Question

If K is an extension field of Qand$\mathit{\sigma }$is an$\mathbf{Q}$ automorphism of K, prove that$\mathit{\sigma }$ is a Q-automorphism.

Short answer

It is proved that $\mathit{\sigma }$ is an Q-automorphsim.

Step-by-step solution

Set of automorphisms

The set of automorphisms on a group forms a group itself, it contain the product the product is composition of homomorphism.

Showing that σ is an Q automorphism.

Here, Kis an extension field of Qand is an automorphismof K . If it automorphism, so it is necessary. It’s also a homomorphism and fulfills the condition of one to one and onto and it also the isomorphism. So, by using the scaling property, we can say that

$$\begin{gathered} \mathit{\sigma }\mathbf{\left(}\mathbf{c}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathit{c}\mathit{\sigma }\mathbf{\left(}\mathbf{x}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{c}\mathit{x} \end{gathered}$$

Assume$\mathit{\sigma }\left(1\right)$ by using$\mathbf{\sigma }\left(x\right)\mathbf{=}\mathbf{x}$ because it is a isomorphism.

Then,

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{\sigma }\mathbf{:}\mathbf{𝆏}\left(n\right)\mathbf{\cong }\mathbf{𝆏}\left(-n\right) \\ \mathit{\sigma }\left(n\right)\mathbf{=}\mathit{n} \end{gathered}$$

Here,$\mathit{n}$is the element of the group $\mathit{Z}$and $\mathit{Z}$is the extension field. Hence, $\mathit{K}$is the extension field of $\mathit{Q}$and is an automorphism of $\mathit{K}$. So, $\mathit{\sigma }$is also a$\mathit{Q}$ automorphism.