Galois Theory - Normal Basis I am studying Galois Theory and have been given the following question:
Find a normal basis of $Q(\sqrt2,\sqrt3)$.
I have found the Galois group to be:
$$
\alpha_1:\sqrt2\mapsto\sqrt2, \sqrt3\mapsto\sqrt3\\
\alpha_2:\sqrt2\mapsto\sqrt3, \sqrt3\mapsto\sqrt2\\
\alpha_3:\sqrt2\mapsto-\sqrt2, \sqrt3\mapsto-\sqrt3\\
\alpha_4:\sqrt2\mapsto-\sqrt3, \sqrt3\mapsto-\sqrt2
$$
Is this correct?
I know what an orbit is and that a basis is said to be normal if it is a single orbit of the Galois group, but don't understand what is meant by $single$.
Any help would be much appreciated!
 A: The Galois group contains the four automorphisms:


*

*$p:a+b\sqrt2+c\sqrt3+d\sqrt6 \mapsto a+b\sqrt2+c\sqrt3+d\sqrt6$

*$q:a+b\sqrt2+c\sqrt3+d\sqrt6 \mapsto a-b\sqrt2+c\sqrt3-d\sqrt6$

*$r:a+b\sqrt2+c\sqrt3+d\sqrt6 \mapsto a+b\sqrt2-c\sqrt3-d\sqrt6$

*$s:a+b\sqrt2+c\sqrt3+d\sqrt6 \mapsto a-b\sqrt2-c\sqrt3+d\sqrt6$


The Galois group $G = \operatorname{Gal}(\Bbb Q(\sqrt2,\sqrt3)/\Bbb Q)$ acts on $L = \Bbb Q(\sqrt2,\sqrt3)$ in an obvious manner. The orbit of an element $x \in L$ is defined as $\{\varphi(x) \mid \varphi \in G\}$, denoted $\mathcal O(x)$.
Therefore, the question is asking you to find $x \in L$ such that $\operatorname{span}_\Bbb Q(\mathcal O(x)) = L$. Since $L$ has basis $\{1,\sqrt2,\sqrt3,\sqrt6\}$, one only need to check if those four basis elements are linear combinations of the orbit.
By inspection, let $x=1+\sqrt2+\sqrt3+\sqrt6$. Then:


*

*$1 = \frac14p(x) + \frac14q(x) + \frac14r(x) + \frac14s(x)$

*$\sqrt2 = \frac14p(x) - \frac14q(x) + \frac14r(x) - \frac14s(x)$

*$\sqrt3 = \frac14p(x) + \frac14q(x) - \frac14r(x) - \frac14s(x)$

*$\sqrt6 = \frac14p(x) - \frac14q(x) - \frac14r(x) + \frac14s(x)$


Therefore, $\mathcal O(x)$ is a normal basis of $L$.

An equivalent condition is that $(p(x),q(x),r(x),s(x))$ be linearly independent.
Taking $\{1,\sqrt2,\sqrt3,\sqrt6\}$ as basis, we write $a+b\sqrt2+c\sqrt3+d\sqrt4$ as $\begin{bmatrix}a&b&c&d\end{bmatrix}^\top$.
Then, writing each element in $(p(x),q(x),r(x),s(x))$ as a row, we obtain the matrix:
$$\begin{bmatrix} a&b&c&d \\ a&-b&c&-d \\ a&b&-c&-d \\ a&-b&-c&d \end{bmatrix}$$
Which we rewrite as:
$$\begin{bmatrix} 1&1&1&1 \\ 1&-1&1&-1 \\ 1&1&-1&-1 \\ 1&-1&-1&1 \end{bmatrix} \begin{bmatrix} a&0&0&0 \\ 0&b&0&0 \\ 0&0&c&0 \\ 0&0&0&d \end{bmatrix}$$
whose determinant is $16abcd$.
It satisfies our condition iff $16abcd\ne0$, which is iff $abcd\ne0$, which is iff none of $a,b,c,d$ is $0$.
So, any $x = a+b\sqrt2+c\sqrt3+d\sqrt6$ works as long as $a$, $b$, $c$, $d$ are all non-zero.
A: Your field is biquadratic, with Galois group $C_2\times C_2$ (and you wrote it down explicitly). As was stressed by Jyrki Lahtonen, a normal basis will consist of the conjugates of a single element. Just try $\sqrt 2 +\sqrt 3$ 
EDIT : I made an error of computation, I shouldn't work too early in the morning ! It is obvious that $\sqrt 6$ cannot be obtained as a $\mathbf Q$-linear combination of the conjugates of $\sqrt 2 +\sqrt 3$, although this is a primitive element of the biquadratic extension. 
