Failure of universal flatification

Question

Raynaud and Gruson proved a beautiful "flatification" theorem (5.2.2): If $S$ is a quasicompact, quasiseparated scheme, and $X$ is a finitely presented $S$-scheme, $M$ is an $\mathcal O_X$-module of finite type, then there is a blowup $f : S' \rightarrow S$ such that the strict transform of $M$ along $f$ is flat over $S'$. (The blowup can be arranged to be an isomorphism over an open subset of $S$ over which $M$ is already flat.)

They observe that when $X$ is projective over $S$, there is in fact a universal birational modification, obtained by resolving the indeterminacy of the rational map from $S$ to the Quot scheme. (If $U$ is an open subset of $S$ over which $M$ is flat, then $M$ induces a map from $U$ to the scheme of quotients of $M$.)

What I would like to have is some intuition for, and an example of, the absence of a universal flatification in the absence of the projectivity hypothesis.

Answer 1

I feel like there should be an example using matrix multiplication, but I tried several variations that did not work (as I commented above). The example below is less naural than matrix multiplication, but it illustrates the key issue.

Let $k$ be a field. Let $S$ be a $3$-dimensional vector space, $V \cong \text{Spec}\ k[x,y,z]$. Thus $\mathbb{P}(V)$ is isomorphic to $\mathbb{P}^2 = \text{Proj} \ k[r,s,t]$. Let $\overline{X}$ be the closed subscheme of $V\times \mathbb{P}(V)$ parameterizing pairs $(v,[w])$ such that $v$ is a multiple over $w$, possibly zero. In coordinates, $((x,y,z), [r,s,t])$, this is the zero scheme of $xs-yr, xt-zr, yt-zs$. Denote by $\overline{\pi}:\overline{X} \to S$ the natural projection.

Fix a line $L\subset \mathbb{P}(V)$, e.g., the zero scheme of $t$, and let $X$ be the open complement in $\overline{X}$ of $\{0\}\times L$. Let $\pi:X\to S$ be the restriction of $\overline{\pi}$ to $X$. Let $M$ be the structure sheaf of $X$. This is not $S$-flat, since the fiber dimension over the origin is $2$, yet the generic fiber dimension is $0$.

The universal flatification of $\overline{\pi}$ is $\overline{\pi}$ itself,, the strict transform in $\overline{X}\times_S \overline{X}$ is the diagonal, and the exceptional set $E=\overline{\pi}^{-1}(\{ 0\})$ is $E\cong \mathbb{P}(V)$. Certainly the strict transform of $\pi$ with respect to $\overline{\pi}$ is flat, but $\overline{\pi}$ is not a universal flatification of $M$.

I can take any finite collection of points inside $L$ inside $E$, and I can "pinch" $\overline{X}$ along those points to form a non-normal scheme $X'$ such that $\overline{\pi}$ factors through $X'$. The morphism from $X'$ to $X$ will also be a flatification of $\pi$. What I would "like" to do in order to make a universal (proper, birational) flatification of $\pi$ is contract / blow down $L$ inside $\overline{X}$. But that is not possible: we cannot even contract a line $L$ in a $2$-plane $E$ (not as a scheme, not even as an algebraic space). Thus, there is no universal (proper, birational) flatification.