Galois groups of $\mathbb{Q}(\alpha)$ for $\alpha = \sqrt{a+b\sqrt{c}}$ where $a,b,c \in \mathbb{Q}$ I'm revising for a Galois Theory exam, and can't manage the last part of a question on a past paper.
I have previously shown that $f(x) = x^4 - 2ax^2 + (a^2-b^2c)$ is a rational polynomial with root $\alpha = \sqrt{a+b\sqrt{c}}$. As usual, I know the other roots of this polynomial are $-\alpha, \beta = \sqrt{a-b\sqrt{c}}$ and $-\beta$
The next part of the question then states:

Suppose now $f$ is irreducible. Show that $G = \text{Gal}(L/ \mathbb{Q})$ (for $L$ the splitting field of $f$) is the Klein four-group $V$ if and only if $\alpha$ can be expressed in the form $\alpha = \sqrt{u} + \sqrt{v}$ for $u,v \in \mathbb{Q}$

I know that if the Galois group is $V$ then the automorphisms must be the identity, $±\sqrt{a±b\sqrt{c}} \mapsto \mp\sqrt{a±b\sqrt{c}}$, $±\sqrt{a±b\sqrt{c}} \mapsto \pm\sqrt{a\mp b\sqrt{c}}$ and $±\sqrt{a±b\sqrt{c}} \mapsto \mp\sqrt{a \mp b\sqrt{c}}$.
How then do I go about showing that $\alpha = \sqrt{u} + \sqrt{v}$? I am unsure as to how I can find a link here, and a hint would be appreciated! 
I am unsure on the other direction also.
 A: If the Galois group is $ V_4 $, then the elements
$$ \sqrt{a + b \sqrt{c}} + \sqrt{a - b \sqrt{c}} $$
$$ \sqrt{a + b \sqrt{c}} - \sqrt{a - b \sqrt{c}} $$
are both contained in a quadratic extension of $ \mathbf Q $, since they are fixed by an index $ 2 $ subgroup of the Galois group. Moreover, they have zero trace. Since an element of the form $ x + y \sqrt{d} $ for $ x, y, d \in \mathbf Q $ and  $ d $ not a perfect square has zero trace iff $ x = 0 $ (and having zero trace is a property that is preserved when passing to a subextension), it follows that we may find $ d_1, d_2 \in \mathbf Q $ such that 
$$ \sqrt{a + b \sqrt{c}} + \sqrt{a - b \sqrt{c}} = \sqrt{d_1} $$
$$ \sqrt{a + b \sqrt{c}} - \sqrt{a - b \sqrt{c}} = \sqrt{d_2} $$
It follows that
$$ \sqrt{a + b \sqrt{c}} = \frac{\sqrt{d_1} + \sqrt{d_2}}{2} = \sqrt{\frac{d_1}{4}} + \sqrt{\frac{d_2}{4}} $$
The other direction has been outlined in the comments, but I provide it here for the sake of completion: $ \sqrt{u} + \sqrt{v} $ can have four conjugates at most, and since its minimal polynomial is assumed to be of degree $ 4 $, it follows that it has exactly four conjugates; and thus the Galois group of $ \mathbf Q(\sqrt{u} + \sqrt{v})/\mathbf Q $ is of order $ 4 $. Now, simply look at where the automorphisms take the primitive element to deduce that the Galois group is $ V_4 $; all of the "obvious" possibilities must yield automorphisms by order counting.
