Suppose that $k$ is a commutative ring and that $A$ is an Azumaya
$k$-algebra. Then there is a well-known morphism from $Aut\_{Alg\_k}(A)$, the group of algebra automorphisms, to the Picard group $Pic(k)$. One way to phrase it is as follows. Every automorphism $\phi$ determines a $k$-linear autoequivalence
$$
\phi^\*: Mod\_A \to Mod\_A
$$
However, since $A$ is Azumaya, every $k$-linear autoequivalence of the category of (right) $A$-modules is of the form $M \mapsto M \otimes\_k J$ for some $J$ in the Picard group of $k$, unique up to isomorphism. Moreover, this assignment takes composition to tensor product. This assignment is part of the Rosenberg-Zelinsky exact sequence
$$
0 \to k^\times \to A^\times \to Aut\_{Alg\_k}(A) \to Pic(k).
$$
It is possible to express $J$ as $HH^0(A,A^\tau)$ where $A^\tau$ is $A$, but with half of its bimodule structure pulled back along $\phi$.


In particular, given an Azumaya $k$-algebra $Q$, let $A = Q \otimes\_k Q$. Then A is also an Azumaya $k$-algebra with automorphism $\phi(a \otimes b) = b \otimes a$, and thus $Q$ determines a (2-torsion) element in $Pic(k)$. This assignment turns out to be Morita invariant and it respects the tensor product; this makes it a (2-torsion) homomorphism from the Brauer group $Br(k)$ to the Picard group $Pic(k)$.


Are there examples of rings $k$ for which this homomorphism is nontrivial?



On a morphism from the Brauer group to the Picard group
It seems to me that the involution of $Q \otimes Q$ that exchanges $a \otimes b$ and $b\otimes a$ is inner, which means that the homomorphism you describe should always be trivial.


Here is the argument, I hope it is correct. Suppose that $Q$ is trivial, that is, there is a projective $k$-module $V$ such that $Q \simeq \operatorname{End}V$. Then $Q \otimes Q \simeq \operatorname{End}(V\otimes V)$. Consider the operator $\tau\colon V\otimes V \simeq V\otimes V$ that exchanges $v \otimes w$ and $w \otimes v$. We can think of $\tau$ as an element of $Q \otimes Q$; conjugation by $\tau$ exchanges $a \otimes b$ and $b\otimes a$.


Now, choose another projective $k$-module $W$ and an isomorphism $Q \simeq \operatorname{End}W$. Consider the induced isomorphism $\phi\colon\operatorname{End}V \simeq \operatorname{End}W$. Then there exists an invertible $k$-module $L$ and an isomorphism $\psi\colon W \simeq L \otimes V$ such that $\psi$ that induces $\phi$. This implies that $\tau$ is independent of the isomorphism $Q \simeq \operatorname{End}V$, and it is a canonical element of $Q\otimes Q$. 


If $Q$ is not trivial, choose a faithfully flat extension $k'/k$ such that $Q\_{k'}$ is trivial. Because it is canonical, by descent theory the element $\tau' \in Q\_{k'}\otimes\_{k'}Q\_{k'}$ constructed above descends to an element $\tau \in Q \otimes\_k Q$ which will induce the involution $Q \otimes Q$ that exchanges $a \otimes b$ and $b\otimes a$.


