# Jacobian computes the Zariski cotangent space at a rational point

In Vakil's Foundations of Algebraic Geometry (July 31, 2023 version), he has the following Exercise (13.1.I):

Suppose $$X$$ is a finite type $$k$$-scheme. Then locally it is of the form $$\operatorname{Spec} k[x_1, \dots , x_n]/(f_1, \dots , f_r)$$. Show that the Zariski cotangent space at a $$k$$-valued point (a closed point with residue field $$k$$) is given by the cokernel of the “Jacobian map” $$k^r\to k^n$$ given by the Jacobian matrix $$J=\begin{pmatrix}\frac{\partial f_1}{\partial x_1}(p) & \cdots & \frac{\partial f_r}{\partial x_1}(p)\\ \vdots & \ddots & \vdots\\ \frac{\partial f_1}{\partial x_n}(p) & \cdots & \frac{\partial f_r}{\partial x_n}(p)\end{pmatrix}$$

and I'm completely stumped trying to solve it. Per a hint Vakil gives, I'm only considering the case where $$p$$ is the origin (and then want to do a change of coordinates for any $$p$$ not the origin) but even in this case I can only prove the statement for $$r = 0$$, where its trivial.

This answer details a way to do something similar (which would immediately give me the answer here), but I cannot follow the proof outline given.

You can just compute. Recall that Vakil defines the cotangent space at a point $$x\in X$$ to be $$\mathfrak{m}/\mathfrak{m}^2$$ for $$\mathfrak{m}\subset\mathcal{O}_{X,x}$$ the maximal ideal.
First, we can compute the cotangent space on any affine neighborhood of a closed point. As $$\mathcal{O}_{X,x}\cong R_m$$ for any closed point, any affine open neighborhood $$\operatorname{Spec} R\subset X$$ of $$x$$, and $$m\subset R$$ the maximal ideal corresponding to $$x$$, we can show that for any ring $$R$$ and maximal ideal $$m\subset R$$ we have $$m/m^2 \cong (m/m^2)_m \cong m_m/m_m^2$$, as $$R\setminus m$$ already acts invertibly on any $$R/m$$-module and localization is exact.
Working with $$k[x_1,\cdots,x_n]/(f_1,\cdots,f_r)$$ at the origin, the quotient $$m/m^2$$ is $$(x_1,\cdots,x_n)/((x_1,\cdots,x_n)^2+(f_1,\cdots,f_r)).$$ Since the the origin belongs to $$V(f_1,\cdots,f_r)$$, we have that all the $$f_i$$ have zero constant term, so we can simplify this quotient to be the quotient of the vector space $$k\langle x_1,\cdots,x_n\rangle$$ by the linear portions of each of the $$f_i$$. But that's exactly the description in your post: the column vector $$\frac{\partial f_i}{\partial x_j}(p)$$ computes the linear term of $$f_i$$, so taking the quotient of $$k^n$$ by the image of this matrix gives you the quotient of vector space $$k\langle x_1,\cdots,x_n\rangle$$ by the linear portions of each of the $$f_i$$.