Given $\mathbb{Q}(\sqrt{2},\sqrt{3})$ be an extension field of $\mathbb{Q}(\sqrt{6})$.
I encounter some confusion on the monic irreducible polynomial for $\sqrt{6}$ over $\mathbb{Q}(\sqrt{2},\sqrt{3})$.
Im thinking the minimal monic irreducible polynomial of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ over $\mathbb{Q}$ is given by $(x^2-2)(x^2-3)$. And for $\mathbb{Q}(\sqrt{6})$ over $\mathbb{Q}$ is $(x^2-6)$.
I cant seems to figure out the minimal monic irreducible polynomial of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ over $\mathbb{Q}(\sqrt{6})$.
any insight is highly appreciated