Given the field of rational functions $\Bbb C(t)$, how do we show that a certain function defines an automorphism. I ask because I read here https://nptel.ac.in/courses/111101001/downloads/problemset8.pdf. That $\sigma(t)=wt$, defines an automorphism but automorphisms must be isomorphisms which implies that $\sigma(u*v)=\sigma(u)*\sigma(v)$. But it seems to me $\sigma(uv)=wuv\neq w^2 uv=\sigma(u)\sigma(v)$.
What I think might be the case :
We're dealing with $\Bbb C(t)=\{\tfrac{f(t)}{g(t)}|f,g \in \Bbb C[x], g(t) \neq0 \}$.
So we should then apply our element as follows :
$\sigma(\tfrac{f(t)}{g(t)})=\tfrac{f(\sigma(t))}{g(\sigma(t))}=\tfrac{f(wt)}{g(wt)}$.
But then in terms of showing this is an automorphism , I'm not sure how to proceed , $f,g$ are just arbitrary functions so I don't understand how could manipulate such functions to show $\sigma(\tfrac{f(t)}{g(t)}\tfrac{h(t)}{k(t)})=\sigma(\tfrac{f(t)}{g(t)})\sigma(\tfrac{h(t)}{k(t)})$.
Could anyone clarify this point for me ?
Note: The link isn't from a course I'm in , I just found it while searching for information on Galois theory.
Presumably $w$ is a non-zero complex number.
The definition of $\sigma$ means that $$\sigma(f)=f\circ\psi,$$where $$\psi(t)=wt.$$That should make it clear that $\sigma(f+g)=\sigma(f)+\sigma(g)$, etc.