If $f$ has more than one root in $K$, then $f$ splits and $K/k$ is Galois? Let $f \in k[x]$ be an irreducible polynomial of prime degree $p$ such that $K \cong k[x]/f(x)$ is a separable extension. How do I see that if $f$ has more than
one root in $K$, then $f$ splits and $K/k$ is Galois?
 A: Let $G=\mathrm{Aut}(K/k)$. Since $f$ has more than one root in $K$, it follows that $|G|>1$. 
Suppose that $|G|<p$, and choose some $1\neq\sigma\in G$ of order $q$. Let $F$ be the subfield of $K$ fixed by $\sigma$. Then by Artin's theorem, it follows that $K/F$ is a Galois extension, with $[K:F]=|\sigma|=q$.
Therefore
$$p=[K:k]=[K:F][F:k]=q[F:k] $$
which is a contradiction because $q<p$ and $p$ is prime.
Hence $|G|=p$, so $K$ is a Galois extension of $k$ and $f$ splits in $K$.
A: Let $L\supseteq K$ be a splitting field of $f$ over $k$, and let $G=\operatorname{Gal}(L/k)$ be the Galois group. As $f$ is separable, it has $p$ distinct roots in $L$. We view $G$ as a group of permutations of the roots, so also as a subgroup of the symmetric group $S_p$. Denote by $H=\operatorname{Gal}(L/K)\le G$ the subgroup associated to the intermediate field $K$.
The order of $G$ is divisible by $[K:k]=p$ so by Cauchy's theorem there exists an element $\sigma\in G$ of order $p$. When viewed as an element of $S_p$ $\sigma$ has to be a $p$-cycle. 
Let $\alpha_1,\alpha_2$ be two roots of $f(x)$ in $K$. By replacing $\sigma$ with its suitable power we can without loss of generality assume that $\sigma(\alpha_1)=\alpha_2$. Number the zeros in such a way that $\sigma(\alpha_i)=\alpha_{i+1}$, for all $i=1,2,\ldots,p-1$, $\sigma(\alpha_p)=\alpha_1$.
The claim is to prove that all the zeros $\alpha_i\in K$. Assume contrariwise that this is not the case. Then we can find three zeros $\alpha_i,\alpha_{i+1},\alpha_{i+2}$ such that $\alpha_i,\alpha_{i+1}\in K$ but $\alpha_{i+2}\notin K$.
This implies that there exists an element $\tau\in H$ such that $\tau(\alpha_{i+2})\neq\alpha_{i+2}$. Consider the automorphism $\delta=\sigma^{-1}\tau\sigma$. As $\tau$ fixes all the elements of $K$
$$
\delta(\alpha_i)=\sigma^{-1}(\tau(\alpha_{i+1}))=\sigma^{-1}(\alpha_{i+1})=\alpha_i.
$$
Therefore $\delta$ fixes the elements of $k(\alpha_i)\subseteq K$. As $[K:k]=p$ is a prime, we can conclude that there are no intermediate fields between $k$ and $K$. Hence $k(\alpha_i)=K$, and $\delta\in H$. 
But,
$$
\delta(\alpha_{i+1})=\sigma^{-1}(\tau(\alpha_{i+2}))\neq\sigma^{-1}(\alpha_{i+2})=\alpha_{i+1}.
$$
So $\delta$ does not fix the element $\alpha_{i+1}$ contradicting the above, and proving the claim.
