Jacobian criterion for smoothness of schemes - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T20:40:16Z http://mathoverflow.net/feeds/question/79956 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79956/jacobian-criterion-for-smoothness-of-schemes Jacobian criterion for smoothness of schemes unknown 2011-11-03T18:09:13Z 2011-11-04T08:21:10Z <p>An affine scheme $X = Spec(A)$ is said to be smooth if for any closed embedding $X\subset\mathbf A^n$, of ideal $I$, it is true that, locally on $x\in X$, the ideal $I$ can be generated by a sequence $f_{r+1},\dots,f_n$ such that their Jacobian has maximal rank.</p> <p>My question is:</p> <ul> <li>Will the Jacobian of ANY set of $n-r$ generators of $I$ be of maximal rank?</li> </ul> http://mathoverflow.net/questions/79956/jacobian-criterion-for-smoothness-of-schemes/79958#79958 Answer by Greg Muller for Jacobian criterion for smoothness of schemes Greg Muller 2011-11-03T18:19:07Z 2011-11-03T18:19:07Z <p>I believe you need to be careful if $A$ is not over a perfect field. When $A$ is over a perfect field, the Jacobian ideal is the $r$th fitting ideal of the module of differentials, and so it is canonical (and independent of the choice of generators).</p> <p>I am getting all of this from Section 16.6 of Eisenbud's Commutative Algebra.</p> http://mathoverflow.net/questions/79956/jacobian-criterion-for-smoothness-of-schemes/79977#79977 Answer by Qing Liu for Jacobian criterion for smoothness of schemes Qing Liu 2011-11-03T20:56:37Z 2011-11-04T08:21:10Z <p>Yes, the rank of the Jacobian matrix doesn't depend on the set of generators of $I$. The Jacobian matrix at $x$ represents the subspace generated by the differentials at $x$ of all $f\in I$. </p> <p>Note that the rank of the Jacobian matrix at $x$ is computed in the fiber where $x$ lives, it has nothing to do with the base scheme.</p> <p><b>Some more explanations</b> We work over a base field $k$. Let $I$ be the ideal defining $X$ in $Y:=\mathbb A^n$. Then we have a canonical exact sequence $$I/I^2 \to \Omega_{Y}|_X \to \Omega_X \to 0$$ where the first map is $\bar{f}\mapsto df\otimes 1$. Tensoring by $k(x)$ we get $$I/I^2 \to \Omega_{Y}\otimes k(x) \to \Omega_X\otimes k(x) \to 0.$$ If $g_1,...,g_m$ are a system of generators of $I$, then $dg_1,...., dg_m$ generate the image of $I/I^2$ in $\Omega_{Y}\otimes k(x)\simeq k(x)^n$. Call this image $C$. Let $J_x$ be the Jacobian matrix associated to $g_1,...,g_m$ at $x$ in a system of coordinates of $Y$. Then the columns of $J_x$ correspond to the images of $dg_1,...,dg_m$ in $C\subseteq \Omega_{Y}\otimes k(x)$. Therefore the rank of $J_x$ is just the dimension of $C$ over $k(x)$, and this is independent on the choice of the system of generators $g_1,...g_m$. </p> <p>By the way, these discussions show that $$\dim_{k(x)} (\Omega_{X}\otimes k(x))=n - \mathrm{rank} J_x.$$ As the smoothness at $x$ is equivalent to $\dim_{k(x)} (\Omega_{X}\otimes k(x))= \dim_x X$, we see that it is also equivalent to $\mathrm{rank} J_x=n-\dim_x X$ which is the Jacobian criterion of smoothness.</p>