I'm trying to prove the following statement:
Let $K|F$ be a finite Galois extension of degree $n$ with Galois group $G = \{\sigma_1,…,\sigma_n\}$. Then $\gamma\in K$ is a primitive element of $K|F$ iff $\sigma_1(\gamma)$, … , $\sigma_n(\gamma)$ are all different.
Here is my shot:
$(\Rightarrow)$ If $\gamma\in K$ is a primitive element of $K|F$, then $K = F(\gamma)$. Since $K$ is Galois, in particular it is normal, which means that $K$ is the splitting field of $f$ over $F$. Consider $f$ the minimal polynomial of $\gamma$ over $F$. Since $K$ is normal, all of the conjugate roots of $\gamma$ belong in $K$.
Now let's consider $id_F:F\to F$, $id_F(a) = a$ for all $a\in F$. Then we can extend $id$ to $\sigma_i:K\to K$, $\sigma_i(\gamma) = \alpha$, with $\alpha$ a root of $f^{id}(x) = f(x)$. Then $\sigma_i(\gamma)$ are all the roots of $f$ ($\gamma$ itself and its $n-1$ conjugate roots). Therefore, $\sigma_1(\gamma)$, … , $\sigma_n(\gamma)$ are $\sigma$ and its $n-1$ conjugate roots, which are all different, q.e.d.
$(\Leftarrow)$ This one is straightforward, since we know (by the primitive element theorem) that every finite and separable extension is simple, which means that has a primitive element. So we got this inclusion for free.
Is this correct? Or should I correct something? Thanks in advance!
Best Answer
The idea in the first direction looks correct, but you should write it more clearly. Assume $K=F(\gamma)$ and let $f\in F[x]$ be the minimal polynomial of $\gamma$ over $F$. It is separable and splits in $K$, so let $\{\alpha_1,...,\alpha_n\}\subseteq K$ be its roots. It is well known that $Gal(K/F)$ acts transitively on the roots. So we can define a map:
$h:\{1,2,...,n\}\to \{\alpha_1,...,\alpha_n\}, \ \ \ \ h(i)=\sigma_i(\gamma)$
Then $h$ is surjective, and since both sets have the same (finite) cardinality, it must be injective. Hence $\sigma_1(\gamma),...,\sigma_n(\gamma)$ are all distinct.
Now, your solution to the second direction is completely wrong. Yes, there is some primitive element, but how do you know that specifically $\gamma$ is primitive? In order to prove it, let $M=F(\gamma)$, this is an intermediate field. By the fundamental theorem of Galois theory we know that $M$ is the fixed field of some subgroup $H\leq Gal(K/F)$. But since $\sigma_1(\gamma),...,\sigma_n(\gamma)$ are all distinct, we know in particular that $\gamma$ is fixed only by the identity. So we must have $H=\{e\}$, and hence $M=K$.