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
The primitive root of unity claim seems to be false.
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 $x^n+a$.