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If $L/K$ is a field extension, then $L$ can be considered as a $K$-vectorspace of dimension $[L:K]$.

If we consider $K$-automorphisms of $L$, they take $\overline\sigma: L\to L$ where $\overline\sigma|_K = \sigma:K\to K$

This means that $\overline\sigma$ must fix all points in $K$ right? I.e. any automorphisms of an infinite field $K$ must be the identity map? $\overline \sigma: a\mapsto a$, $a\in K$ and then $\overline\sigma$ maps all adjoined roots to different roots of their characteristic polynomial.

Can I please have an example of a $K$-automorphism of $L$, and how it permutes the adjoined root to a different adjoined root of its characteristic polynomial?

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One of the easiest examples is to consider $\mathbb{C}$ as an extension of $\mathbb{R}$. Then $\sigma: \mathbb{C} \rightarrow \mathbb{C}$ given by $z \mapsto \overline{z}$ is an automorphism that fixes all of $\mathbb{R}$.

Or for a finite example, adjoining a root of an irreducible polynomial of degree $n$ to a finite field $\mathbb{F}_{q}$ gives an extension field $\mathbb{F}_{q^n}$. Then the Frobenius mapping, which is $\varphi: \mathbb{F}_{q^n}\rightarrow \mathbb{F}_{q^n}$ given by $\alpha \mapsto \alpha^q$ is an automorphism that fixes all the elements of $\mathbb{F}_{q}$

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