# Degree of field extension $\mathbb{Q}\left(\sqrt{1 + \sqrt{3}}\right):\mathbb{Q}$

I think the degree

$$\left[\mathbb{Q}\left(\sqrt{1 + \sqrt{3}}\right):\mathbb{Q}\right]$$

is equal to four, but how do I find the minimum polynomial of such an extension? If I square the term I take care of one square root, but squaring again doesn't help me get rid of the $\sqrt{3}$ term. Any advice is very much appreciated!

• Before squaring again, arrange all non square-root terms on the other side!! Oct 18 '16 at 23:37

## 1 Answer

$x = \sqrt{1 + \sqrt 3} \implies x^2-1 =\sqrt{3} \implies x^4-2x^2-2 = 0$.

This polynomial is irreducible by Eisenstein's criterion, since $2$ divides all the coefficients except the first, and $2^2=4$ does not divide the constant. Hence, the degree of the extension, is the degree of the minimal polynomial, which is $4$.

• Thank you @Bernard I have made the correction. Oct 18 '16 at 23:34