I need to show that the Galois group of $ x^{p^n} − 2 $ over is $F_p$ is a cyclic group of order n.
I need rigorous proof for this.
I know that the Galois group of a polynomial over F is Galois group of its splitting field.
Result1 I know that $F_p$ has precisely $p^n$ elements and the Galois group of polynomial $ x^{p^n} − x $ over $F_p$ is isomorphic to Galois group of order n generated by the frobenius automorphism.
Result 2 Let F be a field of charecteristic not dividing n which contains the nth roots of unity. Then the extension $F(a^\frac{1}{n})$ for $a \in $ F is cyclic over F of degree n. $F(a^\frac{1}{n})$ is the splitting field of $x^n-a$, also $F(a^\frac{1}{n})$ is Galois over F with Gal($F(a^\frac{1}{n})$ /F) isomorphic to the cyclic group of $U_n$ , the group of nth roots of unity of order n."
I guess result 2) can be used here but I am just unable to compare things. Here a=2, what about characteristic of field $F_p$ , what is n here?
Please help . I am getting confused .
This is not true unless $n=1$. Indeed, since $(a+b)^p=a^p+b^p$ in characteristic $p$, $(x-2)^{p^n}=(x-2)^{p^n}$. So this polynomial splits already in $\mathbb{F}_p$, and $\mathbb{F}_p$ is the splitting field, with trivial Galois group. So the Galois group is always cyclic of order $1$, no matter what $n$ is.