Let $f(x) \in F[x]$ be an irreducible separable polynomial of degree p, with distinct roots $\alpha_1, \ldots, \alpha_p$. I want to show that if $F(\alpha_1) = F(\alpha_1, \alpha_2)$, then $F(\alpha_1) = F(\alpha_1, \ldots, \alpha_p)$, ie. $F(\alpha_1)$ is the splitting field of $f(x)$.
So far, what I know is that $F(\alpha_1) \simeq F(\alpha_i)$ for all $1 \leq i \leq p$, can I somehow extend this to show that $(F(\alpha_1))(\alpha_1) \simeq (F(\alpha_1))(\alpha_i)$? Thanks
Let $G = \operatorname{Gal}(f) \subset S_p$ be the galois group of the splitting field. By a standard argument, $G$ contains a cycle of length $p$ (Since any element of order $p$ in $S_p$ is such a cycle).
Let $\sigma \in G$ be a cycle of length $p$. Replacing it by a suitable power, we can assume $\sigma(\alpha_1) = \alpha_2$.
Let $L = F(\alpha_1)$. In particular $\sigma(L) \subset L$ holds, by iterating this we get that $\sigma^n(L) \subset L$ holds for any $n \geq 1$. But since $\sigma$ is a cycle of length $p$, we get that all the $\alpha_i$ occur among $\sigma^n(\alpha_1), n \geq 1$. Hence $L$ contains all roots and is thus the splitting field.
Some more details on the "standard argument":
If $K$ is the splitting field, we have $F \subset F(\alpha_1) \subset K$, hence $$|G| = [K:F] = [K:F(\alpha_1)] \cdot [F(\alpha_1):F] = [K:F(\alpha_1)] \cdot p$$ is a multiple of $p$. By Cauchy's Theorem, $G$ has an element of order $p$.