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

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 $$\mathbb{Q}$$?

• Let $K$ be a field. $\alpha$ is a root of some irreducible polynomial of degree $n$ with coefficients in $K$ if and only if $\{1, \alpha, \dots, \alpha^{n-1}\}$ is a $K$-basis for $K(\alpha)$. If you understand this result, then you will be able to answer both of your questions. Commented Mar 25, 2019 at 8:30

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