Question

Assume $\mathit{[}\mathit{K}\mathit{:}\mathit{F}\mathit{]}$ is finite. Is it true that every F-automorphism of Kis completely determined by its action on a basis of Kover F?

Short answer

Yes. If $\mathit{[}\mathit{K}\mathit{:}\mathit{F}\mathit{]}$is finite, it is true that every F-automorphism of K is completely determined by its action on a basis of K over F.

Step-by-step solution

Basis of extension field definition.

The basis of extension field K over the subfield F is defined when the extension consists the finite elements.

Showing that every  F-automorphism of K is completely determined.

Assume that $\mathit{[}\mathit{K}\mathit{:}\mathit{F}\mathit{]}$ is finite for every Fautomorphism of K is completely get by action on a basis of K over F. Suppose there are basis in given condition, it can be written as $\mathit{f}\mathbf{}\mathbf{:}\mathbf{}\mathit{K}\mathbf{\to }\mathit{K}$.

Then, it is clear that $\mathit{f}\left(x_i\right)\mathbf{=}\left\{x_1,x_2,x_3,...,x_n\right\}$and $\mathit{x}\in \mathit{K}$.

Now using the scaling property to find the every Fautomorphism of K , we have

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{x}\mathbf{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{1}}^{\mathbf{n}}{\mathit{c}}_{\mathbf{i}}{\mathit{x}}_{\mathbf{i}} \\ \mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathit{f}\mathbf{\left(}{\mathbf{c}}_{\mathbf{i}}{\mathbf{x}}_{\mathbf{i}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{f}\mathbf{(}{\mathbf{c}}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}{\mathbf{c}}_{\mathbf{3}}{\mathbf{x}}_{\mathbf{3}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathbf{c}}_{\mathbf{n}}{\mathbf{x}}_{\mathbf{n}}\mathbf{)}\mathbf{} \end{gathered}$$

Now, using the additional property the function becomes:

role="math" localid="1657957826748" $$\begin{gathered} \mathit{f}\mathbf{(}{\mathbf{c}}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathbf{c}}_{\mathbf{n}}{\mathbf{x}}_{\mathbf{n}}\mathbf{)}\mathit{}\mathit{=}\mathit{f}\mathit{\left(}{\mathit{c}}_{\mathit{1}}{\mathit{x}}_{\mathit{1}}\mathit{\right)}\mathit{+}\mathit{f}\mathit{\left(}{\mathit{c}}_{\mathit{2}}{\mathit{x}}_{\mathit{2}}\mathit{\right)}\mathit{+}\mathit{.}\mathit{.}\mathit{.}\mathit{+}\mathit{f}\mathit{\left(}{\mathit{c}}_{\mathit{n}}{\mathit{x}}_{\mathit{n}}\mathit{\right)} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}{\mathit{c}}_{\mathbf{1}}\mathit{f}\mathit{\left(}{\mathit{x}}_{\mathit{1}}\mathit{\right)}\mathit{+}{\mathit{c}}_{\mathbf{2}}\mathit{f}\mathit{\left(}{\mathit{x}}_{\mathit{2}}\mathit{\right)}\mathit{+}\mathit{.}\mathit{.}\mathit{.}\mathit{+}{\mathit{c}}_{\mathbf{n}}\mathit{f}\mathit{\left(}{\mathit{x}}_{\mathit{n}}\mathit{\right)} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{=}\mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{1}}^{\mathbf{n}}{\mathit{c}}_{\mathbf{i}}\mathit{f}\mathit{\left(}{\mathit{x}}_{\mathit{i}}\mathit{\right)} \end{gathered}$$

Hence, yes if it is $\mathit{[}\mathit{K}\mathit{:}\mathit{F}\mathit{]}$ finite then every F automorphism of K is completely determined by the basis of K over F.