Galois extension of degree $2^n$ Let $k$ be a field and $L=k(\sqrt{a_1},\sqrt{a_2},\cdots,\sqrt{a_n})$ with $a_i\in  k$. Suppose $[L:k]=2^n$. Let $a=\sqrt{a_1}+\sqrt{a_2}+\cdots\sqrt{a_n}$. Show that $L=k(a)$.
 A: Let $t=\sqrt{a_1}+\cdots +\sqrt{a_n}$ then we have 
$$k \subseteq k(t)\subseteq k(\sqrt{a_1}, \ldots , \sqrt{a_n})$$
How many conjugates does $t$ have ? Clearly the $2^n$ expressions
$$\pm\sqrt{a_1}\pm\sqrt{a_2}\cdots \pm \sqrt{a_n}$$ are all conjugates, we need only see that they are distinct. If however two were equal by subtracting  them  we get an expression of the form  $\sqrt{a_{i_1}}+\cdots +\sqrt{a_{i_k}}=0$ and this would contradict the assumption. So $t$ has $2^n$ conjugates, so its minimal polynomial must have degree ate least $2^n$.  
A: First, we note that $ L/k $ is a splitting field over a perfect field, and it is therefore a Galois extension and we have $ |\textrm{Gal}(L/k)| = [L : k] = 2^n $. It is evident that the $\sqrt{a_i} $ are linearly independent over $ k $ from the degree of the extension. An automorphism $ \sigma \in \textrm{Gal}(L/k) $ is then uniquely determined by its values at $ \sqrt{a_i} $. However, each of these have only two $k$-conjugates (counting themselves), so there are at most $2^n $ possible values to assign values to each of them, giving rise to $ 2^n $ distinct automorphisms. But the Galois group contains precisely $ 2^n $ automorphisms, therefore any choice of values produces an automorphism.
Now, let $ P $ be the minimal polynomial of $ \alpha = \sum \sqrt{a_i} $ over $ k $, then the Galois group sends a root to another root. However, we can pick an automorphism $ \sigma $ such that
$$ \sigma(\alpha) = \sum \sigma(\sqrt{a_i}) = \sum c_i \sqrt{a_i} $$
for any choice of the $ c_i $ from the set $ \{-1, 1\} $. These values are all distinct (otherwise the square roots would be linearly dependent), and therefore $ P $ has at least $2^n $ distinct roots, implying that it is of degree $ 2^n $ (the field extension cannot contain an element whose degree is greater than the degree of the extension). This means that the set $ S = \{ 1, \alpha, \alpha^2, \ldots, \alpha^{2^n-1} \} $ is a linearly independent set, and since it has $ 2^n $ elements it spans $ L/k $, showing that $ L = k(\alpha) $. (This last argument can also be presented by noting $ [L:k] = [L:k(\alpha)][k(\alpha):k] $ and so $ [L : k(\alpha)] = 1 $.)
