Finding Splitting Field with Minimal Adjoined Elements

Question

I have to find the splitting field of the following in $\mathbb{Q}$:

a) $f(x) = x^6 + 1 $

b) $f(x) = (x^2-3)(x^3+1) $

For a), finding the 6th roots of -1, I concluded that $\pm i, \frac{\pm \sqrt{3} \pm i}{2}$ should be in the splitting field. The minimum adjoined splitting field is thus $\mathbb{Q}[\sqrt{3},i]$.

For b) using similar method, I found that $-1, \sqrt{3}, \frac{1 \pm \sqrt{3}i}{2}$ should be in the splitting field. Is the minimal adjoined field for b) $\mathbb{Q}[\sqrt{3},i]$ as well?

I am not so sure in how to find the minimal elements one must adjoin to $\mathbb{Q}$.

Answer 1

For the first question, note that $x^6+1$ divides $x^{12}-1$, and that all roots of $x^{12}-1$ are contained in $\Bbb{Q}(\zeta_{12})$, where $\zeta_{12}$ denotes a primitive $12$-th root of unity. Because $$[\Bbb{Q}(\zeta_{12}):\Bbb{Q}]=\varphi(12)=4,$$ we see that this is a quartic extension containing all roots of $x^6+1$. You have already shown that any splitting field of $x^6+1$ must contain $i$ and $\sqrt{3}$, from which it easily follows that the degree of a splitting field must be at least $4$. This shows that $\Bbb{Q}(\zeta_{12})$ is a splitting field of $x^6+1$ over $\Bbb{Q}$.

For the second question, a similar argument works, with slightly more effort.

Answer 2

Both splitting fields are equal to $\Bbb Q(i,\sqrt3)=\Bbb Q[i,\sqrt3]$, as you correctly wrote.

You still didn't reply what you mean by "minimal adjoined elements", but if what you want is as few generators as possible, the primitive element theorem tells you that one is sufficient. Anyway, you don't need that theorem: you can take for instance $$\alpha:=i+\sqrt3.$$ Indeed, $$i=\frac{\alpha^3}8\quad\text{and}\quad\sqrt3=\alpha-i.$$