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If $X$ and $Y$ are two varieties and the germs of regular functions $\mathcal O_{x,X}$ and $\mathcal O_{y,Y}$ of two points $x \in X$ and $y \in Y$ are isomorphic as $k$-algebras, then can we find two neighborhoods $x\in U$ and $y \in V$ and an isomorphism $U\simeq V$ sending $x$ to $y$?

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Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. – Julian Kuelshammer Nov 8 '12 at 6:55
The isomorphism induces an isomorphism on their quotient fields, so the conclusion. – Ehsan M. Kermani Nov 8 '12 at 7:42
Why does quotient field isomorphism imply existence of isomorphic neighborhood? – user27126 Nov 8 '12 at 7:56
I think the basic idea is that an isomorphism on the fraction fields induces a birational map between the two varieties. But there may be some conditions attached. – Zhen Lin Nov 8 '12 at 12:03
I feel like the underlying idea of proving the existence of the birational map is the same as this question, which makes it sort of circular. – user27126 Nov 8 '12 at 18:11
up vote 14 down vote accepted

The result follows from two general facts on morphism of varieties ($k$ can be any commutative noetherian ring and $X, Y$ are scheme of finite type over $k$; one can even remove the noetherian hypothesis if $X, Y$ are finitely presented over $k$):

  1. (Extension from local rings to open neighborhoods) Let $\phi: O_{Y,y}\to O_{X,x}$ be a homomorphism of local $k$-algebras. Then there exist open neighborhoods $U\ni x$, $V\ni y$ and a morphism $f : U\to V$ which induces $\phi$ (i.e. $f(x)=y$ and $f_x: O_{Y,y}\to O_{X,x}$ is equal to $\phi$).

  2. (Uniqueness) Let $h_1, h_2 : X\to Y$ be two morphisms such that $h_1(x)=h_2(x)=y$ and $(h_{1})_x=(h_2)_x$. Then there exists an open neighborhood $U$ of $x$ such that $h_1|_{U}=h_2|_{U}$.

Proof of (1): Let $V$ be an affine open neighborhood of $y$ in $Y$. Then $O_Y(V)$ is finitely generated over $k$. Let $$ k[T_1, \dots, T_n]\twoheadrightarrow O_Y(V) $$ be a surjective morphism of $k$-algebras with kernel $I$. Let $b_i\in O_Y(V)$ be the image of $T_i$. As $k[T_1, \dots, T_n]$ is noetherian, $I$ is generated by finitely many polynomials $P_1(\underline{T}), \dots, P_m(\underline{T})$ (here $\underline{T}=(T_1, \dots, T_n)$). There exists an affine open neighborhood $U$ of $x$ and $a_1, \dots, a_n\in O_X(U)$ such that $\phi((b_i)_y)=(a_i)_x$ for all $i\le n$. As $(P_j(\underline{b}))_y=0_y=0$, we have $P_j(\underline{a})_x=0$. So shrinking $U$ if necessary, we can suppose $P_j(\underline{a})=0$. Therefore the $k$-algebra map $$ k[T_1, \dots, T_n]\to O_X(U), \quad T_i\mapsto a_i$$ factorizes throught $\psi: O_Y(V)\to O_X(U)$. By construction we have the following commutative diagram where the vertical arrows are canonical maps to the stalks $$\begin{matrix} O_Y(V) & \stackrel{\psi}{\longrightarrow} & O_X(U)\\ \downarrow&&\downarrow\\ O_{Y,y}&\stackrel{\phi}{\rightarrow}&O_{X,x} \end{matrix} $$ Hence $\psi$ maps the preimage of $m_yO_{Y,y}$ in $O_Y(V)$ to the preimage of $m_xO_{X,x}$ in $O_X(U)$. So the morphism $f: U\to V$ induced by $\psi$ satisfies $f(x)=y$ and $f_x=\phi$.

Proof of (2). This is easier. Shrinking $X$ and $Y$, we can suppose $X$ and $Y$ are affine. Let $\psi_1, \psi_2: O_Y(Y)\to O_X(X)$ be the morphisms of $k$-algebras corresponding to $h_1, h_2$. Let $c_1, \dots, c_m$ be generators of $O_Y(Y)$ as $k$-algebra. Then $$(\psi_1(c_i))_x=(h_1)_x((c_i)_y)=(h_2)_x((c_i)_y)=(\psi_2(c_i))_x.$$ So for some affine open neighborhood $U$ of $x$, we have $\psi_1(c_i)|_U=\psi_2(c_i)|_U$ for all $i\le m$. Thus $\psi_1|_U=\psi_2|_U$ and $h_1|_U=h_2|_U$.

I let you find out how to extend an isomorphism of local rings to an isomorphism of open neighborhoods using (1) and (2). If you need more hints, please tell.

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If QiL allows it, it would be great if some $\TeX$ guru drew the commutative diagram alluded to in this answer. – Georges Elencwajg Nov 9 '12 at 8:32
@GeorgesElencwajg Dear Georges, I finally find a solution by looking at other messages with commutative diagrams. – user18119 Nov 9 '12 at 10:36
Thanks QiL, this makes the answer even better now! – Georges Elencwajg Nov 9 '12 at 11:06

Here is the expanded version of my comment:

Let $f: \mathcal{O}_{x,X} \to \mathcal{O}_{y,Y}$ be the $k$-algebra isomorphism. Since their Krull dimensions are equal, WLOG, we can assume that they are affine varieties of $\mathbb{A}^n.$

For simplicity, assume that $P=Q=0.$ Define a map $g$ from an open subset of $X$ to an open subset of $Y$, by $g(x_1,…,x_n)=(f^{-1}(y_1)(x_1,…,x_n), f^{-1}(y_2)(x_1,…,x_n), …, f^{-1}(y_n)(x_1,…,x_n))$ where $y_i$’s are coordinate functions on $\mathbb{A}^n$ and $g$ is defined on an open subset where $ f^{-1}(y_i)$’s are defined. You can construct the inverse of $g$ on some open subset similarly, therefore, the desired isomorphism.

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Dear ehsanmo, Your solution 1 doesn't quite answer the question, because by passing to quotient fields, you lose any control of what points might be in $U$ and $V$. In particular, there is no way (once you've passed to the quotient fields) to ensure that $x$ is in $U$ or that $y$ is in $V$, let alone that $x$ is taken to $y$. Regards, – Matt E Nov 9 '12 at 0:31
Dear @MattE: thank you for pointing that out, edited. – Ehsan M. Kermani Nov 10 '12 at 21:05

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