TITLE: Prove that $\sqrt{3\pm\sqrt{7}} \not\in \mathbb{Q}(\sqrt{3\mp\sqrt{7}})$.
QUESTION [5 upvotes]: I'm currently solving a fairly long exercise related to Galois theory in which I've come across having to prove that $\sqrt{3+\sqrt{7}} \not\in \mathbb{Q}(\sqrt{3-\sqrt{7}})$ and $\sqrt{3-\sqrt{7}} \not\in \mathbb{Q}(\sqrt{3+\sqrt{7}})$. So far I haven't been able to find an "easy" or simple and understandable way to do so given that this isn't the main part of the problem. 
Any help is appreciated!

REPLY [2 votes]: In general, in this kind of problem, it is better not to mix the two operations + and $\times$. Let me give an illustration here, using only $\times$. Introduce the quadratic field $k=\mathbf Q(\sqrt 7)$. Adopting the notation $x_{\pm}=\sqrt {3 \pm \sqrt 7}$ suggested by @Mindlack, let us write $K_{\pm}=k(x_{\pm})$. These are two extensions of $k$ of degree at most $2$ :

if  $K_{+}$ or $K_{-} =k$, i.e. $(3 \pm\sqrt 7)\in {k^*}^2$, norming down to $\mathbf Q$ shows that $N(3\pm\sqrt 7)=2$ is a square in $\mathbf Q^*$: impossible
if both degrees are 2, $K_{\pm}\subset K_{\mp}$ iff $K_{\pm}= K_{\mp}$, iff $k(x_{+})= k(x_{-})$, iff $2=(3+\sqrt 7)(3-\sqrt 7)\in {k^*}^2$  (no specific calculation, this is rudimentary Kummer theory over $k$), iff $\mathbf Q(\sqrt 2)=\mathbf Q(\sqrt 7)$, iff $2.7$ is a square in $\mathbf Q^*$(again by Kummer): impossible because $\mathbf Z$ is a UFD  ./.