Galois Group Acts Faithfully on Generating Set

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Theorem

Let $G$ be the Galois group of a finite field extension $E/ F$.

Let $\set {\alpha_1, \ldots , \alpha_n}$ be a generating set for field $E$ as an $F$-vector space.

Let $f_1, \ldots, f_n$ be their minimal polynomials.

Let $f = f_1 \dots f_n$ be the product of them.

Then the action of $G$ on roots of $f$ is faithful.

Proof

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Suppose that $\sigma \in G$ fixes each $\alpha_i$.

Then since $\sigma$ is an $F$-linear automorphism of $E$ and since $\set {\alpha_1, \ldots, \alpha_n}$ is a generating set for $E$, we see that $\sigma$ fixes every other element of $E$.

Then $\sigma$ must be the identity map $I_E: E \to E$.

Then the action of $G$ on roots of $f$ is faithful.

$\blacksquare$