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Let $R$ be a field and let $R(x)$ be the field of rational functions in $x$ whose coefficients are in $R$.

Let $g = \left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right) \in GL_2(\mathbb{R})$ and define $\sigma_g(x)$ so that

  • $\forall r \in \mathbb{R}, \sigma_g(r)=r \space\space\space (\sigma_g=Id_R)$

  • $\sigma_g(x) = \frac{ax+b}{cx+d}$

can I conclude $\sigma_g(x) \in Aut(R(x)/R)$?

the opposite direction is clear to me and is detailed in this post.

any help would be appreciated =]

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Daniel
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1 Answers1

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It's important that $(a,b)$ is not a constant multiple of $(c,d)$ so that $\frac{ax+b}{cx+d}$ isn't a constant. The equivalent condition of course is that $\det(\begin{smallmatrix}a&b\\c&d\end{smallmatrix})\ne0$.

Since $\frac{ax+b}{cx+d}$ is algebraic over $R\iff x$ is algebraic over $R$, we know $\frac{ax+b}{cx+d}$ is transcendental.

So there is an isomorphism $R(x)\to R(\frac{ax+b}{cx+d})$. Now think of this as a map $\sigma_g:R(x)\to R(x)$.

Since $\sigma_g$ has an inverse $\sigma_{g^{-1}}$ it is a field automorphism.

anon
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