Equivalence of Definitions of Finite Galois Extension

Theorem
Let $L/K$ be a finite field extension.

1 implies 2
Note that by Finite Field Extension has Finite Galois Group, $G = \Aut {L / K}$ is finite.

Let $\alpha\in L$.

Then its orbit under $G$ is finite.

By:
 * Minimal Polynomial of Element with Finite Orbit under Group of Automorphisms over Fixed Field in terms of Orbit

its minimal polynomial over $K$ splits completely in $L$.

Thus $L/K$ is normal.

By:
 * Finite Orbit under Group of Automorphisms of Field implies Separable over Fixed Field

$L / K$ is separable.

2 implies 3
Since $L / K$ is Galois, it is separable.

Thus, by the Primitive Element Theorem, there exists an $\alpha \in L$ such that $L = \map K \alpha$.

Let $m_\alpha \in K \sqbrk x$ be the minimal polynomial of $\alpha$ over $K$.

Then:
 * $\index L K = \map \deg {m_\alpha}$

Suppose $\sigma \in \Gal {L / K}$.

Then:


 * $\map {m_\alpha} {\map \sigma \alpha} = \map \sigma {\map {m_\alpha} \alpha} = \map \sigma 0 = 0$

since $m_\alpha$ has coefficients in $K$.

Therefore $\map \sigma \alpha$ must be a root of $m_\alpha$.

Every element of $\Gal {L / K}$ is determined by its value at $\alpha$ by our assumption that $L = \map K \alpha$.

Therefore:
 * $\order {\Gal {L / K} } \le \map \deg {m_\alpha} = \index L K$

Next, suppose $\beta$ is a root of $m_\alpha$.

By the normality of $L / K$, we must have $\beta \in L$.

Then, by Abstract Model of Algebraic Extensions:


 * $\map K \alpha \cong K \sqbrk x / \gen {m_\alpha} \cong \map K \beta$

Composing isomorphisms we have an automorphism of $L$ for each $\beta$.

Thus:
 * $\order {\Gal {L / K} } \ge \map \deg {m_\alpha} = \index L K$

from which our result follows.