Determine Gal (K/P)

Question

Let $K$ be a field with 27 elements. Then $K$ has characteristic 3. So $\mathbb{Z}_3$ is the prime field of $K$ and $X^{27}-X$ is the splitting field over $\mathbb{Z}_3$. How do I show that $K$ is the galois extension of the prime field $P$ and how do I determine the galois group $Gal(K/P)$? I just know from a lemma that $Gal(Z_{27}/Z_3)$ is the Galois Group, but would that be the final answer?

Answer 1

Some order in your messy question: in order to avoid confusion, the prime field of characteristic three is denoted $\;\Bbb F_3\;$ .True, we have that $\;\Bbb Z_3:=\Bbb Z/3\Bbb Z\cong\Bbb F_3\;$ ...but that's all there is to it.

Now, $\;K=\Bbb F_{27}\;$, and this field is the splitting field of the polynomial $\;f(x)=x^{27}-x\in\Bbb F_3[x]\;$ over the above mentioned prime field.

Now, let $\;\overline{\Bbb F}\;$ denote an algebraic closure of $\;\Bbb F_3\;$ . then, $\;\overline{\Bbb F}\;$ contains all the roots of any polynomial over $\;\Bbb F_3\;$ .

Prove: The set of elements in $\;\overline{\Bbb F}\;$ that are roots of $\;f(x)\;$ is a field (the operations are carried on $\mod3\;$ , of course). Thus, prove your field $\;K\;$ is this set of elements and deduce $\;K/\Bbb F_3\;$ is a Galois extension.

Check now the map $\;\sigma: K\to K\;,\;\;\sigma x:=x^3\;$. Prove this is a $\;\Bbb F_3\,$- automorphism of $\;K\;$ . What can you say about its order in $\;G:=\text{Gal}\,(K/\Bbb F_3)\;$ ?