Timeline for Why does the field norm on the field extension $ \mathbb C/\mathbb R $ induce a vector space norm?
Current License: CC BY-SA 4.0
11 events
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| Jul 20, 2023 at 16:21 | comment | added | KConrad | I also like a hypothesis of the form "If L does this, then L does this" and that's why I passed to the Galois closure M.:) | |
| Jul 20, 2023 at 15:28 | comment | added | LSpice | Re, also, with further thought, if you want everything in terms of $M$, you could probably show that the manually constructed abs.val.s on the various $\sigma(L)$, any two of which (necessarily) agree on the intersections of their domains, patch together to an abs.val. on $M$. I just like a hypothesis of the form "If $L$ does this, then $L$ does this", rather than "If all extensions do this, then $L$ does this", when possible. But, of course, this is not my answer, so it doesn't need to much matter what I like. | |
| Jul 20, 2023 at 2:32 | comment | added | LSpice | Re, (i) I agree that there's no simplification in the argument, only a weakening of the hypothesis. (ii) Sure, one could still use $\operatorname{Gal}(M/K)$ if desired. (iii) Certainly $\lvert\sigma^{-1}(\cdot)\rvert_L$ is an absolute value on $\sigma(L)$ extending $\lvert\cdot\rvert_K$, so, by uniqueness, it's the only one. (In particular, even though there may be multiple ways of conjugating $L$ to $\sigma(L)$, they all give the same absolute value.) | |
| Jul 20, 2023 at 1:41 | comment | added | KConrad | @LSpice I have two comments and a question. (i) An argument that makes use of all the conjugate fields $\sigma(L)$ is not that far from using the Galois closure $M$ at least implicitly since that's the smallest field extension of $K$ containing every $\sigma(L)$. (ii) Using $\sigma$ in the infinite group ${\rm Gal}(k^{\rm sep}/k)$ is more technical than using $\sigma$ in the finite group ${\rm Gal}(M/K)$ to describe the conjugate fields $\sigma(L)$. (iii) Without working in $M$, what is your argument for why $|\sigma(\alpha)| = |\alpha|$ for all $\sigma$ in ${\rm Gal}(k^{\rm sep}/k)$? | |
| Jul 19, 2023 at 21:38 | history | edited | LSpice | CC BY-SA 4.0 |
Oops, missed a typo
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| Jul 19, 2023 at 21:34 | comment | added | LSpice | … using your notation, if $L$ admits an abs.val. $\lvert\cdot\rvert_L$ extending the abs.val. on $K$, then so does every conjugate $\sigma(L)$ with $\sigma \in \operatorname{Gal}(k^\text{sep}/k)$, and, by uniqueness, the absolute value on $\sigma(L)$ must equal $\lvert\sigma^{-1}(\cdot)\rvert_L$; so we can still compute that $\lvert\alpha\rvert_L^n$ equals $\lvert\operatorname N_{L/K}(\alpha)\rvert_K$ without ever referring directly to the Galois closure $M$. | |
| Jul 19, 2023 at 21:31 | comment | added | LSpice |
While this post is on the front page, I fixed a link that had picked up a stray period. I also changed |…| to \lvert…\rvert, which behaves better in general (e.g., compare $|\sin(z)|$ $|\sin(z)|$ to $\lvert\sin(z)\rvert$ $\lvert\sin(z)\rvert$); I hope it's OK, but, if you don't like it, of course feel free to revert my edit, or just change \newcommand\abs[1]{\lvert#1\rvert} to \newcommand\abs[1]{|#1|}. \\ In your 2nd Theorem, I think we can proceed one extension at a time, rather than the universally quantified hypothesis: …
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| Jul 19, 2023 at 21:29 | history | edited | LSpice | CC BY-SA 4.0 |
Fixed link; name of link; `\abs` and `\operatorname`
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| Apr 11, 2022 at 22:44 | history | edited | KConrad | CC BY-SA 4.0 |
added 2 characters in body
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| Feb 16, 2022 at 19:03 | history | edited | KConrad | CC BY-SA 4.0 |
added 394 characters in body
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| Feb 16, 2022 at 18:54 | history | answered | KConrad | CC BY-SA 4.0 |