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!

  • 1
    $\begingroup$ Before squaring again, arrange all non square-root terms on the other side!! $\endgroup$ – Berci Oct 18 '16 at 23:37

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

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 $4$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.