Prove that the polynomial $f(X)=X^5-9X^3+15X+6$ is irreducible over $\mathbb{Q}\left(\sqrt{2},\sqrt{3}\right)$
Apply Eisenstein's Irreducibility Criterion with $p=3$ we see that $f$ is irreducible over $\mathbb{Q}$. Can conclude that $f$ is irreducible over $\mathbb{Q}\left(\sqrt{2},\sqrt{3}\right)$?
You can't directly conclude that $f$ is irreducible over $\mathbb{Q}(\sqrt{2},\sqrt{3})$ from its irreducibility over $\mathbb{Q}$. But since the degree of $f$ is $5$, you can conclude that for any zero $\alpha$ of $f$,
$$[\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha):\mathbb{Q}(\alpha)]\cdot [\mathbb{Q}(\alpha):\mathbb{Q}]$$
is a multiple of $5$. On the other hand,
$$\begin{align} [\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha):\mathbb{Q}] &= [\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha):\mathbb{Q}(\sqrt{2},\sqrt{3})]\cdot [\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]\\ &= 4[\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha):\mathbb{Q}(\sqrt{2},\sqrt{3})], \end{align}$$
so
$$5 \mid [\mathbb{Q}(\sqrt{2},\sqrt{3},\alpha):\mathbb{Q}(\sqrt{2},\sqrt{3})].$$
It is not hard to see that $[{\mathbb Q}(\sqrt{2},\sqrt{3}):{\mathbb Q}]=4$. Since $4$ and $5$ are coprime we are done, because of the classical result proved here.
