Is it enough to check corank of jacobian matrix at closed points

Question

This is actually exercise 12.2.H of Vakil's notes. In the notes, a k-scheme is defined to be k-smooth of dimension d if there exists a affine open cover(every is of form $A=k[x_1,...,x_n]/(f_1,...,f_r)$) where the Jacobian matrix has corank d at all points. Then 12.2.H says it suffices to check this at all closed points.

The hint says the points satisfying the condition can be described as locus where the Jacobian matrix has corank d can be described in terms of vanishing and nonvanishing of determinants of certain explicit matrices. I guess here he means the minors. I also know if some property is open(if a point x has property P, then there exists an open neiborhood U s.t. every y in U has property P), then it suffices to check it at closed points. Then I am stucked, could some one help me? Thanks!

Answer 1

As you correctly noted, it is about the vanishing and non-vanishing of minors. To be precise, let $A=k[X_1,\ldots,X_n]/(f_1,\ldots, f_r)$ be a finitely generated $k$-algebra and let $x \in \operatorname{Spec}k[X_1,\ldots, X_n]$ containing $f_1,\ldots,f_r$, and let $i \geq 0$ be a fixed integer. Then the Jacobian matrix at $x$ has rank $\leq i$ if and only if all $l \times l$ - minors vanish for all $l > i$, i.e. if and only if $x \in \operatorname{V}(I_i,f_1,\ldots f_r)$, where $I_i$ denotes the ideal generated by all $l \times l$ - minors with $l>i$.

It follows that the points $\operatorname{V}(I_d, f_1,\ldots, f_r) \cap \operatorname{Spec}k[X_1,\ldots,X_n]\setminus \operatorname{V}(I_{d-1},f_1,\ldots,f_r)$ are precisely the points at which the Jacobian matrix has rank $d$.

Now, since $A$ is a finitely generated $k$-algebra, a consequence of Hilbert's Nullstellensatz tells us that the nilradical is equal to the Jacobson radical. But, if the rank at each closed point is equal to $d$, then $I_d$ is contained in the intersection of all maximal ideals containing $(f_1,\ldots, f_r)$, whence $I_d$ is contained in every prime containing $(f_1,\ldots, f_r)$. This proves that the rank at each point has to be less than or equal to $d$.
For the converse inequality, just note that each prime is contained in some maximal ideal.

Answer 2

Let me give a proof more in the spirit of the OP's approach and KReiser's hint.

Suppose that the Jacobian $J$ of $X = \operatorname{Spec}{ k[x_1, \dots, x_n]/(f_1, \dots, f_r)}$ has corank $d$ at all closed points. Thus, $$ V( (n-d)\text{-minors of }J) = \emptyset. $$ This shows that the Jacobian of $X$ has corank $\leq d$ at all points but we still want to show $\geq d$.

In particular, we now know that the Jacobian has rank $\geq n-d$ for all points in $X$. Therefore, now the points with exactly corank $d$ are given by $D((n-d+1)\text{-minors of } J)$ as those indicate that some $n-d+1$ minor does not vanish. But a polynomial not vanishing is an open condition. Since $X$ is quasicompact, this can now be checked on closed points.