I understand that this is a special case of the Jacobian criterion, but I was hoping that there was a simpler argument to prove it than for the criterion itself (I don't fully understand the proof of the criterion given in the book "Commutative Algebra with a view towards Algebraic Geometry" by Eisenbud). Could I get some help understanding this:
Let $k$ be a field and consider the ring $S = k[x_1,...x_n]/I$, where $I=(f_1,...,f_l)$. Prove that the localization of $S$ at the maximal ideal $m=(x_1,...,x_n)$ is a regular ring if and only if
$$rank\left(\left(\frac{\partial{f_i}}{\partial{x_j}}(0)\right)_{i=1,..,l;j=1,...,n}\right)=n-d$$where $d$ is the dimension of the localisation of $S$ at $m$.
EDIT
Here is what I had in mind by a simpler proof, taking into account that I am only looking at localising at $m=(x_1,...,x_n)$:
The matrix given above, I'll call it $M$, represents a mapping $\oplus_{i=1}^{l} Se_i \twoheadrightarrow I/I^2 \rightarrow \oplus_{j=1}^{n} Sdx_i$, and its cokernel is isomorphic to $\Omega_{S/k}$ (from the conormal sequence).
Now from the other conormal sequence: $ m/m^2 \rightarrow S_m/mS_m \otimes \Omega_{S_m/k} \rightarrow \Omega_{(S_m/mS_m)/k} \rightarrow 0$ for the surjective map $S_m \rightarrow S_m/mS_m$, we have that the cokernel of the matrix $M$ taken modulo $m$ (which is equivalent to evaluating at $0$) is $S_m/mS_m \otimes \Omega_{S_m/k}$.
We also know that $\dim(m/m^2)=\dim(k \otimes \Omega_{(S_m)/k})$ - since they are isomorphic, and $\dim(k \otimes \Omega_{(S_m)/k})=\dim(S/m \otimes \Omega_{S_m/k})$. Since the cokernel of the matrix taken modulo $m$, call it $M'$, is $S/m \otimes \Omega_{S_m/k}$ we have rank$(M') = n - \dim(S/m \otimes \Omega_{S_m/k})$. On the other hand $S_m$ is regular if and only if $\dim(m/m^2)=\dim(S_m)$, but $\dim(m/m^2)=\dim(S/m \otimes \Omega_{S_m/k})$.
Could something like this work?