Find a basis for $\mathbb{Q}(\sqrt{1+\sqrt{3}})$ over $\mathbb{Q}$.

Question

My instinct tells me that $\mathbb{Q}(\sqrt{1+\sqrt{3}})=\{a+b\sqrt{3}+c\sqrt{1+\sqrt{3}}+d\sqrt{3}\sqrt{1+\sqrt{3}}: a, b, c, d \in\mathbb{Q}\}$, then it follows that $\{1, \sqrt{3}, \sqrt{1+\sqrt{3}}\}$ is a basis, because $\sqrt{1+\sqrt{3}}\notin \mathbb{Q}(\sqrt{3})$; but I failed to prove the set $\{a+b\sqrt{3}+c\sqrt{1+\sqrt{3}}+d\sqrt{3}\sqrt{1+\sqrt{3}}: a, b, c, d \in\mathbb{Q}\}$ is closed under taking inverses.

Any hint would be appreciated. Many thanks in advance!

Answer 1

Suppose $x=\sqrt{1+\sqrt 3}$. Then $x^2=1+\sqrt 3$ and $(x^2-1)^2=3$. This shows that $\sqrt{1+\sqrt 3}$ is a root of $x^4-2x^2-2$. Eisenstein's criterion shows $x^4-2x^2-2$ is irreducible, so this is the minimal polynomial of $\sqrt{1+\sqrt 3}$.