# Given a field extension with prime degree, if $Aut(K/F) > 1$, then this extension is Galois?

Let $|K:F| = p$ be a field extension of degree $p$. For each $Aut(K/F)\ni\sigma \neq id$, denote $K^\sigma$ the the fixed field of $\sigma$ and we have $$p = |K:F| = |K:K^\sigma||K^\sigma:F|$$ because $p$ is prime and $\sigma \neq id$, we see that $K^\sigma = F$, so indeed $F$ is the fixed field for each non identity element in $Aut(K/F)$, and we have $K$ over $F$ is Galois (this last part is a standard result, but non trivial).

Is this correct? It seems too simple to be true for me... and I didn't even have any assumptions for $Char(F)$.

Pick $\sigma \in \operatorname{Aut}(K/F) \setminus \{\operatorname{id}\}$, so $\sigma(\alpha) \ne \alpha$ for some $\alpha \in K$. Then, $\alpha \in K \setminus F$, and the minimal polynomial of $\alpha$ has degree $p$, and $K = F(\alpha)$.
Let $n$ be the smallest positive integer such that $\sigma^n(\alpha) = \alpha$. Then, $(X-\alpha) (X - \sigma(\alpha)) \cdots (X - \sigma^{n-1}(\alpha))$ is a polynomial that is fixed by $\sigma$, so it is a polynomial with coefficients in $F$. The minimal polynomial of $\alpha$ divides this polynomial, so $n \ge p$. By construction, $\{\alpha, \sigma(\alpha), \cdots, \sigma^{n-1}(\alpha)\}$ are distinct, and they are all conjugates of $\alpha$, so $n \le p$. Therefore we conclude that $n = p$.
And it follows that all conjugates of $\alpha$ are in $K$, so $K$ is the splitting field of the minimal polynomial of $\alpha$, so $K$ is a normal extension of $F$. Also, we found $p$ distinct conjugates of $\alpha$, so $\alpha$ is separable, so $K$ is a separable extension of $F$.
• Maybe a remark: The polynomial $(X-\alpha)\cdots(X-\sigma^{n-1}(\alpha))$ is fixed by $\sigma$ so the coefficients are contained in the fixed field of $\langle \sigma \rangle$. We have the extensions $K - K^{\langle \sigma \rangle} - F$ and because $|K:L|$ is prime, by tower law, we see that $K^\sigma = F$ as $\sigma$ is not identity. – Xiao Aug 12 '18 at 18:59