the Jacobian criterion

How do you tell, by a calculation, whether a point of a variety is smooth or a singular corner? The Jacobian criterion gives a finger-on-the-button test: write down the matrix of partial derivatives of the defining equations, plug in the point, and check its rank. If the rank is as large as it can be — the codimension of the variety — the point is smooth; if the rank drops, the point is singular. It converts a geometric question about smoothness into a piece of linear algebra.

Let X = V(f_1, ..., f_r) be a variety in A^n, and suppose at the point p it has dimension d, so its expected codimension is n - d. Form the Jacobian matrix J(p) = (partial f_i / partial x_j) evaluated at p, an r-by-n matrix. The criterion: p is a smooth (nonsingular) point of X exactly when the rank of J(p) equals n - d. Equivalently, the Zariski tangent space, which is the kernel of J(p), has dimension exactly d. When the rank is full the implicit function theorem analogue applies and X looks locally like affine d-space; when the rank drops below n - d the kernel is too big and p is a singular point. For a hypersurface V(f) in A^n the test is especially simple: p is singular exactly when f(p) = 0 and all partials partial f / partial x_j vanish at p simultaneously.

This is the workhorse for locating singularities, computing where blow-ups are needed, and verifying smoothness in examples and proofs. Two honest caveats. First, in characteristic p the criterion needs care: derivatives can vanish for purely algebraic reasons (the derivative of x^p is zero), so a naive partials-vanish test can wrongly flag or miss singularities, and one must use the right notion of smoothness over the base. Second, you must know or compute the dimension d first, and you must use a set of equations generating the full radical ideal — using non-generating or non-reduced equations gives a Jacobian whose rank lies, falsely reporting singularities that are not there.