Prove that $\sqrt{3} + \sqrt{7}$ is a primitive element of $\mathbb{Q}(\sqrt{3},\sqrt{7})$ I want to find a primitive element of $\mathbb{Q}(\sqrt{3},\sqrt{7})$ over $\mathbb{Q}$. I try with $\sqrt{3} + \sqrt{7}$.
Let $X = \sqrt{3} + \sqrt{7}$ we get
$$ X^4 - 20X^2 + 16 = 0 $$
We can prove that $[\mathbb{Q}(\sqrt{3},\sqrt{7}) : \mathbb{Q}] = 4$ since $\sqrt{3}$ and $\sqrt{7}$ are the roots of $X^4 - 10X^2 + 21$ and it is irreducible over $\mathbb{Q}$ by the Eisenstein criterion with $p = 3$.
Let $P(X) = X^4 - 20X^2 + 16$. Since $\sqrt{3} + \sqrt{7}$ is a root of $P$ then $[\mathbb{Q}(\sqrt{3}+\sqrt{7}) : \mathbb{Q}] \leq 4$.
If we prove that $P$ is irreducible over $\mathbb{Q}$ then the degree is equal to 4. But the Eisenstein criterion does not seem to work...
So, I try another thing. Let $\alpha = \sqrt{3} + \sqrt{7}$, we get $\alpha^2  = 10 + 2 \sqrt{21}$ and $\mathbb{Q} \subset \mathbb{Q}(10 + 2 \sqrt{21}) \subset \mathbb{Q}(\sqrt{3} + \sqrt{7})$. Now
$$ [\mathbb{Q}(10 + 2 \sqrt{21}) : \mathbb{Q}] = 2 \Rightarrow 2 \mid [ \mathbb{Q}(\sqrt{3} + \sqrt{7}) : \mathbb{Q} ] $$
The degree must be 1, 2 or 4.
But  $\mathbb{Q}(10 + 2 \sqrt{21}) \subset \mathbb{Q}(\sqrt{3} + \sqrt{7})$ and $ [ \mathbb{Q}(\sqrt{3} + \sqrt{7}) : \mathbb{Q}(10 + 2 \sqrt{21}) ] > 1$ then the degree must be 2 or 4. I don't know how to conclude with the formula
$$ [ \mathbb{Q}(\sqrt{3} + \sqrt{7}) : \mathbb{Q} ] = [\mathbb{Q}(\sqrt{3} + \sqrt{7})  :\mathbb{Q}(10 + 2 \sqrt{21}) ] [\mathbb{Q}(10 + 2 \sqrt{21}) : \mathbb{Q}]  $$
Thank you. I think I did a mistake somewhere.

Cubing $\alpha=\sqrt 3+\sqrt 7$, you obtain the linear system
\begin{cases}
\alpha=\sqrt 3+\sqrt 7, \\[1ex]
\alpha^3=24\sqrt 3+16\sqrt7,
\end{cases}
from which can easily deduce $\sqrt 3$ and $\sqrt 7$ as a linear combination of $\alpha$ and $\alpha^3$.
