Let $R$ be a commutative ring, let $A$ be an Azumaya algebra over $R$, does its topological Hochschild homology coincide with that of $R$? For example, let $\mathbb{H}$ be the quaternion algebra over real numbers $\mathbb{R}$. Then what are $THH\mathbb(\mathbb{H})$ and $THH\mathbb(\mathbb{R})$？
(I learned that the usual Hochschild homology of $R$ and $A$ coincides, and wondered if the topological version detects the Brauer class.)
The answer is yes, in fact it follows from the Cortiñas-Weibel theorem.
In fact, if $R$ is a commutative connective ring spectrum and $A$ a flat Azumaya algebra over $R$, then the corresponding result follows too. To remove the word "flat" here, one would have to do some extra work, I don't know if the result is true in that generality. I believe if one removes the connectivity assumption on $R$, the result is probably wrong.
Let me explain how to deduce the flat case from Cortiñas-Weibel.
In full generality, $THH(A)$ is invertible over $THH(R)$, in fact, its tensor square is equivalent to $THH(R)$.
Now, for connective commutative $R$, $Pic(R)\to Pic(\pi_0(R))$ is an isomorphism (here, by $Pic$ I mean the "extended" Picard group, where you allow suspensions), and for connective commutative $R$, $THH(R)\to R$ is a $\pi_0$-isomorphism. It follows that $THH(A)\simeq THH(R)$ as $THH(R)$-modules if and only if the same is true after base-change to $\pi_0(R)$. But $THH(A)\otimes_{THH(R)}\pi_0(R)\simeq THH(\pi_0(R)\otimes_R A / \pi_0(R))$.
Now, as $A$ is Azumaya over $R$, $\pi_0(R)\otimes_R A$ is (derived) Azumaya over $\pi_0(R)$. If $A$ is flat, this is furthermore discrete, and so is a classical Azumaya algebra, so we are reduced to the statement: let $R$ be an ordinary commutative ring, $A$ an ordinary Azumaya algebra over $R$, then $THH(A/R)\simeq R$.
Now, using the observation that $THH(A/R)$ has a trivial tensor square and is hence concentrated in degree $0$, this $THH(A/R)$ is equivalent to its $\pi_0$, which is equivalent to ordinary $HH_0(A/R)$, which is isomorphic to $R$ by the Cortiñas-Weibel result.
As I said, I wouldn't be super surprised if one could remove the flatness assumption; it's not unlikely that Cortiñas-Weibel's proof can just be adapted (although the last bit, with the maximal étale subalgebra, doesn't seem to work so well, so I didn't try too hard to do it for this post). An easy consequence of the above is also that if $Pic(\pi_0(R))$ contains no $2$-torsion, then the result also holds, with a much simpler proof.
However, this fails in non-connective settings. For instance, over $\mathbb Q[t^{\pm 1}]$ where $t$ has degree $2$, I believe there is an Azumaya algebra $A$ with $THH(A/\mathbb Q[t^{\pm 1}])\simeq \Sigma \mathbb Q[^{\pm 1}]$.
