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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.

KReiser
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mattematician
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1 Answers1

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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 $f_i$.

KReiser
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