We have the following result:


Let $R=\mathbb{C}[t]\_f$, with $f=(t-a\_1)(t-a\_2)\cdots (t-a\_n)$. Then the automorphism group of $R$ is isomorphic to the group of all Moebius transformations which fix (not necessary pointwise) the set $\{a\_1,\ldots,a\_n,\infty\}$. 


Is it a known result or a direct consequence of some known theorem in algebraic geometry?



A problem on Moebius  transformations
Since $\textrm{Spec}(\mathbb{C}[t]\_f)=\mathbb{A}^1-\{a\_1, \ldots, a\_n\}$, we are reduced to compute $\textrm{Aut}(\mathbb{A}^1-\{a\_1, \ldots, a\_n\})$.


Every automorphism 


$\phi \colon \mathbb{A}^1-\{a\_1, \ldots, a\_n\} \longrightarrow \mathbb{A}^1-\{a\_1, \ldots, a\_n\}$


gives rise to a birational map


$\bar{\phi} \colon \mathbb{P}^1 \dashrightarrow \mathbb{P}^1$


such that $\bar{\phi}(\infty) \in \{a\_1, \ldots, a\_n, \infty \}$. Every birational map of $\mathbb{P}^1$ is an isomorphism, hence $\bar{\phi}$ is a Moebius transformation. 


Since it is induced by $\phi$, such a transformation must fix, not necessarily pointwise, the set $\{a\_1, \ldots, a\_n, \infty \}$, so the claim follows.


