Show that degree of $\mathbb{Q}(\sqrt{1+\sqrt3}):\mathbb{Q}=4$

Question

I am solving an ex.: Find basis of $\mathbb{Q}(\sqrt{1+\sqrt3})$ and degree of $\mathbb{Q}(\sqrt{1+\sqrt3}):\mathbb{Q}$. SO first of all I have showed that $1+\sqrt{3}$ is not a square in $\mathbb{Q}(\sqrt3)$. Also I think that degree in this exercise we find with Short Tower Law. So, degree of $\mathbb{Q}(\sqrt3):\mathbb{Q}$ is $2$. How to show that degree of $\mathbb{Q}(\sqrt{1+\sqrt3}):\mathbb{Q}(\sqrt3)$ is also $2$? And what is the basis of $\mathbb{Q}(\sqrt{1+\sqrt3})$ over $\mathbb{Q}$?

Answer 1

Since $1+\sqrt{3}$ is not a square in $\mathbb{Q}(\sqrt{3})$, it follows that $$ [\mathbb{Q}(\sqrt{1+\sqrt{3}}):\mathbb{Q}(\sqrt{3})]=2 $$ because clearly $\sqrt{1+\sqrt{3}}$ is a root of $x^2-(1+\sqrt{3})\in\mathbb{Q}(\sqrt{3})[x]$.

Now, by general theory, you know that a basis of $\mathbb{Q}(\sqrt{1+\sqrt{3}})$ over $\mathbb{Q}$ is given by $$ 1,\quad\sqrt{3},\quad\sqrt{1+\sqrt{3}},\quad\sqrt{3}\sqrt{1+\sqrt{3}} $$

Answer 2

$$\alpha =\sqrt{1+\sqrt 3}\implies \alpha ^2=1+\sqrt 3\implies \alpha ^4-2\alpha ^2-2=0,$$ this polynomial is irreducible by Eisenstein criterion. Therefore, $$[\mathbb Q(\alpha ):\mathbb Q]=4.$$