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.
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Title  equivalent conditions for normality^{} of a field extension 

Canonical name  EquivalentConditionsForNormalityOfAFieldExtension 
Date of creation  20130322 18:37:56 
Last modified on  20130322 18:37:56 
Owner  azdbacks4234 (14155) 
Last modified by  azdbacks4234 (14155) 
Numerical id  8 
Author  azdbacks4234 (14155) 
Entry type  Theorem 
Classification  msc 12F10 
Related topic  NormalExtension 
Related topic  SplittingField 
Related topic  ExtensionField 
Related topic  AlgebraicExtension 