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Let $X$ be a smooth variety with dimension $ n$ and $ p \in X $ a closed point. By definition, $ \mathscr{O}_p$ is a regular local ring, so we can choose a regular system of parameters $ u_1, \dots, u_n$ to generate the maximal ideal in $ \mathscr{O}_p$. There is a very strong analogy between smooth coordinate charts and a regular system of parameters:

  • The stalk $ \Omega_{\mathscr{O}_p / k} $ of the cotangent bundle is a free $ \mathscr{O}_p$-module with basis $ du_1, \dots, du_n$.
  • The completion $ \widehat{\mathscr{O}_p}$ is the power series ring $ k[[u_1, \dots, u_n]]$.
  • If $ Y \subseteq X $ is a smooth sub-variety with codimension $ d $ passing through $ p $, then there is a regular system of parameters $ t_1, \dots, t_n $ such that in some Zariski open neighborhood around $ p $ we have $Y = V(t_1, \dots, t_d)$. This is an algebraic analogue of the rank theorem from differential geometry

Question: Does this analogy go "all the way" in the Zariski topology? More precisely, is there some Zariski open neighborhood $ p \in U \subseteq X $ on which the morphism $ U \to k[X_1, \dots, X_n]$ defined by $ p \mapsto (u_1(p), \dots, u_n(p))$ is an open immersion? The second bullet point above says that the analogy goes "all the way" in the analytic topology.

EDIT: OK, for an elliptic curve the morphism $ U \to k[X_1, \dots, X_n]$ cannot be an open immersion. Can anything be said about this morphism?

DBr
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  • Well, now that I think about it, an elliptic curve is an example that this analogy does not go all the way because it is not rational... – DBr Dec 24 '13 at 19:25

1 Answers1

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OK I am going to answer my own question. The morphism $ U \to k[X_1, \dots, X_n] $ is exactly an etale map. The details can be found in Mumford's red book.

DBr
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