$X^2-q \in \mathbb{Q}(\sqrt{p})[X]$ is irreducible and $[\mathbb{Q}(\sqrt{p}, \sqrt{q}): \mathbb{Q}]=4$ I don't know how to prove the following:
Let $p \neq q$ be prime numbers, then $\sqrt{p}, \sqrt{q} \in \mathbb{R}$ and $\mathbb{Q}(\sqrt{p}),\mathbb{Q}( \sqrt{q}),\mathbb{Q}(\sqrt{p}, \sqrt{q}) \subset \mathbb{R}$.
Show that $X^2-q \in \mathbb{Q}(\sqrt{p})[X]$ is irreducible, $[\mathbb{Q}(\sqrt{p}, \sqrt{q}): \mathbb{Q}]=4$ and that $1, \sqrt{p}, \sqrt{q}, \sqrt{pq}$ is a basis for $\mathbb{Q}(\sqrt{p}, \sqrt{q})$.
I keep finding posts about Galois groups but we didn't introduce that concept yet.
Thanks in advance for any help.
 A: Notice that $X^2 - q$, $X^2 - p$ are irreducible over $\mathbb{Q}$ because they have no roots, since any root would imply that $\sqrt{p}$ and $\sqrt{q}$ are rational numbers. Using the same proof as for the irrationality of $\sqrt{2}$, we know this isn't true.
We then just need to show that $X^2 - q$ remains irreducible over $\mathbb{Q} (\sqrt{p})$. Again, it suffices to show there are no roots. A general element of $\mathbb{Q} (\sqrt{p})$ looks like $a + b \sqrt{p}$ for $a, \ b \in \mathbb{Q}$; plugging this in:
$$(a+ b \sqrt{p})^2 - q = 0 \implies a^2 + b^2 p - q + 2 ab \sqrt{p} = 0$$
which tells us $ab = 0$ and $a^2 + b^2 p = q$. This means $a=0$ or $b=0$. In the first case, we find that $p$ divides $q$, a contradiction. In the second case, we find that $a^2 = p$ for some rational number, again, impossible. This means that $[\mathbb{Q} ( \sqrt{p} , \sqrt{q} ) : \mathbb{Q} ( \sqrt{p} ) ] =2$. Using the tower law,
$$[\mathbb{Q} ( \sqrt{p} , \sqrt{q} ) : \mathbb{Q} ] = [\mathbb{Q} ( \sqrt{p} , \sqrt{q} ) : \mathbb{Q} ( \sqrt{p} ) ] \cdot [ \mathbb{Q} ( \sqrt{p}) : \mathbb{Q} ] = 4$$
Then to show that the set mentioned in the original question is a basis, it suffices to show $\mathbb{Q}$-linear independence. I leave that to you.
