# Understanding a Solution (Splitting Fields)

Consider the following set-up:

We have a polynomial $f(x)=x^6+3$. Define $L$ to be the simple extension of $\mathbb{Q}$ defined by $f$.

I want to prove the following claim:

Claim: L is a splitting field for $f$ over $\mathbb{Q}$

I have a solution that has been provided, but I'd like help in understanding why it works. It proceeds as follows:

Let $\alpha$ be a root of $f$ and note that $\beta=\frac{\alpha+1}{2}$ is a primitive sixth root of unity. Define $L=\mathbb{Q}(\alpha)$

Note that $f$ has roots $\pm\alpha, (\pm\alpha\pm\alpha^4)/2$. These roots are all distinct, and so $L$ is the splitting field.

So I have two main questions coming from this solution:

1. How can we simply "note" that $\beta$ as defined is a primitive sixth root of unity. Am I missing something here? Also, more generally, given an arbitrary polynomial in the form $x^n+\gamma$, is it possible to generate primitive roots of unity in terms of $\gamma$?

2. How do we know that the roots have this form? I can't immediately see why these elements would necessarily be roots. I have done the algebra to make sure they are, but where is the initial insight coming from?

Thanks

Of course, we should first look at $\beta^6$ to see if it reduces to 1. But when I looked at this, there did not seem to be any reason it should reduce that way.
So I built a root of f in wolframalpha, namely $\alpha=3^{\frac16}(\cos(\pi/6)+i\sin(\pi/6))$, and checked to see if half of 1 plus that number is a 6th root of unity. It does not appear to be true, if this calculator is to be trusted.
However, $\frac{\alpha^3+1}{2}$ is a 6th root of unity, so it may be a dropped superscript. Armed with this fact it would be easy to see that $\frac{\alpha+\alpha^4}{2}=\alpha\frac{\alpha^3+1}{2}$ is a root of $x^6+3$.
I think the strategy is not exposed very well by this "note that this happens" write-up. The idea for polynomials of the form $x^n+a$ is that once you pick a single root $\alpha$, the other roots are just $\omega^k \alpha$ for some primitive $n$th root of unity $\omega$. (This computation is obvious for you, right?)
So, if you demonstrate that $\alpha$ already furnishes that primitive $n$th root of unity in its extension, then the extension must contain all the roots for $x^n+a$.