Let $R$ be a commutative ring with $1$ (standard setting for commutative algebra). Consider the ring of polynomials $R[T_1,…,T_n]$.
For $m \le n$ let $f_1 ,…,f_m \in R[T_1,…,T_n]$.
Define $J(f_1,…,f_m):= \det(\partial f_i/\partial T_i)_{ij}$
Assume that for every maximal ideal $\mathfrak{m}$ of $R$ for $J$ in the case $m < n$ is, that if $J$ is the ideal of the $m \times m$ minors of $M = (\partial f_i / \partial T_j)_{ij}$ and $I=(f_1,\ldots,f_m)$ then $J + I = R[T_1,\ldots,T_n] = A$.
$J(f_1,…,f_m)$ doesn't vanish in $R[T_1,…,T_n] \otimes_R R/\mathfrak{m}=(R/\mathfrak{m})[T_1,…,T_n]$.
How to conclude that in that case the induced ring $B:=R[T_1,…,T_n]/(f_1,…,f_m)$ is flat over $R$?
Remark: This statement was mentioned in J. Milne's Etale Cohomology (page 23) and the argument was only sketched as follows:
Use iteratively Prop. 2.5: Let $A$ be a flat $R$-algebra and $a \in A$. Then $A/(a)$ is a flat $R$-algebra if for every maximal prime ideal $\mathfrak{m}$ of $R$ the image of $a$ under $A \to A/\mathfrak{m}$ is a non zero divisor in $A/\mathfrak{m}$.
I don't understand how this argument helps:
Applying this proposition to $J$ gives (since $(R/\mathfrak{m})[T_1,…,T_n]$ has no zero divisors other then $0$, and $J$ by assumption is not zero) that $R[T_1,…,T_n]/J$ is flat over $R$. But why does this imply that $R[T_1,…,T_n]/(f_1,…,f_m)$ is flat over $R$?
Best Answer
$\newcommand{\ideal}[1]{\mathfrak{#1}}\newcommand{\tensor}{\otimes}$ The correct condition for $J$ in the case $m < n$ is, that if $J$ is the ideal of the $m \times m$ minors of $M = (\partial f_i / \partial T_j)_{ij}$ and $I=(f_1,\ldots,f_m)$ then $J + I = R[T_1,\ldots,T_n] = A$.
If $B = A/I$ we want to show for all maximal ideals $\ideal{n} \subseteq A$ and $\ideal{p} = \ideal{n} \cap R$ the extension $B_\ideal{n}/R_\ideal{p}$ is flat. For $\ideal{n} \not\supseteq I$ this is true, as $B_\ideal{n} = 0$. So only the $\ideal{n} \supseteq I$ remain.
Calling $k = k(\ideal{p}) = R_\ideal{p}/\ideal{p} R_\ideal{p}$ we have to consider the rings $B_p = A/(f_1,\ldots,f_p)$ and their base changes $C_p = (B_p)_\ideal{n} \tensor_R k(\ideal{p}) = (k[T_1,\ldots,T_n]/(\bar{f}_1,\ldots,\bar{f}_p))_\ideal{n}$ where $\bar{f}$ is the image in $A \tensor_R k$ of $f \in A$.
Now, the key to the further argument is the observation, that because of the condition on the minors of $M$, the polynomials $\bar{f}_1,\ldots,\bar{f_p}$ define a nonsingular variety in $\mathrm{spec}(k[T_1,\ldots,T_n])$ (the Jacobian $(\partial f_i/\partial T_j)_{{i=1,\ldots,p}\atop {j=1,\ldots,n}} \tensor_R k$ has full rank $p$ at $\ideal{n}$). So $C_p$ is an integral domain and $\bar{f}_{p+1}$ is a non-zero divisor in $C_p$ because it is not contained in the ideal $(\bar{f}_1,\ldots,\bar{f}_p)$.
So by the proposition 2.5 cited, as $(B_0)_\ideal{n} = A_\ideal{n}$ is $R_\ideal{p}$-flat, so is $(B_1)_\ideal{n} = (B_0)_\ideal{n}/(f_1)$ and finally, by induction also $(B_m)_\ideal{n} = B_\ideal{n}$.
The correct version of proposition 2.5 is:
Let $B$ be a flat $A$-algebra and consider $b \in B$. If the image of $b$ in $B \tensor_A k(\ideal{n} \cap A)$ for all $\ideal{n} \subseteq B$, maximal, is a non-zero-divisor, then $B/bB$ is a flat $A$-algebra.