Given $u=e^{i2\pi\over{7}}$, determine whether $\mathbb{Q}(u)/\mathbb{Q}$ is a Galois extension

Question

I'm a bit confused because I tried this:

$u=e^{i2\pi\over{7}}=(e^{i2\pi})^{1\over7}=(\cos2\pi+i\sin2\pi)^{1\over7}=1$

Answer 1

Your confusion comes from the fact that $1^\frac{1}{7}$ has many solutions, $1$ being just one of them. The complete set of solutions is given by the roots of the polynomial $f(x) = x^7 - 1$.

Now, there are two ways to show that the extension is a Galois extension depending on where you are in the material. One way is to show that the degree of the field extension is equal to the number of $\mathbb{Q}$-automorphisms of $\mathbb{Q}[u]$. The other way is to show that $\mathbb{Q}[u]$ is a splitting field of some polynomial (which will be $f(x) = x^7 - 1$).

If you want to take the former approach, first we need to find the degree of the extension $\mathbb{Q}[u]$. Well, $f(x) = x^7 - 1 = (x-1)(x^6 + x^5 + x^4 + x^3 + x + 1)$. The sixth-degree factor of $f$ is irreducible, so $u$ is the root of a degree-$6$ irreducible polynomial. Thus, $[\mathbb{Q}[u]:\mathbb{Q}]=6$. From here, we want to show that there are six $\mathbb{Q}$-automorphisms of $\mathbb{Q}[u]$. This isn't so bad. Hint: the automorphism group is cyclic; try to find its generator.

Answer 2

The "equality" $$e^{\frac{2\pi i}7}=(e^{2\pi i})^{1/7}$$ is not true.

The usual properties of the exponentiation are valid only when the base is a positive real number, as you have just proved. If the base is negative or not real, you need to consider complex exponentiation, which have some weird properties. For instance: $i^{\sqrt 2}$ has infinitely many values, or more correctly, you have to specify which branch of the complex logarythm you are working with.

If you still need help to prove that the extension is of Galois, just say it.