# Automorphisms and Splitting Fields

Note: This question comes from a non-examined question sheet from an undergrad maths course.

I want to find the splitting fields of the following polynomials:

1. $x^3-1$ over $\mathbb{Q}$
2. $x^3-2$ over $\mathbb{Q}$
3. $x^3-2$ over $\mathbb{F}_5$
4. $x^3-3$ over $\mathbb{F}_{13}$

In each case, I also want to show that the number of automorphisms of the splitting field is at most the degree of the extension.

Thoughts so far:

1. The splitting field is $\mathbb{Q}(i, \sqrt{3})$, which has degree 4.
2. The splitting field is $\mathbb{Q}(\sqrt[3]{3},e^{\frac{2 \pi i}{3}})$, which has degree 6.
3. I note that $(x^3-2)$ has $3$ as a root in $\mathbb{F}_5$, and so $x^3-2=(x-3)(x^2+3x+4)$ in this field. We need a $\sqrt{5}$ to find the routes, but I'm not sure whether this can just be added to the field $\mathbb{F}_5$ in the same way that we add it to the field $\mathbb{Q}$. Help here would be appreciated.
4. No thoughts on this yet other than to try each number from $1$ to $12$ and see if any work in $\mathbb{F}_{13}$ and to then try and reason as in (3).

Regarding the automorphisms, I would really like it if somehow could demonstrate how these calculations can be made, perhaps using one of the examples above as a template. I think I struggle conceptually and walk as methodologically with this part.

Any answers are much appreciated. Thanks.

General fact: automorphisms over the 'starting' field $F$ are as many as the degree of the extension over that field. Besides, if the extension is a splitting field, they are a group by composition (Galois group), a subgroup of the permutations of the roots. So, as an example, the first two polynomial discussion:
1. The poly. factors in $(x-1)(x^2+x+1)$: the roots of the second factor are $\frac{-1\pm i\sqrt{3}}{2}$, so the splitting field is $K=\mathbb Q(i\sqrt 3)$, the poly. $(x^2+x+1)$ is irreducible over $\mathbb Q$ (has no linear factors) so $[K:\mathbb Q]=2$. So, the Galois group has to be $C_2$.
2. The splitting field is $K=\mathbb Q(\sqrt[3]2,i\sqrt 3)$; the poly. is irreducible because of Eisenstein's criterion so $\mathbb Q(\sqrt[3]2)$ has degree 3 over $\mathbb Q$, and $K$ has degree 2 over $\mathbb Q(\sqrt[3]2)$ because you're adding a root of a 2nd degree poly (->degree 1 or 2) but $\mathbb Q(\sqrt[3]2)\subset \mathbb R$ and $K$ doesn't, so degree is at least 2. The splitting field thus has degree 6 and the Galois group is $D_3$, the dihedral group.