Jacobian criterion

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 $0 \ne 1 - 0$.

Going deeper

For a student

The criterion turns the abstract notion of a regular local ring 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, the matrix presents $\Omega_{A/k}$. The criterion says that $A$ is smooth 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.