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How can I show $\mathbb{Q}(\alpha)\cap\mathbb{R}=\mathbb{Q}$ where $\alpha=\sqrt{\frac{3+\sqrt{7}i}{2}}$?

$\alpha$ is a root of a degree 4 irreducible polynomial over $\mathbb{Q}$, so $\mathbb{Q}(\alpha):\mathbb{Q}$ is a degree 4 extension and has as basis $1,\alpha,\alpha^2,\alpha^3$. We can write an arbitrary non-rational element of $\mathbb{Q}(\alpha)$ as a $\mathbb{Q}$-linear combination of these basis elements, but after that I'm not sure how to proceed.

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  • $\begingroup$ Work in the Galois closure. Find its Galois group. Find the subgroups of the Galois group. Find their fixed fields. Identify which corresponds to $\Bbb Q(\alpha)\cap\Bbb R$. $\endgroup$ – Lord Shark the Unknown Apr 16 at 3:34
  • $\begingroup$ I'm actually trying to use this in order to determine the galois group. This is the only missing step I have, so how can I prove this directly? $\endgroup$ – user500144 Apr 16 at 3:41
  • $\begingroup$ $\beta\in\Bbb Q(\alpha)$. Indeed $\beta=\pm2/\alpha$. $\endgroup$ – Lord Shark the Unknown Apr 16 at 4:35
  • $\begingroup$ Nevermind, got it. $\endgroup$ – user500144 Apr 16 at 4:49
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If you are really interested in determining the Galois group, observe that the Galois closure is $L=K(\alpha,\beta)$, where $K=\Bbb Q(i\sqrt7)$ and $\alpha^2=\frac12(3+i\sqrt7)$ and $\beta^2=\frac12(3-i\sqrt7)$. Thus $L/K$ is a Kummer extension, and its degree is the order of the subgroup of $K^\times/(K^\times)^2$ generated by the $\frac12(3\pm i\sqrt7)$.

The norm of $\frac12(3+i\sqrt7)$ is $4$ which suggests it may be a square of a norm $2$ algebraic integer. But $\left[\frac12(1-i\sqrt7)\right]^2= \frac12(-3-i\sqrt7)$. Thus $\alpha=\pm i\frac12(1-i\sqrt7)$. Likewise $\beta=\pm i\frac12(1+i\sqrt7)$.

Then $L=\Bbb Q(\alpha)=\Bbb Q(i,\sqrt7)$. This has Galois group $(C_2)^2$. Also $\Bbb Q(\alpha)\cap\Bbb R=\Bbb Q(\sqrt7)$.

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To find the Galois group, observe that $\alpha^2 = \frac{3+\sqrt{7}i}{2}$, so that $(2\alpha^2 - 3)^2 + 7 = 0$, which can be rewritten as $\alpha^4 - 3\alpha^2 + 4 = 0$. The roots of this will then be $\alpha, -\alpha, \bar{\alpha}, -\bar{\alpha}$. Since $\alpha\bar{\alpha} = 2$, these are $\alpha, -\alpha, {2 \over \alpha}, -{2 \over \alpha}$. Hence $\mathbb{Q}(\alpha)$ is the splitting field of $\alpha^4 - 3\alpha^2 + 4$ and is thus a Galois extension of degree 4.

So the Galois group is either $\mathbb{Z_2} \times \mathbb{Z_2}$ or $\mathbb{Z_4}$. Since the Galois group permutes the roots, to verify it's $\mathbb{Z_2} \times \mathbb{Z_2}$ you just have to verify that each possible automorphism of $\mathbb{Q}(\alpha)$ has order two. This follows from just listing the possible images of $\alpha$ and observing that repeating the automorphism always gives you back $\alpha$.

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