The Stacks project

Proof. Choose generators $\alpha _1, \ldots , \alpha _ n$ of $E$ over $F$. Let $P_1, \ldots , P_ n$ be the minimal polynomials of $\alpha _1, \ldots , \alpha _ n$ over $F$. Set $P = P_1 \ldots P_ n$. Observe that $(x - \alpha _1) \ldots (x - \alpha _ n)$ divides $P$, since each $(x - \alpha _ i)$ divides $P_ i$. Say $P = (x - \alpha _1) \ldots (x - \alpha _ n)Q$. Let $K/E$ be the splitting field of $P$ over $E$. We claim that $K$ is the splitting field of $P$ over $F$ as well (which implies that $K$ is normal over $F$). This is clear because $K/E$ is generated by the roots of $Q$ over $E$ and $E$ is generated by the roots of $(x - \alpha _1) \ldots (x - \alpha _ n)$ over $F$, hence $K$ is generated by the roots of $P$ over $F$.

Uniqueness. Suppose that $K'/E$ is a second smallest extension such that $K'/F$ is normal. Choose an algebraic closure $\overline{F}$ and an embedding $\sigma _0 : E \to \overline{F}$. By Lemma 9.10.5 we can extend $\sigma _0$ to $\sigma : K \to \overline{F}$ and $\sigma ' : K' \to \overline{F}$. By Lemma 9.15.3 we see that $\sigma (K) \cap \sigma '(K')$ is normal over $F$. By minimality we conclude that $\sigma (K) = \sigma (K')$. Thus $\sigma \circ (\sigma ')^{-1} : K' \to K$ gives an isomorphism of extensions of $E$. $\square$