One result I really like about Brauer groups is that $\mathrm{Br}(\mathbb{P}_k^n)\cong\mathrm{Br}(k)$ for every possible field $k$. So the Azumaya algebras on projective space are not too wild, and completely controlled by the base field.

Yesterday I started wondering about how this result is actually proved (my knowledge of proofs of arithmetic nature is abysmal), and what would happen for twisted forms of projective space, also known as Brauer-Severi varieties. This is a very silly question, in the sense that you take a representative $A$ of an element $[A]$ of the Brauer group of the base field, construct its Brauer-Severi variety $\mathrm{BS}(A)$, and then take its Brauer group $\mathrm{Br}(\mathrm{BS}(A))$. It is like Xzibit became a mathematician and said something silly involving dogs and Brauer groups.

The Brauer group of projective space

There are different ways of proving that $\mathrm{Br}(\mathbb{P}_k^n)=\mathrm{Br}(k)$. One can take a look at the corresponding MathOverflow question or these notes of Tetsuya Uematsu.

I will sketch an approach I learnt from Colliot-Thélène's notes, which will be useful when discussing Brauer-Severi varieties. So consider the Leray spectral sequence for $\mathbb{P}_k^n\to\mathrm{Spec}\,k$ and the sheaf $\mathbb{G}_{\mathrm{m}}$. Denote a separable closure of $k$ by $k^s$. We get a long exact sequence $$ 0\to\mathrm{Pic}(\mathbb{P}_k^n)\to\mathrm{Pic}(\mathbb{P}_{k^s}^n)\to\mathrm{Br}(k)\to\mathrm{Br}(\mathbb{P}_k^n)\to\mathrm{H}^1(k_{\mathrm{et}},\mathrm{Pic}(\mathbb{P}_{k^s}^n))\to\ldots $$ where $\mathrm{Br}_1(\mathbb{P}_k^n)=\mathrm{ker}\left(\mathrm{Br}(\mathbb{P}_k^n)\to\mathrm{Br}(\mathbb{P}_{k^s}^n)\right)$ is the algebraic Brauer group. To get the desired result we need the following facts:

  1. Because $\mathbb{P}_k^n$ has a $k$-rational point, the morphism $\mathrm{Pic}(\mathbb{P}_{k^s}^n)\to\mathrm{Br}(k)$ is zero, because we can obtain a retract to the first inclusion.
  2. By some other method one needs to prove that $\mathrm{Br}(\mathbb{P}_{k^s}^n)=0$, hence the algebraic Brauer group of projective space is equal to the whole Brauer group.
  3. To prove surjectivity, we need that $\mathrm{H}^1(k_{\mathrm{et}},\mathrm{Pic}(\mathbb{P}_{k^s}^n))=0$. As explained in Colliot-Thélène's notes, one can appeal to the short exact sequence of Galois modules $$0\to\mathrm{Pic}_{\mathbb{P}_k^n/k}^0(k^s)\to\mathrm{Pic}(\mathbb{P}_{k^s}^n)\to\mathrm{NS}(\mathbb{P}_{k^s}^n)\to 0$$ for this, where in our situation we have that the first term is zero, and the third term is $\mathbb{Z}$ with trivial action, hence we indeed get vanishing, and surjectivity.

The Brauer group of Brauer-Severi varieties

A first misconception that I had is that for every variety $X$ defined over a field $k$ there exists an inclusion $\mathrm{Br}(k)\hookrightarrow\mathrm{Br}(X)$. Such a thing needs the existence of a $k$-rational point, which of course is not the case for any non-trivial Brauer-Severi variety.

I haven't found a general statement for Brauer group of Brauer-Severi varieties, I will now discuss what I know for sure in dimension 1. But I'll pose a precise question after the sketch of the proof, which might also be a proof for the general question.

The following is based on section 5.0.3 in Colliot-Thélène's notes. So let $C/k$ be a conic, where $\mathrm{char}(k)\neq 2$ to be on the safe side. Writing it as $\mathbb{V}(x_0^2-ax_1^2-bx_2^2)$ after an appropriate change of coordinates we know that it actually is the Brauer-Severi variety for the quaternion algebra $(a,b)_k$.

In this situation we still have the vanishing of $\mathrm{H}^1(k_{\mathrm{et}},\mathrm{Pic}(\mathbb{P}_{k^s}^n))$, in precisely the same way.

On the other hand, if there is no rational point (i.e. the quaternion algebra is not split), the morphism $\mathrm{Pic}(\mathbb{P}_k^n)\to\mathrm{Pic}(\mathbb{P}_{k^s}^n)$ turns out to have image $2\mathbb{Z}$. Hence we get a short exact sequence $$ 0\to\mathbb{Z}/2\mathbb{Z}\to\mathrm{Br}(k)\to\mathrm{Br}(C)\to 0 $$

and the non-zero element in $\mathrm{Br}(k)$ is precisely the quaternion algebra whose Brauer-Severi variety we are looking at. E.g. if we take the Brauer-Severi variety for the Hamilton quaternions over the field, we see that any non-split conic (they are all isomorphic anyway) over the reals has trivial Brauer group!

Based on this proof, I have the following question.

Question Is the Brauer group of a Brauer-Severi variety of a central simple algebra always the Brauer group of the base field modulo the subgroup generated by the central simple algebra?

It seems to me that the proof outlined above should give precisely this result. Am I overlooking something, or is this indeed the case?