# Show that $[\mathbb{Q}(\sqrt[4]{2}, \sqrt{3}):\mathbb{Q}]=8$

Show that $[\mathbb{Q}(\sqrt[4]{2}, \sqrt{3}):\mathbb{Q}]=8$

$[\mathbb{Q}(\sqrt[4]{2}, \sqrt{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[4]{2}),\mathbb{Q}(\sqrt{3})][\mathbb{Q}(\sqrt{3}),\mathbb{Q}]=(4)(2)=8$. For $x^4-2$ and $x^2-3$ are the minimal polynomials of $\sqrt[4]{2}$ in $\mathbb{Q}(\sqrt{3})$ and $\sqrt{3}$ in $\mathbb{Q}$ respectively. Is this correct?

• You should justify why $x^4-2$ is irreducible over $\Bbb{Q}(\sqrt3)$. It is irreducible over $\Bbb{Q}$ by Eisenstein, but that no longer applies (without selected pieces of algebraic number theory) in this extension field. – Jyrki Lahtonen Sep 4 '17 at 3:47
• You could try and mimick these ideas and show that $\sqrt3\notin\Bbb{Q}(\root4\of2)$. – Jyrki Lahtonen Sep 4 '17 at 3:51
• I think it is easier to say $x^4-2= (x^2-\sqrt{2})(x^2+\sqrt{2})= (x-\sqrt[4]{2}) h(x)$ is irreducible over $\mathbb{Q}(\sqrt{3})$ because $\sqrt{2} \not \in \mathbb{Q}(\sqrt{3})$. @JyrkiLahtonen – reuns Sep 4 '17 at 3:59
• That's a good idea @reuns. The quadratic factors where a real root is paired up with a non-real obviously don't have coefficients in $\Bbb{Q}(\sqrt3)$ :-) – Jyrki Lahtonen Sep 4 '17 at 4:10