This question asks for a proof of the following fact: Let $R$ be a commutative ring and $f_1,\dots,f_k\in R[x_1,\dots,x_n]$ polynomials such that the Jacobian $J=(\partial f_{j}/\partial x_i)_{i,j}$ has total rank in every residue field $k(P)$ for $P\in\text{Spec}(R)$ Then the following $R$-algebra is flat $$A=R[x_1,\dots,x_n]/(f_1,\dots,f_k).$$
I am not sure if I understand it correctly, so I ask:
Question 1: Is it true that if we take $R=F$ a field (of characteristic zero), $k=n$, and the Jacobian of $f_1,\ldots,f_n$ is a non-zero scalar, then the claim says that $$A=F[x_1,\dots,x_n]/(f_1,\dots,f_n).$$ is flat as an $F$-module? (or as an $F[x_1,\ldots,x_n]$-module?)
EDIT: Actually, what I really wish to ask is:
Question 2: Is the quoted result dealing with $k \leq n$ only, or is it dealing with $k > n$ also?
I am confused: On the one hand, no restriction on $\{k,n\}$ is mentioned in the quoted question, but on the other hand, the two references i ii in its comment seem to deal with $k \leq n$ (if I am not wrong).
Examples for question 2 (see this question):
(1) $f_1=x^2+y,f_2=y$, $R=k[x]$, $n=1$. $\frac{k[x][y]}{\langle x^2+y,y \rangle}= \frac{k[x]}{\langle x^2 \rangle}$ is not flat over $k[x]$, by the result quoted here. However, the it is not possible to apply the above result (assuming that $k > n$ is ok), since the Jacobian is the one-row matrix $(1 1)$ (partial derivatives of $f_1,f_2$ with respect to $y$), so the rank is one (the rank is not total= not equals two).
(2) $f_1=x^2+y,f_2=x$, $R=k[x]$, $n=1$. $\frac{k[x][y]}{\langle x^2+y,x \rangle}= \frac{k[x,y]}{\langle x,y \rangle}=k$. Actually, it is not possible to apply the above result (assuming that $k > n$ is ok), since the Jacobian is the one-row matrix $(1 0)$ (partial derivatives of $f_1,f_2$ with respect to $y$), so the rank is one (the rank is not total= not equals two).
(3)!!! $f_1=x+y^2,f_2=y+x+y^2$, $R=k[x]$, $n=1$. Again, $\frac{k[x][y]}{\langle x+y^2,y+x+y^2 \rangle}= \frac{k[x,y]}{\langle x,y \rangle}=k$. Here the Jacobian is the one-row matrix $(2y, 1+2y)$ (partial derivatives of $f_1,f_2$ with respect to $y$), and the rank is total (two). It seems that this example shows that it is NOT possible to take $k > n$, since $k$ is not a flat $k[x]$-module, by the flatness criterion presented in the first answer to this question, with the action of $k[x]$ on $k$ defined as follows: $x \lambda=0$, for all $\lambda \in k$, and scalars are computed as ususal, in $k$.
Remarks: (1) This MO question seems relevant. (2) This paper seems relevant, but it deals with $n=1$ and instead of taking $(f_1)$ it takes $I$ and shows that it must be a principal ideal. (We can take $R=k[y]$ and then deal with $k[y][x]$. But it is not possible to apply the stronger results, since $R$ is required to be quasi-local).
Thank you very much!
