$p(x) \in F[x]$ Has One Root in Normal Field Extension $E$ $\implies$ $p(x)$ Splits Entirely in $E$

Question

Problem: Suppose that $E$ is normal over $F$. Let $p(x) \in F[x]$ have one root in $E$. Show that $p(x)$ therefore splits entirely in $E$.

EDIT: As per the comment below, assume also $p(x)$ is irreducible.

Attempt:

1. Let $\deg(p(x)) = n$. Furthermore let $n>1$ else the problem be trivial.

2. I see no (at least obvious) connection between this theorem and the Fundamental Theorem of Galois Theory (in particular, we aren't assuming $p(x)$ is separable).

3. If we let $\alpha$ be the root in $E$ the problem refers to, we might be able to make use of $F(\alpha)$. But it's unclear how to proceed from this observation.

Answer 1

Let $\alpha \in E$ such that $p(\alpha) = 0$, as you defined it. We want to show that if $p(\beta) = 0$ for some $\beta \neq \alpha$, then $\beta \in E$ as well.

Since $\alpha, \beta$ are distinct roots of $p$, and $p$ irreducible, then $F(\alpha) \cong F(\beta)$. By the normality of $E / F$, we know that $E$ is the splitting field of some polynomial in $F[x]$. Hence $E(\alpha) \cong E(\beta)$. Then $[E(\beta) \colon E] = [E(\alpha) \colon E] = 1$ because $\alpha \in E$. Thus, the minimal polynomial of $\beta$ over $E$ has degree $1$, so $\beta \in E$.