Timeline for Is a complete intersection satisfying Jacobian matrix smooth criterion a smooth variety?

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May 13, 2013 at 8:29	history	edited	Olivier	CC BY-SA 3.0	Fixed the title
May 13, 2013 at 6:47	comment	added	Jack Huizenga		This is really differential geometry, not algebraic geometry (although it is true in the algebraic setting as well, when the proof is written appropriately). Your $r$ functions give a map $f:(\mathbb{C}^*)^n \to \mathbb{C}^r$. The variety $X$ is $f^{-1}(0)$, and the condition on the matrix says that $0$ is a regular value of $f$. So $X$ is a smooth manifold. Near any point $p\in X$, $X$ is the transverse intersection of the $r$ hypersurfaces $f_1=f_2=\cdots f_r = 0$, and these hypersurfaces are smooth at $p$.
May 13, 2013 at 5:20	comment	added	user29283		You say "of course" it is true when $I$ is radical, but how is the radical condition relevant in any justification? It is true without the radical property, due to the functorial criterion for smoothness -- e.g., see Proposition 7(c) in 2.2 of "Neron Models" -- and the fact that the functorial criterion implies the power series property for completed local rings at closed points when working with finite type schemes over an algebraically closed field.
May 13, 2013 at 3:12	history	asked	Li Yutong	CC BY-SA 3.0