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Say that an algebraic number $\alpha\in \mathbb{C}$ is totally real iff $\mathbb{Q}(\alpha)$ is a totally real number field.

Why does the set of all totally real numbers form a subfield of $\mathbb{R}$?

What is an example of a totally real number that is neither totally positive nor totally negative, i.e. which is positive with respect to some ordering and negative with respect to another?

anomaly
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mathematick
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    Please focus on one question per post, and include your attempts. Quick beginner guide for asking a well-received question. – Anne Bauval May 10 '23 at 10:26
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    As for your 2nd question, try $\sqrt2.$ – Anne Bauval May 10 '23 at 10:43
  • (1) Take any two totally real fields $\Bbb Q(a)$, $\Bbb Q(b)$, and for each fix one field embeeding into $\Bbb C$. Well, we land in $\Bbb R$. Let $A,B\in\Bbb R$ be the corresponding images in $\Bbb R$ of $a,b$, then there is a natural map from $\Bbb Q(a,b)$ to $\Bbb Q(a,b)$, which is a subfield of $\Bbb R$. Also, each embeeding of $\Bbb Q(a,b)$ into $\Bbb C$ is of this shape. (Take one, restrict it to $\Bbb Q(a)$, $\Bbb Q(b)$. We land in $\Bbb R$.) So $a\pm b$, $ab$, $a/b$ (if it make sense) are also totally real. (2) Let us take a look at $\sqrt 2$. Is it totally real / positive / negative? – dan_fulea May 10 '23 at 10:44
  • @dan_fulea You mean "then there is a natural map from Q(a,b) to Q(A,B)" [sending $a$ to $A$ and $b$ to $B$] but that is not true. E.g. $a=b=A=-B=\sqrt2.$ – Anne Bauval May 10 '23 at 11:57
  • @AnneBauval Yes, there was no place to arrange for the conditions on $a,b$ in the comment, one has to arrange for compatibility of embeedings in the lattice of fields. (One has to take the intersection of the fields, which is also totally real, and take two embeedings which are compatible on this intersection. Only such embeedings matter. I was starting in fact with an embeeding of the bigger field $\Bbb Q(a,b)$, then restricted it... It is what we need when trying to show that the field operations are not leaving totally real fields.) – dan_fulea May 10 '23 at 12:55
  • @dan_fulea Not quite: I think we must neither start with "for each [$\Bbb Q(a)$ and $\Bbb Q(b)$] fix one field embeeding into $\Bbb C$", nor "start with an embeeding of the bigger field $\Bbb Q(a,b)$", but rather with an embedding of $\Bbb Q(c)$ (where $c=a+b$ or $ab$), then extend it to $\Bbb Q(a,b),$ and restrict the latter to $\Bbb Q(a)$ and $\Bbb Q(b),$ to derive it is real. Btw, why is "the intersection of the fields also totally real"? (but we don't seem to need it). – Anne Bauval May 10 '23 at 13:46
  • For what is worth: I just read that "the field of all totally real algebraic numbers is the fixed field of all involutions in the absolute Galois group of $\Bbb Q$" (M. D. Fried, Dan Haran, H. Völklein, Real Hilbertianity and the field of totally real numbers, in Arithmetic Geometry, Contemp. Math. 174). I do not understand yet why it is true, but it would suffice to prove it is indeed a field. For a more pedestrian proof, see either just above Lemma 1 in this paper, or my previous comment. – Anne Bauval May 10 '23 at 15:38
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    @AnneBauval I've been working with the paper of M. D. Fried, Dan Haran and H. Völklein as well. However, it is not clear to me why the definition from above (i.e. the field containing all totally real numbers) and their definition (i.e. the fixed field of all involutions in the absolute Galois group) are the same. – mathematick May 11 '23 at 10:02
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    @AnneBauval Thank you for the nice reference! – mathematick May 11 '23 at 10:03

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