# equivalent conditions for normality of a field extension

###### Theorem.

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

1. 1.

$K$ is normal over $F$;

2. 2.

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

3. 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 $\{f_{x}:x\in X\}$ 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 $\sigma(u_{i})\in\{u_{1},\ldots,u_{n}\}$ 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 $\sigma|_{K}:K\rightarrow\overline{K}$ 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. ∎

Title equivalent conditions for normality of a field extension EquivalentConditionsForNormalityOfAFieldExtension 2013-03-22 18:37:56 2013-03-22 18:37:56 azdbacks4234 (14155) azdbacks4234 (14155) 8 azdbacks4234 (14155) Theorem msc 12F10 NormalExtension SplittingField ExtensionField AlgebraicExtension