Let f : X → Y be a dominant, finite type morphism of integral Noetherian schemes. We assume X is normal and Y regular. In this and the next blog post we define Sing(f) to be the closed set of points of X where f isn’t smooth and we define Disc(f) to be the image of Sing(f) in Y.

In this setting a purity result is one that gives an upper bound on the codimension of Sing(f) in X or a lower bound on the codimension of Disc(f) in Y.

When the relative dimension of f is 0, then the Stacks project has Zariski-Nagata purity of branch locus. If f is generically finite, then the Stacks project has purity of ramification locus and over fields we even have that the complement of Sing(f) is affine in X (Tag 0ECA).

But what if the relative dimension is > 0?

For the rest of this blog post, let’s say f has relative dimension m ≥ 0 in the sense that every irreducible component of every fibre has dimension m.

The simplest case is when Y and X are smooth over a field k with Y of dimension n and X of dimension n + m. In fact, let’s assume Y and X are both affine spaces over k. Picking coordinates x_1, …, x_{n + m} on X and coordinates y_1, …, y_n on Y and say f = (f_1, …, f_n) is given in coordinates by y_i = f_i(x_1, …, x_{n + m}) for some polynomials f_i. Then of course the locus Sing(f) is the set of points of X where the n × (n + m) matrix of partial derivatives (d f_i / d x_j) does not have maximal rank. We conclude immediately that codim Sing(f) ≤ m + 1. Taking the image in Y we find codim Disc(f) ≤ m + 1.

In fact, this is all you can say! Here are some examples to show this is true. (We assume the characteristic of k is not 2.)

Relative dimension m = 0.

(A) For n ≥ 1 we have the example (f_1, …, f_n) = (x_1, …, x_{n – 1}, x_n^2). Then Sing(f) = V(x_n) and Disc(f) = V(y_n).

Relative dimension m = 1.

(A) For n ≥ 1 we have the example (f_1, …, f_n) = (x_1, …, x_{n – 1}, x_nx_{n + 1}). Then Sing(f) = V(x_n, x_{n + 1}) and Disc(f) = V(y_n). Here Disc(f) is a divisor in Y and the map Sing(f) → Disc(f) is an isomorphism. This is the usual example of a family of nodal curves.

(B) For n ≥ 2 we have the example (f_1, …, f_n) = (x_1, …, x_{n – 1}, x_1x_n + x_{n + 1}^2). Then Sing(f) = V(x_1, x_{n + 1}) and Disc(f) = V(y_1, y_n). Here Disc(f) has codimension 2 in Y, the fibres over points of Disc(f) are nonreduced, and the map Sing(f) → Disc(f) has fibres of dimension 1. Thanks to Will Sawin for finding this example.

Relative dimension m = 2.

(A) For n ≥ 1 we have the example (f_1, …, f_n) = (x_1, …, x_{n – 1}, x_n^2 + x_{n + 1}^2 + x_{n + 2}^2). Here Sing(f) = V(x_n, x_{n + 1}, x_{n + 2}) and Disc(f) = V(y_n). Here Disc(f) is a divisor in Y and the map Sing(f) → Disc(f) is an isomorphism. This is the usual example of a family of nodal surfaces.

(B) For n ≥ 2 we have the example (f_1, …, f_n) = (x_1, …, x_{n – 1}, x_1 x_n + x_{n + 1}^2 + x_{n + 2}^2). Here Sing(f) = V(x_1, x_{n + 1}, x_{n + 2}) and Disc(f) = V(y_1, y_n). Here Disc(f) has codimension 2 in Y, the fibres of f over points of Disc(f) are planes meeting in lines, and the nap Sing(f) → Disc(f) has fibres of dimension 1.

(C) For n ≥ 3 we have the example (f_1, …, f_n) = (x_1, …, x_{n – 1}, x_1 x_n + x_2 x_{n + 1} + x_{n + 2}^2). Here Sing(f) = V(x_1, x_2, x_{n + 2}) and Disc(f) = V(y_1, y_2, y_n). Here Disc(f) has codimension 3 in Y, the fibres of f over points of Disc(f) are nonreduced, and the map Sing(f) → Disc(f) has fibres of dimension 2.

It seems pretty clear that we can keep going in this manner for arbitrary m. Enjoy!

Edit 3/8/2020 Will Sawin gives this general example: the map from A^{2n+m} with coordinates (x_1,…,x_n, y_1,…,y_n,z_1,….z_m) to A^{n+1} given by (x_1,…,x_n , x_1y_1+ …. + x_ny_n + z_1^2 + …. + z_m^2) is a map with (n+m-1)-dimensional fibers which has a codimension n+1 locus on the base where the codimension of the singularities of the fiber is exactly m-1 (the locus where the x_i and z_i all vanish), showing that all such pairs appear sharply.