Proposition 4

Tag 00IO, permanent. Cite it as [Bourbaki, Tag 00IO].

Algebra, VIII, p. 293, § 16. Other Descriptions of the Brauer Group

The product of $\tau$ -extensions endows the set Ex $_{\tau}(G,F)$ with the structure of an abelian group. Its identity element is the class of the semitrivial extension $\mathscr{I}_{\tau}$ . The inverse of the class of a $\tau$ -extension $\mathscr{E}$ is the class of $s_*(\mathscr{E})$ .

The associativity of the law follows from the commutativity of the diagrams

\begin{array}{ccc} G & \xrightarrow{\delta} & G \times G \\ \downarrow \delta & & \downarrow \operatorname{Id} \times \delta \\ G \times G & \xrightarrow{\delta \times \operatorname{Id}} & G \times G \times G \end{array}

and

\begin{array}{ccc} F \times F \times F & \xrightarrow{m \times \operatorname{Id}} & F \times F \\ \downarrow \operatorname{Id} \times m & & \downarrow m \\ F \times F & \xrightarrow{m} & F \end{array}

and from Corollary 2 of VIII, p. 288; Corollary 2 of VIII, p. 291; and Proposition 3 of VIII, p. 292.

Let $\Delta : F \to F \times F$ be the diagonal mapping $f \mapsto (f, f)$ . Let $\mathscr{E} = (\Gamma, \iota, \pi)$ be a $\tau$ -extension. Let $\tilde{\Delta} : \Gamma \to \Gamma \times_G \Gamma$ be the group homomorphism given by $\gamma \mapsto (\gamma, \gamma)$ . The following diagram commutes:

\begin{array}{ccccc} F & \xrightarrow{\iota} & \Gamma & \xrightarrow{\pi} & G \\ \downarrow \Delta & & \downarrow \tilde{\Delta} & & \parallel \\ F \times F & \longrightarrow & \Gamma \times_G \Gamma & \longrightarrow & G. \end{array}

By Proposition 2 of VIII, p. 290, it follows that the $(\tau \times \tau) \circ \delta$ -extension $\delta^*(\mathscr{E} \times \mathscr{E})$ is isomorphic to $\Delta_*(\mathscr{E})$ .

Denote by $c : F \to F$ the constant homomorphism $f \mapsto 1$ . by Example 1 of VIII, p. 292, the fact that $\mathscr{I}_\tau$ is an identity element for this law of composition follows from the isomorphism from $\delta^*(\mathscr{E} \times \mathscr{E})$ to $\Delta_*(\mathscr{E})$ and the commutative diagram

\begin{array}{ccc} F & \xrightarrow{(\operatorname{Id}_F \times c) \circ \Delta} & F \times F \\ (c \times \operatorname{Id}_F) \circ \Delta & & \downarrow \operatorname{Id}_F \\ & & \downarrow m \\ F \times F & \xrightarrow{m} & F. \end{array}

The last assertion follows from the commutative diagram

\begin{array}{ccc} F & \xrightarrow{(\operatorname{Id}_F \times s) \circ \Delta} & F \times F \\ (s \times \operatorname{Id}_F) \circ \Delta & & \downarrow c \\ & & \downarrow m \\ F \times F & \xrightarrow{m} & F. \end{array}

Let $\mathscr{E}_1 = (\Gamma_1, \iota_1, \pi_1)$ and $\mathscr{E}_2 = (\Gamma_2, \iota_2, \pi_2)$ be $\tau$ -extensions. The group isomorphism $\Gamma_1 \times \Gamma_2 \to \Gamma_2 \times \Gamma_1$ given by $(\gamma_1, \gamma_2) \mapsto (\gamma_2, \gamma_1)$ restricts to a group isomorphism $\sigma : \Gamma_1 \times_G \Gamma_2 \to \Gamma_2 \times_G \Gamma_1$ . Because of the relations

\sigma(\iota_1(f), \iota_2(f)^{-1}) = (\iota_2(f^{-1}), \iota_1(f^{-1})^{-1})

for $f\in F$ , the group homomorphism $\sigma$ induces, by passing to the quotients, a morphism of $\tau$ -extensions from $\mathscr{E}_1\mathscr{E}_2$ to $\mathscr{E}_2\mathscr{E}_1$ . So the law of composition is commutative.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Nothing.

Proposition 4

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