How to prove that if $\sigma\in \operatorname{Gal}(k(x)/k)\Leftrightarrow \sigma(x)=\frac{ax+b}{cx+d}$? [duplicate]

Solution 1:

The formula $\sigma(x)=\frac{ax+b}{cx+d}$ is only supposed to tell you directly what $\sigma$ does to one particular element of $k(x)$, namely $x$ itself, not what it does to other elements like $0\in k(x)$. For example, say we consider $\mathbb{Q}(e)/\mathbb{Q}$. The equation $\sigma(e)=e+1$ is just a single equation, it doesn't tell you directly what $\sigma(0)$ is. It doesn't make any sense to "evaluate" $2.71828\dots$ to $0$.

You can however indirectly determine what $\sigma$ does to any element of $k(x)$ using the knowledge of $\sigma(x)$ and the fact $\sigma$ is a field automorphism. Since any element of $k(x)$ is built from scalars of $k$ and $x$ using the basic arithmetic operations, and $\sigma$ preserves these scalars and operations, we can say if $f(x)$ is any rational expression in $x$ (an element of $k(x)$) then $\sigma(f(x))$ must equal $f(\sigma(x))$. For example, if $\sigma(x)=x+1$ then $\sigma(x^2-1)=\sigma(x)^2-1=(x+1)^2-1$ and $\sigma(0)=0$.