Let $A$ be an Azumaya algebra over the commutative ring $R$ and let $\newcommand{\tr}{\operatorname{tr}} \tr\colon A \to R$ be the reduced trace as defined (for example) in Section IV.2 of M.-A. Knus, M. Ojanguren, Théorie de la Descente et Algèbres d'Azumaya (doi:10.1007/BFb0057799). Let $g$ be an automorphism of $A$ as ring, but not necessarily as $R$-algebra. Since we may identify $R$ with $Z(A)$, $g$ induces an automorphism of $R$. I need a reference/proof that the following holds:

$\tr(a^g) = \tr(a)^g $ for all $a\in A$.

In trying to prove this myself, I came up with the following.

  1. There is a unique element $t=\sum x_i \otimes y_i \in A\otimes A $ such that $\tr(a) = \sum x_iay_i $ for all $a\in A$. This element has the properties $t^2 = 1$ and $t (a\otimes b) = (b\otimes a) t$. The element $t^g$ has the same property and so $t^g = rt $ for some $r\in R$ and $r^2=1$. This means that $\tr(a^g)= r\tr(a)^g$. It remains to show $r=1$.

  2. Another attempt: It is not difficult to reduce to the case where $A$ has constant rank. Let $S$ be a faithfully flat extension such that $A\otimes S \cong \mathbf{M}_n(S)$. If $g_{|R}$ extends from $R$ to a ring automorphism of $S$, then we can extend $g$ to $A\otimes S$ in an obvious way. Since trace and scalar extensions commute, this reduces the question to the case where $A$ is a matrix ring. In that case we can compose $g$ with the automorphism of $A\cong \mathbf{M}_n(R)$ that acts as $g^{-1}$ on the entries of a matrix. We get an $R$-algebra automorphism of $A$, and for these the result follows from Lemme IV.2.2 in Knus-Ojanguren.
    However, I do not know whether there always is a faithfully flat extension $S$ that splits $A$ and such that $g_{|R}$ extends to $S$. (In my application, $g$ has finite order, if that helps.)


I would break this into two steps. Step 1. Let $A$ and $B$ be Azumaya algebras over $R$, and let $f:A\to B$ be an isomorphism of $R$-algebras. Then for every $a$ in $A$, $\text{tr}_{B/R}(f(a))$ equals $\text{tr}_{A/R}(a)$. Step 2. Let $A$ be an Azumaya algebra over $R$, and let $e:R\to R'$ be a homomorphism of commutative, unital rings. Denote by $E:A\to A'$ the induced homomorphism, where $A'$ equals $R'\otimes_R A$. Then $\text{tr}_{A'/R'}(E(a))$ equals $e(\text{tr}_{A/R}(a))$. Combining these two, given a homomorphism $e:R\to R'$, and given a homomorphism of Azumaya algebras over $R'$, $f:A'\to B$, then $\text{tr}_{B/R'}(f(E(a)))$ equals $e(\text{tr}_{A/R}(a))$. Apply this now in the special case that $R'$ equals $R$, $e$ is the restriction to $R$ of $g$, $B$ equals $A$, and the homomorphism $f:R\otimes_{e,R}A \to A$ is the unique $R$-algebra homomorphism induced by $g$.

Of course it remains to prove Step 1 and Step 2. I claim that each of these follows easily from the usual construction of $\text{tr}_{A/R}$ as the composition of the natural $R$-algebra homomorphism $L_{A/R}:A\to \text{Hom}_{R-\text{mod}}(A,A)$ coming from left multiplication with the trace homomorphism $\text{tr}:\text{Hom}_{R-\text{mod}}(A,A) \to R$. For $a$ in $A$, the $R$-module homomorphism $$f\circ L_{A/R}(a):A\to B$$ equals the $R$-module homomorphism $$L_{B/R}(f(a))\circ f:A\to B.$$ Indeed, this is equivalent to the identity $f(a\cdot x) = f(a)\cdot f(x)$. Because of this, $L_{B/R}(f(a))$ equals $f\circ L_{A/R}(a)\circ f^{-1}$. Thus, by the usual identities for trace, $\text{tr}_{B/R}(f(a))$ equals $\text{tr}_{A/R}(a)$.

Step 2 is similar, the $R'$-module homomorphism $$L_{A'/R'}(E(a)):R'\otimes_R A \to R'\otimes_R A$$ is the unique $R'$-module homomorphism induced by the $R$-module homomorphism $$L_{A/R}(a):A\to A.$$ Thus, taking any $R$-module basis $(x_i)$ for $A$ (which you can do Zariski locally, at least), the matrix representative of $L_{A'/R'}(E(a))$ with respect to $(1\otimes x_i)$ is the matrix of $R'$-entries obtained by applying $e$ to the entries of the matrix representative of $L_{A/R}(a)$. Thus the identity holds.

  • $\begingroup$ Thanks for your answer. Your're not using the reduced trace of $A$, however. To get the reduced trace, you have to chose a faithfully flat extension $S$ of $R$ such that $A\otimes S \cong \text{End}_S(P)$ for some projective $S$-module $P$. That said, I think your steps remain correct for the reduced trace, too. I see this for Step 1. As for Step 2, this probably follows from the same kind of argument one uses to show that the reduced trace is well-defined. So I think I'll have another look at the definition of the reduced trace. $\endgroup$ – Frieder Ladisch Aug 12 '13 at 19:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.