Let $d$ be a real positive algebraic integer of degree $3$ or $5$. Assume that $\mathbb{Q}(d)$ and $\mathbb{Q}(\sqrt{d})$ are totally real number fields. Is there a possible $d$ which makes that $\mathbb{Q}(d)=\mathbb{Q}(\sqrt{d})$?
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4Would $d=(1+\sqrt2)^2=3+2\sqrt2$ fit the bill? – Jyrki Lahtonen Sep 05 '12 at 08:11
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Yes. You are right. I edited my question though. What if degree of $d$ is $3$ or $5$? – user39519 Sep 05 '12 at 08:17
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Let $c=2\cos(2\pi/7)$. then $\mathbb{Q}(c)$ is a totally really cubic field. Let $d=c^2\notin\mathbb{Q}$. Then $\mathbb{Q}(d)$ cannot be a proper subfield of $\mathbb{Q}(c)$, because then the latter would have an even degree.
In the quintic case the example $c=2\cos(2\pi/11)$, $d=c^2$, works for much the same reason.
Jyrki Lahtonen
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Hmm. In the odd degree case, wouldn't just about any non-rational square within a totally real field work? The even degree is a more interesting case. Gotta go now. Hopefully somebody can pick this up. – Jyrki Lahtonen Sep 05 '12 at 08:24
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1+1. It's amazing what a difference in obviousness you can get from such a tiny change as considering $\mathbb{Q}(c)$ and its subfield $\mathbb{Q}(c^2)$ rather than $\mathbb{Q}(d)$ and its extension field $\mathbb{Q}(\sqrt{d})$ – Sep 05 '12 at 08:25
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Jyrki, if you can stand the excitement: https://math.stackexchange.com/questions/3926383/does-there-getting-from-1-to-sqrt42-using-sqrt-alpha2-1 is about totally real fields, although I think that OP knows less than I do about that aspect. – Will Jagy Nov 29 '20 at 02:26