# totally real number field generated by square root of an algebraic integer

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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Would $d=(1+\sqrt2)^2=3+2\sqrt2$ fit the bill? – Jyrki Lahtonen Sep 5 '12 at 8:11
Yes. You are right. I edited my question though. What if degree of $d$ is $3$ or $5$? – user39519 Sep 5 '12 at 8:17

## 1 Answer

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.

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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 5 '12 at 8:24
+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 $\mathbb{Q}(\sqrt{d})$ – Hurkyl Sep 5 '12 at 8:25