Timeline for Multiplication law in a central simple algebra of dimension 9 over a global field
Current License: CC BY-SA 4.0
6 events
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| Feb 5, 2023 at 22:50 | comment | added | GreginGre | @MikhailBorovoi: Note that, since all the local invariants are killed by 3, your $D$ has exponent (and then index ) dividing $3$. Hence $D\simeq M_3(D')$, where $D'$ is a central simple algebra of degree $3$. I assume that not all your local invariants are $0$, so $D'$ is actually division. It might simplify your computation... | |
| Mar 16, 2021 at 17:52 | history | edited | Mikhail Borovoi | CC BY-SA 4.0 |
added 5 characters in body
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| Mar 16, 2021 at 17:50 | comment | added | Mikhail Borovoi | @WillSawin: Thank you, this is very helpful! | |
| Mar 16, 2021 at 15:48 | comment | added | Will Sawin | This will amount to a congruence condition at the finitely many primes with nontrivial invariants and the finitely many ramified primes of $L$, plus the condition that at every non-split unramified prime $v$ with $i_v=0$, the $v$-adic valuation of the element is a multiple of $3$. I don't think one necessarily has a formula for such an element in terms of the invariants so I don't know how explicit one can make it, but one can certainly make it more explicit than this. | |
| Mar 16, 2021 at 15:46 | comment | added | Will Sawin | I don't know the exact answer but I can tell you the form it will probably have. Take a cyclic cubic field $L$ over $k$ which is not split (either ramified or inert) at every place $v$ where $i_v \neq 0$. The algebra will be generated by $L$ and $\sigma$ where conjugation by $\sigma$ acts on $L$ as a generator of the Galois group, and $\sigma^3 \in L$ is a particular element. The main game is to choose this element to get the right invariants. | |
| Mar 16, 2021 at 11:17 | history | asked | Mikhail Borovoi | CC BY-SA 4.0 |