Fundamental Theorem of Galois Theory

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Theorem

Let $L / K$ be a finite Galois extension.

Let $\Gal {L / K}$ denote the Galois group of the extension $L / K$.

Let $H$ denote a subgroup of $\Gal {L / K}$ and $F$ denote an intermediate field.

The mappings:

$H \mapsto L_H$, and

$F \mapsto \Gal {L / F}$

are inclusion-reversing and inverses.

Moreover, these maps induce a bijection between the normal subgroups of $\Gal {L / K}$ and the normal, intermediate extensions of $L / K$.

Proof

First, we show that the maps are inclusion-reversing.

Let $K \subset F_1 \subset F_2 \subset L$.

Let $G_i = \Gal {L / F_i}$.

Let $\sigma \in G_2$.

Then $\sigma$ is an automorphism of $L$ which fixes $F_2$.

Since $F_1 \subset F_2$, it follows that $\sigma$ fixes $F_1$ and consequently $\sigma \in G_1$.

Let $H_1 \subset H_2 \subset \Gal {L / K}$.

Let $F_i = L_{H_i}$.

Let $x \in F_2$.

Then $\map \sigma x = x$ for all $\sigma \in H_2$.

Since $H_1\subset H_2$, the same equality holds for each element of $H_1$ and thus $x \in F_1$.

For the remainder of the proof:

let $G$ denote $\Gal {L / K}$

for any field $K \subset F \subset L$ let $G_F$ denote $\Gal {L / F}$.

Next, we demonstrate that the two functions described are inverses.

That is:

For any intermediate field $K \subset F \subset L$:

$F = L_{G_F}$

For any subgroup $H \subset G$:

$H = G_{L_H}$

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For the first equality, $F$ is fixed by every element of $G_F$.

By definition $L_{G_F}$ includes all elements of $L$ which are fixed by $G_F$.

Thus we have $F \subseteq L_{G_F}$.

Aiming for a contradiction, suppose there exists $\alpha \in L_{G_F} \setminus F$.

Then:

$\index {\map F \alpha} F > 1$

where $\index {\map F \alpha} F$ is the degree of $\map F \alpha / F$.

We can express the minimal polynomial of $\alpha$ in terms of $G_F$ as:

$\ds \map {m_\alpha} x = \prod_{\sigma \mathop \in G_F} \paren {x - \map \sigma \alpha}^\frac 1 {\index L {\map F \alpha} }$

However, by our assumption, $\map \sigma \alpha = \alpha$ for each $\sigma$.

Thus:

\(\ds \map {m_\alpha} x\)

\(=\)

\(\ds \paren {x - \alpha}^\frac {\index L F} {\index L {\map F \alpha} }\)

\(\ds \)

\(=\)

\(\ds \paren {x - \alpha}^{\index {\map F \alpha} F}\)

Degree of Field Extensions is Multiplicative

Since $\index {\map F \alpha} F > 1$, this contradicts the separability of $L / F$.

Therefore, the first equality holds.

For the second equality, it is immediate that $H \subset G_{L_H}$.

Aiming for a contradiction, suppose $H$ were a proper subset of $G_{L_H}$.

By Primitive Element Theorem, there exists an $\alpha \in L$ such that $L = \map {L_H} \alpha$.

Consider the polynomial:

$\ds f = \prod_{\sigma \mathop \in H} \paren {x - \map \sigma \alpha}$

The coefficients of $f$ are evidently elements of $L_H$ and $f$ is monic by construction.

However:

$\index L {L_H} = \map \deg f = \order H < \order {G_{L_H} } = \index L {L_H}$

by definition of Galois extension.

This is a contradiction and it follows that $H = G_{L_H}$.

Finally, we demonstrate the correspondence between normal subgroups of $G$ and the intermediate normal extensions of $K$.

Suppose $K \subset F \subset L$ is an intermediate field and $F / K$ is a normal extension.

We let $H = \Gal {F / K}$ denote the Galois group of interest.

Let $\sigma \in G$ and $\tau \in H$.

We want to show that $\sigma^{-1} \tau \sigma \in H$ to conclude that $H$ is normal.

We have that $F \subset L$.

Thus $\sigma$ restricts to an embedding of $F$ in $\overline K$.

However, since $F / K$ is a normal extension, the image of every embedding of $F$ is again $F$.

Thus, $\sigma$ restricts to an automorphism of $F$.

Let $x \in F$.

Then $\sigma \left({x}\right) \in F$.

We have that $\tau$ fixes $F$.

Thus:

$\map \tau {\map \sigma x} = \map \sigma x$

Therefore:

$\map {\sigma^{-1} } {\map \tau {\map \sigma x} } = x$

and we conclude that:

$\sigma^{-1} \tau \sigma \in H$

Next, suppose $H$ is a normal subgroup of $G$ and $F = L_H$.

Let $\tau \in H$ and $\sigma: F \mapsto \overline K$ be an embedding of $F$.

By Extension of Isomorphisms, we extend $\sigma$ to $\overline \sigma$, an automorphism of $L$.

Consider the composition $\hat \sigma^{-1} \tau \hat \sigma = \hat \tau \in H$ by our assumption of normality.

Then:

$\map {\hat \sigma^{-1} \tau \hat \sigma} x = \map {\hat \tau} x = x$

which implies that:

$\map \tau {\map {\hat \sigma} x} = \map {\hat \sigma} x \in F$

Since $x \in F$:

$\map {\hat \sigma} x = \map \sigma x \in F$

which was required to be shown.

$\blacksquare$

Source of Name

This entry was named for Évariste Galois.