Computing the minimal polynomial over $\mathbb{Q}(\sqrt{3})$

Question

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?

Answer 1

Your first minimum polynomial is correct and your method for showing containment is fine. In regards to finding the minimum polynomial over $\mathbb Q(\sqrt{3})$ the usual method is to write the equation

$$x=\sqrt{1+\sqrt{3}}$$

in terms of elements in the base field. Squaring both sides we have

$$x^2=1+\sqrt{3}$$

notice that $1+\sqrt{3} \in \mathbb Q(\sqrt{3})$. In order to show this polynomial is irreducible you have to be a little clever. Notice that $[\mathbb Q(\sqrt{3}):\mathbb Q]=2$ and $[\mathbb Q(\alpha) : \mathbb Q]=4$ so that tells us the degree of the minimum polynomial of $\alpha$ over $\mathbb Q(\sqrt{3})$ is either $2$ or $4$. Hence our polynomial is irreducible because it cannot be of a lower degree.

Answer 2

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.