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?

  • $\begingroup$ 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 $\sqrt{2} = \sqrt{-i} * (1+i)$. $\endgroup$ – Losyres Oct 24 '18 at 15:57

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