Timeline for why are subextensions of Galois extensions also Galois?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 26, 2023 at 23:48 | comment | added | Will Sawin | @ZachTeitler Yes, good point. | |
| Aug 26, 2023 at 23:48 | history | edited | Will Sawin | CC BY-SA 4.0 |
edited body
|
| Aug 26, 2023 at 23:36 | comment | added | Zach Teitler | You wrote "$x$ is fixed by every automorphism" but then refer to $s$. It seems like a typo and I wanted to ask if that's what it was; sorry, I didn't mean to be so terse... | |
| Aug 26, 2023 at 22:12 | comment | added | Zach Teitler | In this proof $x=s$? | |
| Feb 1, 2013 at 16:41 | comment | added | Will Sawin | I can't think of an easy proof. There is a proof that is standard in Galois theory, but I think that just makes this the regular proof in new clothing. | |
| Feb 1, 2013 at 14:23 | comment | added | Eric Wofsey | Is it obvious that if there are "enough" automorphisms then there must be $[L:K]$ of them? | |
| Feb 1, 2013 at 3:31 | comment | added | Will Sawin | Yes. Obviously it being a tensor product does not say much. | |
| Feb 1, 2013 at 3:23 | comment | added | David Benjamin Lim | Dear Will, when you say <<$L \otimes_K L$ is a product of copies of L>>, by product you mean for example $L \times L$? Thanks. | |
| Feb 1, 2013 at 0:58 | history | answered | Will Sawin | CC BY-SA 3.0 |