Smoothness implies a condition on the Jacobian in every affine open In chapter 12 of FOAG, Ravi Vakil defines smoothness in the following way

A k-scheme is k-smooth of dimension d, or smooth of dimension d over
k, if it is of pure dimension d, and there exists a cover by affine
open sets $\operatorname{Spec} k[x_1, ... , x_n]/(f_1,..., f_r)$ where
the Jacobian matrix has corank d at all points. (In particular, it is
locally of finite type.) A k-scheme is smooth over k if it is smooth
of some dimension.

(NB: Vakil defines the Jacobian as the transpose of the usual Jacobian.) A few paragraphs down, there is the following discussion of the necessity that the condition hold for all open affine covers:

12.2.7. You can check that any open subset of a smooth k-variety is also a smooth k- variety. With what we know now, we could show that
this implies that k-smoothness is equivalent to the Jacobian being
corank d everywhere for every affine open cover (and by any choice of
generators of the ring corresponding to such an open set). Indeed, you
should feel free to do this if you cannot restrain yourself. But the
cokernel of the Jacobian matrix is secretly the space of differentials
(which might not be surprising if you have experience with
differentials in differential geometry), so this will come for free
when we give a better version of this definition in Definition 21.3.1.
The current imperfect definition will suffice for us to work out
examples. And if you don’t want to wait until Definition 21.3.1, you
can use Exercise
12.2.I below to show that if k algebraically closed, then smoothness can be checked on any open cover.

This paragraph is a little confusing.  He starts by saying a result can be proved by the reader, and then ends the paragraph by suggesting that the special case when $k$ is algebraically closed can be proved by the reader. In any case the suggestion is clearly that proving that smoothness implies the Jacobian condition on every affine open subset is not wholly trivial, else such an important result would be left as an exercise and not delayed for 9 chapters.  This makes me feel that I'm missing something, because it does seem basically trivial to me.  Namely, a few pages back we did the following exercise:

12.1.H. EXERCISE (THE CORANK OF THE JACOBIAN IS INDEPENDENT OF THE PRESENTATION). Suppose $A$ is a finitely-generated k-algebra, generated
by $x_1, . . . , x_n$, with ideal of relations $I$ generated by $f_1,...,
 f_r$. Let $p$ be a point of $\operatorname{Spec}A$. (a) Suppose $g \in I$. Show that appending
the column of partials of $g$ to the Jacobian matrix (12.1.6.1) does not
change the corank at $p$. Hence show that the corank of the Jacobian
matrix at $p$ does not depend on the choice of generators of $I$. (b)
Suppose $q(x_1,..., x_n)\in k[x_1,...,x_n]$. Let h be the polynomial
$y-q(x_1,...,x_n) \in k[x_1,..., x_n, y]$. Show that the Jacobian
matrix of $(f_1,..., f_r, h)$ with respect to the variables $(x_1, ..., x_n, y)$ has the same corank at $p$ as the Jacobian matrix of $(f_1,...
, f_r)$ with respect to $(x_1,..., x_n)$. Hence show that the corank of
the Jacobian matrix at $p$ is independent of the choice of generators
for $A$.

