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If it is wrong, then how about restrict the varieties to smooth ones?

In order to investigate it I also asked whether Aut(X) always acts transitively over a projective variety X. But I don't know how to prove/disprove this either... This latter question is also one thing I want to ask in this post.

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3 Answers 3

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As PseudoNeo observes, the group of automorphisms of a compact Riemann surface (i.e. smooth projective curve over $\mathbb{C}$) of genus $> 1$ is always finite (and the order can in fact be bounded). This implies that the local rings are generically non-isomorphic. Indeed, if $C$ is such a curve, and $p,q$ two points such that $\mathcal{O}_{C, p} \simeq \mathcal{O}_{C, q}$ as $\mathbb{C}$-algebras, then the associated isomorphism $\mathrm{Spec} \mathcal{O}_{C, p} \simeq \mathrm{Spec} \mathcal{O}_{C, q}$ (which automatically extends to a birational map $C \dashrightarrow C$ taking $p$ to $q$) would extend to an isomorphism of $C$ with itself, because $C$ is a smooth and projective curve (to see that a birational map between smooth projective curves is an isomorphism, one can use the valuative criterion). It follows that if $p$ is a fixed point of the curve, then there are only finitely many points $q$ such that $\mathcal{O}_{C, p} \simeq \mathcal{O}_{C, q}$.

The intuition is that the local ring at a point on an algebraic variety (unlike the case with smooth manifolds or analytic spaces, for instance) is not a truly local invariant in that it remembers far too much about the variety: for instance, it is enough to reconstruct the variety up to birational equivalence. This is one of the reasons that the algebraic process of completion is sometimes more relevant. When one completes the local ring at a smooth point on an algebraic variety, one gets a power series ring over $\mathbb{C}$, so the completion only remembers the dimension of the variety.

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Indeed! But by brutal force is my way of checking the extension to a birational map, and that birational map between smooth projective curves is an isomorphism. It seems you have another way to look at them(for eg. why do you pass to the Spec of the two local ring first? And what is the valuative criterion?) I'm curious about it. –  Honglu Nov 23 '11 at 2:14
    
@lethe: The valuative criterion states essentially the following. Let $k$ be a field. A variety $X$ over $k$ is proper if and only if a rational map into $X$ from a $k$-discrete valuation ring (where $k$ is contained in the group of units) extends to a regular map. (At least, that's roughly how I think of it; the statement is more general, and applies to a morphism of schemes. It's in Hartshorne.) The extension to a birational map is automatic once you have an isomorphism between the two local rings. The language about $\mathrm{Spec} $ is simply convenient here... –  Akhil Mathew Nov 23 '11 at 2:47
    
...as it implies that, for instance, given varieties $X, Y$, then a map between their local rings at two points $p \in X, y \in Y$ induces a rational map of varieties in the opposite direction. –  Akhil Mathew Nov 23 '11 at 2:48

It is not true that Aut(X) acts transitively. It's for example false for all Riemann surfaces of genus > 1, as the automorphism group is then finite.

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Ahh thanks, it seems to ring a bell! Is there a reference for the proof of this result? –  Honglu Nov 22 '11 at 11:24
    
The proof I know is through Weierstrass points (en.wikipedia.org/wiki/Weierstrass_point): it might interest you, as Weierstrass points are special points on a Riemann surface of genus >1, so their existence proves immediately that Aut(X) cannot act transitively. I know a proof of Hurwitz's theorem through this route is included in Farkas and Kra's book Riemann surfaces. –  PseudoNeo Nov 22 '11 at 12:46

If $X$ is an algebraic group over an algebraic closed field, then $X$ acts transitively on itself by translations, so Aut$(X)$ acts also transitively on $X$.

In general, if Aut$(X)$ (which is an algebraic group if $X$ is projective) acts transitively on $X$, and if the stabilizer $H$ at some point of $X$ is a normal subgroup (e.g. if Aut$(X)$ is abelian), then $X$ is the quotient of $G/H$ and is an algebraic group. So in some sens, what you require can only happen for abelian varieties (or algebraic groups if you don't ask $X$ to be projective). (Well, this happens also for some rational varieties as the projective spaces).

In a more positive direction, if $x, y$ are smooth points in $X, Y$ with the same dimension over a perfect base field, then the strict henselizations of their local rings are isomorphic (''strict'' can be dropped if you are over an algebraically closed field). This says that for the étale topology the local ring at a smooth point is uniquely determined by the dimension.

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Hmm. Do you have a reference on the statement on strict henselianizations? It sounds interesting. –  Akhil Mathew Nov 23 '11 at 19:18
    
@AkhilMathew, the strict henselization represents the universal étale cover of Spec of the local ring, so it doesn't change with finite étale cover. Now Zariski locally at $x$ and $y$, $X, Y$ are étale over an open subvariety of the same affine space, so we are reduced to the situation of two (closed) points in an affine space. While writing I see that I should suppose the base field $k$ be perfect... So base change to a finite separable extension of $k$ (which is an étale morphism), we can suppose $x, y$ are rational, then they have isomorphic local rings. –  user18119 Nov 23 '11 at 20:10
    
Thanks! That's what I was looking for. –  Akhil Mathew Nov 26 '11 at 15:21

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