where $R = \mathcal{O}_{B, \overline{b}}$, $A = \mathcal{O}_{X, \overline{x}}$, $B = \mathcal{O}_{Y, \overline{y}}$, and $C = \mathcal{O}_{X \times _ B Y, (\overline{x}, \overline{y})}$.

**Proof.**
The formation of the cotangent complex commutes with taking stalks and pullbacks, see Lemmas 91.18.9 and 91.18.3. Note that $C$ is a henselization of $A \otimes _ R B$. $L_{C/R} = L_{A \otimes _ R B/R} \otimes _{A \otimes _ R B} C$ by the results of Section 91.8. Thus the stalk of $E$ at our geometric point is the cone of the map $L_{A/R} \otimes C \to L_{A \otimes _ R B/R} \otimes C$. Therefore the results of the lemma follow from the case of rings, i.e., Lemma 91.15.2.
$\square$

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