# What does the tensor product of two central simple algebras correspond to geometrically?

Let $k$ be a field, assumed to have characteristic $0$ for simplicity (though this probably isn't necessary).

Let $A$ be a central simple algebra over $k$ of dimension $n^2$. Then the collection of left ideals of rank $n$ in $A$ may be given the structure of a variety $B(A)$, which turns out to be a Brauer-Severi variety (i.e. a form of projective space) of dimension $n-1$. Moreover a classical theorem states that every Brauer-Severi variety of dimension $n-1$ arises in this way.

Next recall that given two central simple algebras $A_1$ and $A_2$ over $k$, their tensor product $A_1 \otimes_k A_2$ is also a central simple algebra over $k$. My question concerns how the corresponding Brauer-Severi varieties are related.

How may one visualise $B(A_1 \otimes_k A_2)$ in terms of $B(A_1)$ and $B(A_2)$? In particular, is there a geometrical operation that one may perform on $B(A_1)$ and $B(A_2)$ to obtain $B(A_1 \otimes_k A_2)$?

This question is slightly vague; so let me give you an example of the kind of thing I want, but which unfortunately does not work. Namely, the only natural construction which I can think of here is the fibre product. But $B(A_1) \times B(A_2)$ is not isomorphic to $B(A_1 \otimes_k A_2)$ for two obvious reasons:

• It has the wrong dimension.

• It is isomorphic to a product of projective spaces, and not projective space, over an algebraic closure $\overline{k}$ of $k$.

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As a first guess, I would think that $B(A_1 \otimes_k A_2)$ corresponds to the form of projective space the product $B(A_1) \times B(A_2)$ Segre embeds into. –  Michael Stoll Apr 8 '13 at 20:21
Your guess is clearly correct. A maximal left ideal of $A_1$ and a maximal left ideal of $A_2$ combine to give a maximal left ideal of $A_1 \otimes_k A_2$, and for $k$ algebraically closed this is the Segre embedding. –  Will Sawin Apr 8 '13 at 22:29
@Michael and Will: Thanks for your comments. This shows that there is always a morphism $B(A_1) \times B(A_2) \to B(A_1 \otimes_k A_2)$. It seems quite possible that $B(A_1 \otimes_k A_2)$ is in some respects determined by this morphism, i.e. this morphism should satisfy some kind of universal property. Perhaps $B(A_1 \otimes_k A_2)$ is the Brauer-Severi variety of smallest dimension that $B(A_1) \times B(A_2)$ embeds into? –  Daniel Loughran Apr 9 '13 at 7:33
I'd guess you take the product of the cones over the varieties and divide out by the diagonal action of $\mathbb{G}_m$. –  anon Apr 10 '13 at 14:01