TITLE: Computing the minimal polynomial over $\mathbb{Q}(\sqrt{3})$
QUESTION [6 upvotes]: Let $F:=\mathbb{Q}(\sqrt{1+\sqrt{3}})$ and $\alpha=\sqrt{1+\sqrt{3}}$. I have to find the minimal polynomial of $\alpha$ over $\mathbb{Q}$ and over $\mathbb{Q}(\sqrt{3})$. I also need to show that $\mathbb{Q}(\sqrt{3})$ is contained in $F$.
My attempt, the minimal polynomial over $\mathbb{Q}$ is $x^4-2x^2-2$, which is irreducible by Eisenstein's criterion. How to find the minimal polynomial over $\mathbb{Q}(\sqrt{3})$?
And for the second one $\sqrt{3}=\alpha^2-1$, that's how we can show the containment. Right?

REPLY [2 votes]: Here’s an approach that’s more advanced and less satisfactory than @Jacob’s. It’s known (and easily checked) that $\mathbb Z[\sqrt3]$ is a principal ideal domain, ramified only at $2$ and $3$. And that the units are generated by $-1$ and $\sqrt3+2$. Since $2=(\sqrt3+1)^2/(\sqrt3+2)$, your number $1+\sqrt3$ is prime, so $X^2-(1+\sqrt3)$ is irreducible.