# Does $\delta$-invariant give a sufficient condition for flatness for plane curve singularities?

Let $$\pi:\mathcal{C}\to S$$ be a morphism of schemes such that $$\mathcal{C} \subset \mathbb{C}^2 \times S$$ with the inclusion map commuting with the natural projection to $$S$$ and for all $$s \in S$$, $$\pi^{-1}(s)$$ is a curve in $$\mathbb{C}^2$$ (under the restriction of the inclusion of $$\mathcal{C}$$ into $$\mathbb{C}^2 \times S$$). Suppose further that $$S$$ is an integral scheme. If for all $$s \in S$$, the $$\delta$$-invariant of the corresponding curve $$\mathcal{C}_s:=\pi^{-1}(s)$$ is constant, can we then say that $$\pi$$ is flat? I think this is true if $$S$$ is smooth. Here, I weaken it to integrality.

Any hint/reference will be most welcome.

Without further hypotheses on $$S$$, there are counterexamples. For instance, begin with $$\widetilde{S}$$ equal to the affine line $$\mathbb{A}^1_k = \text{Spec} \ k[s]$$. Inside of the affine space $$\mathbb{A}^2_S = \text{Spec}\ k[s,t,u]$$, consider the closed subscheme $$\widetilde{\mathcal{C}}$$ equal to the zero scheme of $$t(1-(1-s)u)$$. This is flat over $$\widetilde{S}$$, yet the fibers do not have constant $$\delta$$-invariant. For every $$s\neq 1$$, the $$\delta$$-invariant equals $$1$$. For $$s=1$$, the $$\delta$$-invariant equals $$0$$.

Notice, the fiber over $$s=1$$ is a closed subcurve of the fiber over $$s=-1$$. Thus, denote the coproduct of $$s=1$$ and $$s=-1$$ as follows, $$\nu:\widetilde{S} \to S, \ \ S = \text{Spec}\ k[y,z]/\langle z^2-y^2(1+y) \rangle, \ \ \nu^*y = s^2-1, \ \nu^*z = s(s^2-1).$$ This morphism is an isomorphism away from $$s=1$$ and $$s=-1$$, which both map to a common point $$p$$ of $$S$$. Thus, define $$\mathcal{C}\subset \mathbb{A}^2_S$$ to be the closure of the image of $$\widetilde{\mathcal{C}}$$ away from $$p$$. This is most certainly not flat, yet the fiber over every point of $$S$$ is a reduced plane curve with $$\delta$$-invariant equal to $$1$$.

One important feature of this example is that $$S$$ is not normal.

Additional Hypothesis. Let $$S$$ be an integral Noetherian scheme that is normal.

Under the additional hypothesis, there is a positive answer without any assumption on the $$\delta$$-invariant.

Proposition. For every projective, smooth $$S$$-scheme $$X$$, for every closed subscheme $$\mathcal{C}$$ of $$X$$ such that every associated point of $$\mathcal{C}$$ dominates $$S$$ and has codimension $$1$$ in $$X$$, then $$\mathcal{C}$$ is $$S$$-flat if and only if every $$S$$-fiber of $$\mathcal{C}$$ has codimension $$1$$ in the corresponding $$S$$-fiber of $$X$$.

Proof. When $$\mathcal{C}$$ is empty, this is trivial. Thus, assume that $$\mathcal{C}$$ is not empty. One direction is straightforward: flatness implies constancy of fiber dimension. It remains to prove the other direction.

The $$S$$-flat locus of $$\mathcal{C}$$ is open in $$\mathcal{C}$$. Since the local ring of $$S$$ at each codimension $$1$$ point is a DVR, and since a finitely presented scheme over a DVR is flat if and only if every associated point of the domain dominates the generic point of the DVR, the flat locus contains the fiber over every codimension $$1$$ point. Thus, the closure $$Z$$ of the image in $$S$$ of the non-$$S$$-flat locus of $$\mathcal{C}$$ has codimension $$\geq 2$$ everywhere.

Although it is a bit of a sledgehammer, perhaps the "fastest" way to finish the argument is to use the existence and properties of the relative Hilbert scheme of $$X/S$$. (Definitely there is also an argument via local algebra and / or using Raynaud-Gruson.)

Over $$S\setminus Z$$, the family $$\mathcal{C}$$ defines a section of the relative Hilbert scheme of $$X/S$$. Since the Hilbert scheme is a disjoint union of projective $$S$$-schemes, the closure $$S'$$ of the image of this section is projective over $$S$$. After base change to $$S'$$, there is a flat extension $$\mathcal{C}'$$. Since $$X$$ is $$S$$-smooth, also $$\mathcal{C}'$$ is a Cartier divisor in $$X'=X\times_S S'$$. However, for every point $$s$$ of $$S$$, for each point $$s'$$ of $$S'$$ over $$s$$, the Cartier divisor $$\mathcal{C}'_{s'}$$ in $$X_{s'}$$ is a positive linear combination of prime Cartier divisors that equal the irreducible components of $$\mathcal{C}_s$$. Since the total degree (with respect to some ample invertible sheaf) is constant over $$S'$$, also the positive coefficients of this linear combination are bounded. Thus, the morphism $$S'\to S$$ has finite fibers.

Every proper morphism with finite fibers is a finite morphism. Since $$S$$ is normal, every finite, birational morphism from an integral scheme $$S'$$ to $$S$$ is an isomorphism. Indeed, the local ring $$\mathcal{O}_{S',s'}$$ is intermediate between $$\mathcal{O}_{S,s}$$ and every localization $$\mathcal{O}_{S,t}$$ at a codimension $$1$$ point $$t$$ that specializes to $$s$$. Since $$S$$ is normal, the local ring $$\mathcal{O}_{S,s}$$ equals the intersection of all such local rings $$\mathcal{O}_{S,t}$$ in the fraction field. Thus, $$S'\to S$$ is an isomorphism. In other words, already $$\mathcal{C}$$ is $$S$$-flat. QED

Back to the setting of the original question, for an open subscheme $$U$$ of $$X$$ that is dense in every $$S$$-fiber, let $$\mathcal{C}_U$$ be a closed subscheme of $$U$$ whose associated points dominate $$S$$ and are codimension $$1$$ in $$U$$, and such that every $$S$$-fiber of $$\mathcal{C}_U$$ has codimension $$1$$ in the corresponding $$S$$-fiber of $$U$$. Denote by $$\mathcal{C}$$ the closure of $$\mathcal{C}_U$$ in $$X$$. Then $$\mathcal{C}$$ satisfies the hypotheses of the proposition. Thus, $$\mathcal{C}$$ is $$S$$-flat. Therefore, also $$\mathcal{C}_U$$ is $$S$$-flat.