Find degree of extension $Q$($\sqrt{1+\sqrt{-3}}$ + $\sqrt{1-\sqrt{-3}}$) over $Q$

Question

I tried solving this textbook problem.Any hint how to simplify or find the degree of extension in this case ?I guess maximum degree can be 4.

Answer 1

Hint:

Let $x=\sqrt{1+\sqrt{-3}}+\sqrt{1-\sqrt{-3}}$. Then, observe that $$ x^2=1+\sqrt{-3}+2\sqrt{(1+\sqrt{-3})(1-\sqrt{-3})}+1-\sqrt{-3}=2+2\sqrt{1-(-3)}=6. $$ Therefore, this satisfies $x^2-6=0$.

What do you think now? Are there any rationals that satisfy $x^2=6$?

Answer 2

Hint: If we let $\alpha=\sqrt{1+\sqrt{-3}}+\sqrt{1-\sqrt{-3}}$ then what is $\alpha^2$?