Question. $\mathbb{Q}(\sqrt{2},\sqrt{3})$ is isomorphic to $\mathbb{Q}[x]/((x^2-2)(x^2-3))$? I know that $\mathbb{Q}(\sqrt{2})\simeq \mathbb{Q}[x]/(x^2-2)$ with $\varphi:\mathbb{Q}[x]\to \mathbb{Q}: \varphi(f(x))=f(\sqrt{2}), \ker(\varphi)=(x^2-2)$ and $im(\varphi)=\left\{a+b\sqrt{2}\right\}$.
Is the following correct? $\phi:\mathbb{Q}[x,y]\to \mathbb{Q}:\phi(f(x,y))=f(\sqrt{2},\sqrt{3})$ and $\ker(\phi)=((x^2-2)(x^2-3))$, $im\phi=\left\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}:a,b,c,d\in\mathbb{Q}\right\}$?