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?

  • $\begingroup$ Is the homomorphism you wrote down the same as the map $\mathrm{Br}(k) = \pi_0 \mathbf{Br}(k) \to \pi_1 \mathbf{Br}(k) = \mathrm{Pic}(k)$ determined by $\eta\in \pi_1(S^0)$ where $\mathbf{Br}(k)$ is the Brauer space of $k$? $\endgroup$
    – skd
    Jun 17, 2019 at 18:53
  • $\begingroup$ @skd Yes, it is. (In fact, that's my original motivation for asking.) $\endgroup$ Jun 17, 2019 at 20:30

2 Answers 2


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 $b\otimes a$.

  • 1
    $\begingroup$ Wonderful, thank you! What I find curious about this is that it uses that the twist $L \otimes L \to L \otimes L$ is the identity for $L$ invertible; I had verified this separately and it was what led me to wonder if the same were true of the twist in the Brauer group. $\endgroup$ Jun 19, 2019 at 6:26
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    $\begingroup$ Very nice! The element $\tau$ is known as the Goldman element of $Q$, see Propostion 5.1 in Saltman's "Lectures on Division Algebras", which also proves it is canonical. $\endgroup$ Jun 19, 2019 at 6:46

Here is another way to see that the given map should be zero. It corresponds to a map of sheaves of grouplike $\mathbb{E}_\infty$-spaces $K(\mathbb{G}_m,2)\rightarrow K(\mathbb{G}_m,1)$. All such maps are nullhomotopic (for example by delooping to sheaves of spectra and using a $t$-structure argument).

I would also guess that the map is given by $A\mapsto\mathrm{HH}(A/k)$. Since Hochschild homology is symmetric monoidal, $\mathrm{HH}(A/k)$ is a line bundle. The argument above shows that it is trivial.

On the other hand, I suspect there are derived Azumaya algebras where this is non-trivial. If we look at the derived Brauer sheaf (on the étale site of an ordinary commutative ring) it is an extension $$K(\mathbb{G}_m,2)\rightarrow\mathbf{dBr}\rightarrow K(\mathbb{Z},1).$$ Taking Hochschild homology gives a map of sheaves of grouplike $\mathbb{E}_\infty$-spaces $\mathbf{dBr}\rightarrow\mathbf{dPic}$, where $\mathbf{dPic}$ fits into a fiber sequence $$K(\mathbb{G}_m,1)\rightarrow\mathbf{dPic}\rightarrow K(\mathbb{Z},0).$$ The map $\mathbf{dBr}\rightarrow\mathbf{dPic}$ thus canonically factors through a map $K(\mathbb{Z},1)\rightarrow K(\mathbb{G}_m,1)$ of grouplike $\mathbb{E}_\infty$-spaces. I believe this is the map that sends $1$ to $-1\in\mathbb{G}_m(\mathbb{Z})$. Indeed, the copy of $\mathbb{Z}$ is coming from suspension in the derived category.

To make this concrete, one needs a ring $k$ where the induced map $H^1(k,\mathbb{Z})\rightarrow\mathrm{Pic}(k)$ is non-zero. I don't have an example, but I'd be surprised if this didn't exist.

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    $\begingroup$ I believe that the ring $\mathbb C[x,y](y^2 - x^2(x-1))$ (the ring of the standard affine nodal cubic) should have this property. $\endgroup$
    – Angelo
    Jun 19, 2019 at 16:05

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