Equivalence of Definitions of Normal Extension

Theorem
Let $L/K$ be an algebraic field extension. Then the following are equivalent:


 * $(1): \quad$ For every irreducible polynomial $f\in K[x]$ with a root in $L$, $f$ splits completely in $L$.


 * $(2): \quad$ For every embedding $\sigma$, of $L$ in the algebraic closure $\overline{K}$ which fixes $K$ pointwise, $\sigma(L) = L$.

Proof
$1\implies 2$

Suppose $\alpha\in L$ is an arbitrary element and $\sigma: L\mapsto \overline{K}$ is an arbitrary embedding of $L$ fixing $K$.

We wish to show that $\sigma(\alpha)\in L$.

Let $m_\alpha$ be the minimal polynomial of $\alpha$ over $K$, which exists because $L/K$ is algebraic.

Since $\sigma$ fixes $K$, $\sigma(\alpha)$ must also be a root of $m_\alpha$.

By our assumption, $\alpha\in L$ implies that all roots of $m_\alpha$ are in $L$ and consequently $\sigma(\alpha)\in L$.

$2\implies 1$

Again, let $\alpha\in L$ and let $m_\alpha\in K[x]$ be its minimal polynomial over $K$.

We must show that for every root $\beta$ of $m_\alpha$, there exists an embedding $\sigma_\beta$, of $L$ in $\overline{K}$ such that $\sigma_\beta(\alpha) = \beta$.

Consider the intermediate field $K[\alpha]\subset L$.

By Abstract Model of Algebraic Extensions, we have an automorphism $\tau_\beta$ for each root $\beta$ of $m_\alpha$ such that $\tau_\beta(\alpha) = \beta$ and $\tau_\beta$ fixes $K$.

By Extension of Isomorphisms, each $\tau_\beta$ can be extended to an embedding $\sigma_\beta$ of $L$ in $\overline{K}$ such that $\sigma_\beta|_{K[\alpha]}=\tau_\beta$.

By our assumption, $\sigma_\beta(L)=L$ for each $\beta$ and consequently every root of $m_\alpha$ is in $L$.