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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.

My question is:

  • Will the Jacobian of ANY set of $n-r$ generators of $I$ be of maximal rank?
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    $\begingroup$ Qing Liu's proof is great, but I think you can prove this without any algebraic geometry. If you add a "useless" generator $f_{n+1}$ to the given list, then the corresponding row of the Jacobian is a linear combination of the other rows, so adding such "useless" generators doesn't change the rank. Thus, if $f_{r+1}, \dots, f_n$ and $g_{r+1}, \dots, g_m$ are two different lists of generators, the associated Jacobians have the same rank as the Jacobian associated to the combined list $f_{r+1}, \dots, f_n, g_{r+1}, \dots, g_m$. $\endgroup$
    – Ashvin Swaminathan
    Commented May 14, 2018 at 3:36

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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 $f\in I$.

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.

Some more explanations 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 $g_1,...g_m$.

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.

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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).

I am getting all of this from Section 16.6 of Eisenbud's Commutative Algebra.

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  • $\begingroup$ Thanks for your answer. I am interested in the case where the base could be a general ring, not necessarily a field. $\endgroup$
    – Nicolás
    Commented Nov 3, 2011 at 19:54
  • $\begingroup$ Eisenbud states that, for general Noetherian rings not over a perfect field, the set of singular primes may not even be Zariski closed, so there will be no ideal which defines it. $\endgroup$
    – Greg Muller
    Commented Nov 3, 2011 at 20:00
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    $\begingroup$ If $A$ is finitely generated over any field $k$, it is excellent, so its singular locus is closed. If $A$ is not finitely generated over $k$, but $k$ has characteristic $0$, then $A$ is still excellent. If $k$ has positive characteristic, $A$ might not be excellent even if $k$ is perfect (say finite). $\endgroup$
    – Qing Liu
    Commented Nov 3, 2011 at 21:02
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    $\begingroup$ I think, Greg, you are confusing the regular locus of a scheme $X$ with the smooth locus of a morphism of schemes $X \to S$. The latter one is always open, but the first one might be not open, although this can never happen for schemes of finite type over a field as Quing Liu points out. If $S$ is the spectrum of a field $k$ and $X$ is of finite type over $S$, the smooth locus is always contained in the regular locus, but in general you have equality only if $k$ is perfect. An example would be if $X$ is the spectrum of a finite inseparable extension of $k$. Then $X$ is regular but not smooth. $\endgroup$
    – Torsten Wedhorn
    Commented Nov 4, 2011 at 9:28

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