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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?

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1 Answer 1

up vote 3 down vote accepted

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.

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1  
Dear Francesco, this is not quite correct. The automorphism $\phi \colon \mathbb{A}^1-\{0\}\longrightarrow \mathbb{A}^1-\{0\}$ defined by $\phi(z)=z^{-1}$ does indeed extend to an automorphism $\bar{\phi} \colon \mathbb{P}^1 \dashrightarrow \mathbb{P}^1$ but this latter automorphism does not send $\infty$ to itself –  Georges Elencwajg Mar 14 '11 at 12:38
    
Dear George, you are completely right :-) The point is, of course, that either $\infty$ goes to itself or it goes to some of the $a_i$ (as in your example). I will edit the answer, thank you for the remark. –  Francesco Polizzi Mar 14 '11 at 13:32
    
We prove the result algebraically(using Aut(C(t))). And i am not familiar with algebraic geometry. Would please tell the exact theorem (or reference book) used in the proof so that i can known which proof is more direct: (1)The 1-1 correspondence of Aut(C[t]_f) and Aut(A 1 −{a 1 ,…,a n }). and (2)Any automorphism ϕ:A 1 −{a 1 ,…,a n }⟶A 1 −{a 1 ,…,a n } gives rise to a birational map ϕ ¯ :P 1 ⇢P 1 Thanks a lot! – ren l 0 secs ago –  ren l Mar 14 '11 at 14:00
    
(1) In general, if $A$ and $B$ are rings then any (iso)morphism of locally ringed spaces $\textrm{Spec}(B) \to \textrm{Spec}(A)$ is induced by a (iso)morphism of rings $A \to B$, and conversely. See [Hartshorne, Algebraic Geometry, Prop. 2.3 p. 73]. (2) This is just the definition of birational map: a birational map of $\mathbb{P}^1$ is a self-map which is an isomorphism over a Zariski open set (i.e., over $\mathbb{P}^1$ minus a finite number of points). It is well known that the birational maps of curves are biregular, so for $\mathbb{P}^1$ they are exactly the Moebius transformations. –  Francesco Polizzi Mar 14 '11 at 17:03
    
Thanks. I will look for the proof of the well known result to see whether it depend on the result of Aut(C(t)). If we replace $R=C[t]_f$ by $R=S^{-1}C[t]$, where $S$ be a muliplicative closed set. Let $P=\{a\in \C| f(a)=0 for some f\in S\}$. We will have the automorphism group of R is isomorphic to the group of moebius transformations which fix the set $P\cup \{\infty\}$. This times A^1-P is possiblly not a zarisk open set. How to deal with it in an algebra geometry way? –  ren l Mar 14 '11 at 23:25

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