$\mathbb{Q}(\alpha)\cap\mathbb{R}=\mathbb{Q}$, 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.
 A: 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 $\Bbb Q(\alpha)\cap\Bbb R=\Bbb Q(\sqrt7)$.
A: 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 $\alpha$.
