# Degree of the field of rational numbers extended by a complex number

So I want to calculate the degree of the following fields over the field of the rational numbers: $$\mathbb{Q} \left(e^{\frac{2\pi i}{3}}\right),$$

$$\mathbb {Q} \left(\sqrt{2},\sqrt{1+i}\right)$$

I know that if I extend the rational field by irrational numbers I can use the degrees of the minimal polynomials and the multiplication formula for field extensions. But with the complex numbers I don't know how to figure out the degree of the minimal polynomials.

Remember that $$e^{2\pi i}=1$$, now start with $$\alpha =e^{\frac{2\pi i}{3}}$$ and see if you can get a polynomial in $$\alpha$$ with rational coefficients.
Let me help you with $$\alpha=\sqrt{1+i}$$. $$\alpha =\sqrt{1+i}\\\alpha^2=1+i\\\alpha^2-1=i\\(\alpha^2-1)^2=-1\\\alpha^4-2\alpha^2+2=0$$ So the minimal polynomial of $$\alpha=\sqrt{1+i}$$ is $$x^4-2x^2+2$$. Can you take it from here?
• Yes thanks! So I have degree 2 for the first one. And degree 4 for the second, because the minimal polynomials of $\sqrt{1+i}$ has degree 4 and $\sqrt{2} = \sqrt{-i} * (1+i)$. – Losyres Oct 24 '18 at 15:57