Timeline for What does the tensor product of two central simple algebras correspond to geometrically?
Current License: CC BY-SA 3.0
6 events
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| Apr 10, 2013 at 14:01 | comment | added | anon | I'd guess you take the product of the cones over the varieties and divide out by the diagonal action of $\mathbb{G}_m$. | |
| Apr 10, 2013 at 11:05 | history | edited | Daniel Loughran | CC BY-SA 3.0 |
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| Apr 9, 2013 at 7:33 | comment | added | Daniel Loughran | @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? | |
| Apr 8, 2013 at 22:29 | comment | added | Will Sawin | 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. | |
| Apr 8, 2013 at 20:21 | comment | added | Michael Stoll |
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.
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| Apr 8, 2013 at 18:19 | history | asked | Daniel Loughran | CC BY-SA 3.0 |