It seems to me one has only to combine this exercise with the other equally trivial fact, allowing one to equate the codimension at $p$ of the Jacobian for an affine open with the codimension at $p$ of the Jacobian in a distinguished open subset containing $p$. Namely:
Lemma: For $$[\mathfrak{p}]\in \operatorname{Spec} k[x_1, ... , x_n]/(f_1,..., f_r),$$ and
$$NG.\text{  }g \in k[x_1, ... , x_n]\backslash \mathfrak{p},$$
the Jacobian of $(f_1,..., f_r)$ at $[\mathfrak{p}]$ has the same codimension as the Jacobian of $(f_1,..., f_r, y\cdot g-1)$ at $$[\mathfrak{p}]\in \operatorname{Spec} k[x_1, ... , x_n,y]/(f_1,..., f_r, y\cdot g-1)=  \operatorname{Spec} \big( k[x_1, ... , x_n]/(f_1,..., f_r)\big)_g$$
So my question is, don't these simple facts immediately give that smoothness can be checked in any open affine cover, or am I missing something?
In case it's not clear, the argument I have in mind goes like this:  Assume $X$ is a $k$-variety which is smooth of dimension $d$, and let $$U=\operatorname{Spec} k[x_1,...,x_n]/(f_1,...,f_r)\subset X$$ be an affine open.  Let $p\in U$. We want to check that the Jacobian of the $f_i$'s has corank $d$ at $p$.  Since $X$ is smooth, $p$ is contained in some $$V=\operatorname{Spec} k[x_1,...,x_m]/(g_1,...,g_s)\subset X$$ such that the $g_i$'s have corank $d$ at $p$.  Now, we can find an open $W\subset U\cap V$ such that $W$ is distinguished in both $W$ and $V$.  Now, by the lemma above, since $W$ is distinguished in $ V$ and the $g_i$'s have corank $d$ at $p$, any relations for the ring of $W$ have Jacobian of corank $d$ at $p$.  Then by another application of the lemma, the Jacobian of the $f_i$'s has corank $d$ at $p$, QED.
 A: Answer: I use the notation of Hartshorne, Chapter I. If $(A, \mathfrak{m})$ is a noetherian local ring with residue field $k$ it follows (HH.I.Prop.5.2A)
$$KR.\text{  }krdim(A) \leq dim_k(\mathfrak{m}/\mathfrak{m}^2).$$
There is a formula (see the proof of HH.Thm.I.5.1) saying
$$rk(J_p)=n-dim_k(\mathfrak{m}_p/\mathfrak{m}^2_p)$$
Here we assume $k$ is algebraically closed and $p$ a closed point (in the sense of HH, Chapter II). Assume $Y:=V(I) \subseteq \mathbb{A}^n_k$ (in the notation of chapter I, HH) is an algebraic variety with coordinate ring $A(Y):=k[x_1,..,x_n]/I$ and maximal ideal $\mathfrak{m}_p$ corresponding to the point $p\in Y$. Localize $A:=A(Y)$ at $\mathfrak{m}:=\mathfrak{m}_p$ you get a noetherian local ring $(A_{\mathfrak{m}}, \tilde{\mathfrak{m}})$ with
$$krdim(A)=krdim(A_{\mathfrak{m}}) \leq dim_k(\tilde{\mathfrak{m}}/\tilde{\mathfrak{m}}^2)=dim_k(\mathfrak{m}/\mathfrak{m}^2).$$
Here we have used that there is an equality
$$dim_k(\mathfrak{m}/\mathfrak{m}^2)=dim_k(\tilde{\mathfrak{m}}/\tilde{\mathfrak{m}}^2).$$
Hence
$$dim(V):=krdim(A) \leq dim_k(\mathfrak{m}/\mathfrak{m}^2) =n-rk(J_p(I))=dim_k(T_p(V)).$$
Hence for any point $p\in V$ it follows
$$dim(V)≤dim_k T_p(V)$$
and
$$dim_k(T_p(V))=n−rank(J_p(I)):=corank(J_p(I)).$$
Hence the rank $rk(J_p(I))$ (and $corank(J_p(I))$) is independent of choice of generators $f_i$ of the ideal $I$ since it depends on the maximal ideal $\mathfrak{m}$ of $p$.
Question: "So my question is, don't these simple facts immediately give that smoothness can be checked in any open affine cover, or am I missing something?"
Answer: Yes, smoothness depends only on the local ring $A_{\mathfrak{m}}$ of $A$ at the maximal ideal $\mathfrak{m}$ corresponding to the point $p$, and this local ring is an intrinsic invariant - it is independent of choice of affine open neighborhood containing $p$.
Note: We define the point $p$ to be regular/non-singular iff there is an equality
$$(*) \text{  }dim(V) = dim_k(T_p(V))=n-rank(J_p(I))$$
and $V$ to be smooth/regular iff $(*)$ holds for all points $p\in V$. Hence to define non-singularity/smoothness me may use the Jacobian criterion and the Jacobian matrix or equivalently the local ring at $p$.
Note: Smoothness/regularity is usually studied for schemes of finite type over a field or a Dedekind domain. If $A:=k[x_1,x_2,..,x_n....]$ is a polyomial ring on a countably infinite number of variables and if $I⊆A, I:=(f_1,f_2,f_3,..,)$ has an infinite set of generators the "scheme" $S:=Spec(A/I)$ is (in some sense) "well defined". The jacobian matrix $J_p(I)$ at a closed point $p∈S$ is an "infinite matrix" and you may not use the Jacobian criterion to define "non-singularity" at $p$. The local ring $A_{\mathfrak{m}_p}$ at $p$ is "well defined" but $\mathfrak{m}_p/\mathfrak{m}^2_p$ has infinite dimension over $\kappa(p)$, hence you will have difficulties in defining "regularity". You cannot use the "standard definition" in $KR$. Hence a problem with "schemes not of finite type over a field" is that it is difficult to even define the notion "smoothness".
Note: The proof in HH.Prop.I.5.1 proves that you may check non-singularity at a point $p\in Y$ using the local ring $A:=\mathcal{O}_{Y,p}$ : $p$ is non-singular iff $A$ is a regular local ring. The local ring $A$ does not depend on an affine open subscheme $V:=Spec(B) \subseteq Y$ containing $p$. By definition
$$\mathcal{O}_{Y,p}:=lim_{p\in U}\mathcal{O}_Y(U) \cong lim_{p\in U}\mathcal{O}_V(U)\cong \mathcal{O}_{V,p}$$
for any open subscheme $p\in V \subseteq Y$. In particular it holds for any basic open subscheme $D(g):=Spec(B_g) \subseteq V$ with $p\in D(g)$:
$$\mathcal{O}_{V,p} \cong \mathcal{O}_{D(g),p} \cong (B_g)_{\mathfrak{p}},$$
where $\mathfrak{p}$ is the prime ideal of $p$.
This argument proves that you may choose any open subscheme $p\in V:=Spec(B)$ and any set of generators $x_1,..,x_n$ for $B$ as $k$-algebra and any set of generators $I:=(f_1,..,f_l) \subseteq k[x_1,..,x_n]$ for the ideal defining $Y$ in $V$.
