# $\mathbb{Q}(\alpha)\cap\mathbb{R}=\mathbb{Q}$, where $\alpha=\sqrt{\frac{3+\sqrt{7}i}{2}}$

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.

• Work in the Galois closure. Find its Galois group. Find the subgroups of the Galois group. Find their fixed fields. Identify which corresponds to $\Bbb Q(\alpha)\cap\Bbb R$. Apr 16, 2019 at 3:34
• 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? Apr 16, 2019 at 3:41
• $\beta\in\Bbb Q(\alpha)$. Indeed $\beta=\pm2/\alpha$. Apr 16, 2019 at 4:35
• Nevermind, got it. Apr 16, 2019 at 4:49

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 $$\Bbb Q(\alpha)\cap\Bbb R=\Bbb Q(\sqrt7)$$.
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 $$\alpha$$.