I encountered a question which asks the following;
Find a splitting field for $f(x) = x^6 − 2 $ over $GF(7)$. Is $f(x)$ irreducible over $GF(7)$? If not, factorise $f(x)$ into irreducibles over $GF(7)$.
Here is what I did, and I just wanted to see if I'm doing it right ? Any help is much appreciated :).
$f(x)=x^6-2$ over $GF(7)$.
Let $\alpha$ be a root of $f(x) \Rightarrow \alpha^6=2 \Rightarrow\alpha^7=2\alpha$
$2\alpha\neq\alpha$ as $\alpha \neq 0$ and $2\neq1$ in $GF(7) \Rightarrow \alpha \notin GF(7)$.
So next we try
$\alpha^{7^{2}}=(\alpha^7)^7=(2\alpha)^7=128\alpha^7=5\alpha^7=10\alpha=3\alpha$
But $3\alpha \neq \alpha$ as $\alpha \neq0$ and $3\neq1$ in $GF(7) \Rightarrow \alpha \notin GF(7^{2})$.
So next we try
$\alpha^{7^3}=((\alpha^7)^7)^7=(3\alpha)^7=4\alpha^7=8\alpha=\alpha$ as $8=1$ in $GF(7)$
So $\alpha \in GF(343)$. which means that the splitting field for $\alpha$ is $GF(343)$ over $GF(7).$
What's more $f(x)$ is irreducible over $GF(7)$ as it needs the extension $GF(7)(\alpha)\cong GF(343)$ before it can be split into linear factors.