Invariance of reduced trace of Azumaya algebras

Question

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.)

Answer 1

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.