Automorphism group of $F=\mathbb F_3[x]/(x^3+2x-1)$.

Question

Let $F=\mathbb F_3[x]/(x^3+2x-1)$, where $\mathbb F_3$ is the field with $3$ elements. Which one of the following are true:

1. $F$ is a field with $27$ elements.
2. $F$ is separable but not a normal extension of $\mathbb F_3$.
3. The automorphism group of $F$ is cyclic.
4. The automorphism group of $F$ is abelian but not cyclic.

I know that $F$ has $27$ elements so $1$ is true. But I am confused with rest of the options. Can some conclusion be drawn from the order of the automorphism group? If so. how can i determine the order of the automorphism group of $F$? Any hints please.

Answer 1

1) It is true , since $x^3+2x-1$ is irreducible over $F_3$.

2) It is not true, as $F$ is a normal extension over $F_3$.

Explanation why $F$ is normal

(i) $F$ is a finite extension over $F_3$.

(ii) $F$ is the splitting field of the polynomial $x^{27} - x$ over $F_3$.

These two conditions are sufficient for an extension to be normal.

3) It is true .

The automorphism group of $F$ consists of 3 mappings namely

$x \to x$ , $x \to x^3$ and $x \to x^9$

Hence the automorphism group is of order 3 and thus it is cyclic. Also one can refer the link automorphisms of a finite field in order to understad the construction of these maps.

4) It is not true by 3)

Answer 2

($2$) holds since $F$ is the splitting field of $x^{27}-x$. You can verify the automorphism group is generated by $x\mapsto x^3$. The fact that the degree is $3$ along with separability and normality tell you that the order of the automorphism group is exactly $3$, but then basic group theory tells you it must be cyclic. The fact that you can name the generator, is a nice bonus though.