Equivalence of Definitions of Finite Galois Extension
Theorem
Let $L / K$ be a finite field extension.
The following definitions of the concept of Finite Galois Extension are equivalent:
Definition 1
$L / K$ is a (finite) Galois extension if and only if the fixed field of its automorphism group is $K$:
- $\map {\operatorname{Fix}_L} {\Gal {L / K} } = K$
Definition 2
$L / K$ is a (finite) Galois extension if and only if it is normal and separable.
Definition 3
$L / K$ is a (finite) Galois extension if and only if the order of the automorphism group $\Aut {L / K}$ equals the degree $\index L K$:
- $\order {\Aut {L / K} } = \index L K$
Proof
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:
its minimal polynomial over $K$ splits completely in $L$.
Thus $L / K$ is normal.
By:
$L / K$ is separable.
$\Box$
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.
$\Box$
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