About splitting fields and primitive elements. Question 1: Find the splitting field $K$ of $x^4 + x^2 + 1$ over $\mathbb{Q}$. What is the degree of $K/\mathbb{Q}$? ($\textbf{Hint:}$ Recall that $\xi = e^{2\pi i/3}$ is such that $\xi^2 + \xi + 1 = 0$ and $cos(\pi/3) = 1/2$ and $sin(\pi/3) = \sqrt{3}/2)$
Question 2: Find a primitive element $\xi$ for $\mathbb{Q}(\sqrt{2},\sqrt{3})$. Prove your claim.
My proof: 
Let $x^2 = \xi$, then $\xi^2 + \xi + 1= 0$. Now the roots of this polynomial can be written as
            $$
   \psi_1 = e^{2\pi i/3},\psi_2 = e^{5\pi i/3}
   $$
            Hence,
            $$
   x^2 = e^{2\pi i/3}, \mbox { or } x^2 = e^{5\pi i/3}
   $$
            and
            $$
   x = \pm e^{\pi i/3}, x = \pm e^{5\pi i/6}
   $$
            Therefore, we can write the solutions of the equation as
            $$\begin{align*}
   x_1 =& \frac{1}{2} + i\frac{\sqrt{3}}{2}\\
   x_2 =& - \frac{1}{2} - i\frac{\sqrt{3}}{2} \\
   x_3 =& -\frac{\sqrt{3}}{2} + i\frac{1}{2} \\
   x_4 =& \frac{\sqrt{3}}{2} - i\frac{1}{2}
   \end{align*}$$
Now the desired splitting field is $K = \mathbb{Q}(\sqrt{3},i)$. The degree of this extension $\mathbb{Q}(\sqrt{3},i)/\mathbb{Q}$ can be calculated using this well-known result 
            $$
   [\mathbb{Q}(\sqrt{3},i):\mathbb{Q}] = [\mathbb{Q}(\sqrt{3},i):\mathbb{Q}(i)][\mathbb{Q}(i):\mathbb{Q}] = 2\times2 = 4.
   $$
Alternatively, we can see that 
            $$
   \mathbb{Q}(\sqrt{3},i) = \{a1 + b\sqrt{3} + ci + d\sqrt{3}i: a,b,c,d \in \mathbb{Q} \}
$$
Hence, $\{1,\sqrt{3},{i},\sqrt{3}i\}$ is a basis for $\mathbb{Q}(\sqrt{3},i)$ and the degree of this extension is the number of elements in it's basis which is four. $\hspace{5pt}\square$
I just want to know if my solution for question 1 is correct.
About question 2: I know that $\alpha = \sqrt{2} + \sqrt{3}$ must be a primitive element but I did not know how to prove it. Maybe the primitive element theorem can help since this extension is finite? Or I have to find some minimal polynomial to prove that $Q(\alpha) = Q(\sqrt{2},\sqrt{3})$? I was looking in Wikipedia and I found this:
"If the minimal polynomial of $\alpha$ in $Q[x]$ is $a(x) = x^4 − 10x^2 + 1 = (x - \sqrt{3} - \sqrt{2})(x + \sqrt{2} - \sqrt{3})(x - \sqrt{2} + \sqrt{3})(x + \sqrt{2} + \sqrt{3})$.
The minimal polynomial in Q[x] of the sum of the square roots of the first n prime numbers is constructed analogously, and is called a Swinnerton-Dyer polynomial."
This is very interesting for me but in my class they never discuss this theorem.
Any other way to prove this?

We want to prove that $\mathbb{Q}(\sqrt{2} + \sqrt{3}) = \mathbb{Q}(\sqrt{2}, \sqrt{3})$. Obviously, we have $\mathbb{Q}(\sqrt{2} + \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3})$, so it's enough to prove the opposite inclusion.
Consider $\mathbb{Q}(\sqrt{2} + \sqrt{3}) \ni (\sqrt{2} + \sqrt{3})^2 = 2 + 3 + 2 \sqrt{6}$. Therefore, $\sqrt{6} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$. 
Now, $\sqrt{6}(\sqrt{2}+\sqrt{3}) = \sqrt{12} + \sqrt{18} = 2\sqrt{3} + 3 \sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$. Therefore, $(2\sqrt{3} + 3 \sqrt{2}) - 2 \cdot (\sqrt{2} + \sqrt{3}) = \sqrt{3} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$, and so also obviously $\sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$.
