Theorem   If $K/F$ 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 $\sigma:K\rightarrow\overline{F}$ 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 $\set{f_x:x\in X}$ 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 $u,u_1,\ldots,u_n\in K$ because $\sigma$ fixes $F$ pointwise, we have $\sigma(u_i)\in\set{u_1,\ldots,u_n}$ for $1\leq i\leq n$ and since $\sigma$ is injective, it must simply permute the roots of $f$ Thus $u_1,\ldots,u_n\in\sigma(K)$ 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 $\tau:F(u)\rightarrow F(v)$ 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 $\sigma:\overline{K}\rightarrow\overline{K}$ It follows that $\sigma\vert_K:K\rightarrow\overline{K}$ 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.