Timeline for Subfields of division rings of degree $2$ which are not invariant
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| S Sep 23, 2023 at 15:49 | history | bounty ended | THC | ||
| S Sep 23, 2023 at 15:49 | history | notice removed | THC | ||
| Sep 23, 2023 at 15:49 | vote | accept | THC | ||
| Sep 19, 2023 at 9:06 | history | edited | Tom De Medts |
Added ra.rings-and-algebras tag
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| Sep 18, 2023 at 9:10 | comment | added | Tom De Medts | A small remark about terminology: the word "degree" for central simple algebras is used for the square root of the dimension (because the dimension over the center is always a square), so it is better to use "dimension" instead. For instance, quaternion algebras have degree $2$. | |
| Sep 18, 2023 at 9:07 | answer | added | Tom De Medts | timeline score: 4 | |
| Sep 18, 2023 at 5:16 | comment | added | YCor | Just a remark: you're considering the normalizer $N_{A^*}(B^*)$ of $B^*$ in $A^*$, and asking whether there are examples with $N_{A^*}(B^*)/B^*=\{1\}$. In the case of $\mathbf{C}\subseteq\mathbf{H}$ you are mentioning, this quotient is a 2-element group. | |
| S Sep 17, 2023 at 21:01 | history | bounty started | THC | ||
| S Sep 17, 2023 at 21:01 | history | notice added | THC | Authoritative reference needed | |
| Sep 15, 2023 at 11:46 | comment | added | THC | @Eoin: can you make your remark into an answer (with more details) ? | |
| Sep 15, 2023 at 3:03 | comment | added | Kimball | What are you working over? If you are working over a field, isn't B forced to be a quadratic field extension? | |
| Sep 14, 2023 at 18:04 | comment | added | Eoin | At least going by your non-example, any commutative and separable $B$ will have a nontrivial automorphism which is stable under conjugation by an element of $A\setminus B$ by Skolem-Noether (stacks.math.columbia.edu/tag/074P).\\ On the other hand, if $B$ is a purely inseparable extension of degree 2 inside, say $A$ a quaternion division algebra over characteristic 2 field, then every automorphism of $B$ is trivial. If some element $x\in A$ conjugated $B$ to $B$ then it would be the identity on $B$, hence in the centralizer of $B$, which is $B$ in this case. | |
| Sep 14, 2023 at 17:40 | history | asked | THC | CC BY-SA 4.0 |