Timeline for Is any/every order on a number field forced by some finite extension?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Jan 27, 2014 at 17:33 | vote | accept | Colin McLarty | ||
| Jan 27, 2014 at 17:01 | history | edited | Emil Jeřábek | CC BY-SA 3.0 |
added 324 characters in body
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| Jan 27, 2014 at 16:37 | comment | added | Emil Jeřábek | All right. Thanks for the additional information. | |
| Jan 27, 2014 at 16:33 | comment | added | David E Speyer | I realized that a little after writing the comment; hence the edit to my comment (which I made before seeing your reply). | |
| Jan 27, 2014 at 16:32 | comment | added | Emil Jeřábek | The (more elementary, I believe) result I intended to use at the end is that for any field, elements positive under all its orderings are exactly those that are sums of squares. | |
| Jan 27, 2014 at 16:30 | comment | added | David E Speyer | Moreover, you only need four squares. Siegel proved the following (see mathoverflow.net/a/14473/297 ): Let $E$ be a number field and $x \in E$. Then we have $x = a^2+b^2+c^2+d^2$ with $a$, $b$, $c$, $d \in E$ if and only if $\sigma(x) \geq 0$ for every embedding $\sigma: E \to \mathbb{R}$. | |
| Jan 27, 2014 at 16:27 | history | answered | Emil Jeřábek | CC BY-SA 3.0 |