equivalent conditions for normality of a field extension

Theorem.

If $K\mathrm{/}F$ is an algebraic extension of fields, then the following are equivalent:

1. $K$ is normal over $F$;

2. $K$ is the splitting field over $F$ of a set of polynomials in $F[X]$;

3. if $\overline{F}$ is an algebraic closure $F$ containing $K$ and $\sigma :K\to \overline{F}$ is an $F$-monomorphism, then $\sigma (K)=K$.

Proof.

(1)$\Rightarrow$(2) Let $X$ be an $F$-basis for $K$, and for each $x\in X$, let $f_x$ be the irreducible polynomial of $x$ over $F$. By hypothesis, each $f_x$ splits over $K$, and because we evidently have $K=F(X)$, it follows that $K$ is a splitting field of $\{f_x:x\in X\}$ over $F$.
(2)$\Rightarrow$(3) Assume that $K$ is a splitting field over $F$ of $S\subseteq F[X]$. Given $f\in S$, we may write $f(X)=u{\prod }_{i=1}^n(X-u_i)$ for some $u,u_1,\mathrm{\dots },u_n\in K$; because $\sigma$ fixes $F$ pointwise, we have $\sigma (u_i)\in \{u_1,\mathrm{\dots },u_n\}$ for $1\le i\le n$, and since $\sigma$ is injective, it must simply permute the roots of $f$. Thus $u_1,\mathrm{\dots },u_n\in \sigma (K)$. As $K$ is generated over $F$ by the roots of the polynomials in $S$, we obtain $K=\sigma (K)$.
(3)$\Rightarrow$(1) Let $\overline{K}$ be an algebraic closure of $K$, noting that, since $K$ is algebraic over $F$, that same is true of $\overline{K}$, and consequently $\overline{K}$ is an algebraic closure of $F$ containing $K$. Now suppose $f\in F[X]$ is irreducible and that $u\in K$ is a root of $f$, and let $v$ be any root of $F$ in $\overline{K}$. There exists an $F$-isomorphism $\tau :F(u)\to F(v)$ such that $\tau (u)=v$. Because $\overline{K}$ is a splitting field over both $F(u)$ and $F(v)$ of the set of irreducible polynomials in $F[X]$, $\tau$ extends to an $F$-isomorphism $\sigma :\overline{K}\to \overline{K}$. It follows that ${\sigma |}_K:K\to \overline{K}$ is an $F$-monomorphism, so that, by hypothesis, $\sigma (K)=K$, hence that $v=\sigma (u)\in K$. Thus $f$ splits over $K$, and therefore $K/F$ is normal.∎