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.

I don't know what is your background in Galois theory, but essentially, this theory was conceived to replace tedious (and most of the time blind) polynomial calculations by natural conceptual manipulations of groups. Here we deal with quadratic and biquadratic fields. General quadratic fields are of the form $\mathbf Q(\sqrt a)$, with $a \in \mathbf Q^*, \notin \mathbf {Q^*}^2$, and it is straightforward to show that to quadratic fields $\mathbf Q(\sqrt a),\mathbf Q(\sqrt b)$ are equal iff $ab \in \mathbf {Q^*}^2$. It follows that, if $ab \notin \mathbf {Q^*}^2$, then the extension  $K=\mathbf Q(\sqrt a,\sqrt b)$, which is the compositum of two distinct quadratic fields, is actually biquadratic, i.e. galois with group $G\cong C_2 \times C_2$. Obviously $G$ contains exactly three subgroups of order two, isomorphic to $C_2$. Then the first fundamental theorem of Galois theory  gives us all the subfields of $K$, which are $\mathbf Q, K$ and  three quadratic subfields, $\mathbf Q(\sqrt a),\mathbf Q(\sqrt b),\mathbf Q(\sqrt ab)$ exactly. It follows that the subfield $\mathbf Q(\sqrt a +\sqrt b)$ must coincide with $K$.
Of course we can apply this to $a=3, b=7$, but the galois approach gives much more. For instance it answers one recurrent question in this thread, which is the determination of all the subextensions of $\mathbf Q(\sqrt p_1),...,(\sqrt p_n)$, where the $p_i$ are $n$ distinct primes, see e.g. https://math.stackexchange.com/a/3444776/300700.
