---
tag: H1CZXMZ6HE
kind: statement
role: theorem
name: Jacobian criterion
msc: [13H05, 14B05, 13N05]
uses: [0PF6TM8TWT, 0SDC934R52, 00MHHZ9SNE, MVGBS3J1P7, 4C1AE84N5N, 63XXTF6B5M, 047NDTEPCV, 0ZC9YTXZRY, 0EGN1E1JPB]
about: [63XXTF6B5M, 047NDTEPCV]
sources:
  - url: https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant
---

## Statement

Let $k$ be an [[algebraically closed field|0PF6TM8TWT]] and let $P$ be a [[prime ideal|00MHHZ9SNE]] of the [[polynomial ring|0SDC934R52]] $k[x_1, \ldots, x_n]$, generated by $f_1, \ldots, f_t$.
Let $A = k[x_1, \ldots, x_n]/P$ and let $d$ be its [[Krull dimension|MVGBS3J1P7]].
Let $a \in k^n$ be a point with $f_1(a) = \cdots = f_t(a) = 0$, and let $\mathfrak m$ be the maximal ideal of $A$ generated by the classes of $x_1 - a_1, \ldots, x_n - a_n$.
Let $J(a)$ be the $t \times n$ [[Jacobian matrix|047NDTEPCV]] $\left( \frac{\partial f_i}{\partial x_j}(a) \right)$ of formal partial derivatives.
Then $\rank J(a) \le n - d$, and the [[localization|4C1AE84N5N]] $A_{\mathfrak m}$ is a [[regular local ring|63XXTF6B5M]] if and only if $\rank J(a) = n - d$.

## Remarks

Proof idea: let $M$ be the maximal ideal of $a$ in $k[x_1, \ldots, x_n]$.
By Taylor expansion at $a$, the image of $P$ in the $n$-dimensional space $M/M^2$ is spanned by the linear parts of the $f_i$, so it has dimension $\rank J(a)$.
Hence $\dim_k \mathfrak m/\mathfrak m^2 = n - \rank J(a)$.
Since $A$ is a finitely generated domain over $k$, the local ring $A_{\mathfrak m}$ has dimension $d$, and $\dim_k \mathfrak m/\mathfrak m^2 \ge d$ with equality exactly when $A_{\mathfrak m}$ is regular.

Example: for the cusp $f = y^2 - x^3$ in $k[x, y]$, one has $d = 1$ and $J = (-3x^2, \; 2y)$.
At $a = (0, 0)$ the rank is $0 \ne 1$, so the local ring at the origin is not regular.
At $a = (1, 1)$ the matrix is $(-3, 2)$, which has rank $1$ in every characteristic, so the local ring there is regular.

The hypothesis that the $f_i$ generate $P$ is needed, not only that they have the same zeros.
For $f = x^2$ in $k[x]$, the zero set is the point $0$ and the ring $k[x]/(x)$ is a field, hence regular, but $J(0) = (0)$ has rank $0 \ne 1 - 0$.

## Explanations

### L1 human

A point of a curve or surface defined by polynomials is smooth when the equations cut it out cleanly to first order.
The test is linear algebra: compute the matrix of partial derivatives at the point and its rank.
If the rank is as large as possible, namely the number of variables minus the dimension, the point is nonsingular.
For the cusp $y^2 = x^3$, all partial derivatives vanish at the origin, which is the sharp point.

### L2 human

The criterion turns the abstract notion of a [[regular local ring|63XXTF6B5M]] into a computation.
It identifies the Zariski cotangent space $\mathfrak m/\mathfrak m^2$ with the cokernel of the transposed Jacobian matrix at the point.
The set of singular points is therefore cut out by the minors of size $n - d$ of the Jacobian matrix together with $P$, so it is a closed subset.
In the language of [[Kähler differentials|7HCFARHG7D]], the matrix presents $\Omega_{A/k}$.
The criterion says that $A$ is [[smooth|286G0P1B69]] over $k$ near $a$ exactly when this module is free of rank $d$ near $a$.
Over a field that is not perfect, regularity and smoothness differ, and the rank condition then characterizes smoothness rather than regularity.