Lemma 9.16.1. Let $F$ be a field. Let $P \in F[x]$ be a nonconstant polynomial. There exists a smallest field extension $E/F$ such that $P$ splits completely over $E$. Moreover, the field extension $E/F$ is normal and unique up to (nonunique) isomorphism.

**Proof.**
Choose an algebraic closure $\overline{F}$. Then we can write $P = c (x - \beta _1) \ldots (x - \beta _ n)$ in $\overline{F}[x]$, see Lemma 9.10.2. Note that $c \in F^*$. Set $E = F(\beta _1, \ldots , \beta _ n)$. Then it is clear that $E$ is minimal with the requirement that $P$ splits completely over $E$.

Next, let $E'$ be another minimal field extension of $F$ such that $P$ splits completely over $E'$. Write $P = c (x - \alpha _1) \ldots (x - \alpha _ n)$ with $c \in F$ and $\alpha _ i \in E'$. Again it follows from minimality that $E' = F(\alpha _1, \ldots , \alpha _ n)$. Moreover, if we pick any $\sigma : E' \to \overline{F}$ (Lemma 9.10.5) then we immediately see that $\sigma (\alpha _ i) = \beta _{\tau (i)}$ for some permutation $\tau : \{ 1, \ldots , n\} \to \{ 1, \ldots , n\} $. Thus $\sigma (E') = E$. This implies that $E'$ is a normal extension of $F$ by Lemma 9.15.5 and that $E \cong E'$ as extensions of $F$ thereby finishing the proof. $\square$

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