Galois extension $\mathbb Q(1/2+\sqrt{-3}/2)/\mathbb Q$

Question

I have this exercise:

Determine whether the extension $\mathbb Q(1/2+i\sqrt{3}/2)/\mathbb Q$ is Galois.

If we call $\alpha=1/2+i\sqrt{3}/2$, then we find $\alpha^2-\alpha+1=0$. Its roots are $1/2+i\sqrt{3}/2$ and $1/2-i\sqrt{3}/2$ and therefore this extension is separable and the splitting field of this polynomial with coefficients in $\mathbb Q$. Now I think I should prove both roots are inside $\mathbb Q(1/2+i\sqrt{3}/2)$.

The first root is obvioulsly in; now if we square $((1/2+i\sqrt{3}/2)^2=-1/2+i\sqrt{3}/2=-(1/2-i\sqrt{3}/2)$, so the second root also is in and then the extension is Galois.

Could we correct or improve this proof?

Thanks

Answer 1

Your solution is correct, but don't forget that when obtaining one root from another, you can also add and multiply by the rational numbers.

In your case, let $\alpha=\frac{1}{2}+\frac{i\sqrt{3}}{2}$. Then $-\alpha+1=\frac{1}{2}-\frac{i\sqrt{3}}{2}$. You don't necessarily have to square $\alpha$ every time.