# Flat scheme-theoretic closure

Suppose $$R$$ is a discrete valuation ring with fraction field $$K$$. Let $$X\subset \mathbf{P}^n_{C_K}$$ be a closed subscheme, flat over $$C_K$$, a smooth projective curve over $$K$$.

Let $$C_R$$ be a flat regular proper model for $$C$$ over $$R$$ and take $$\mathcal{X}$$ the scheme-theoretic closure of $$X$$ in $$\mathbf{P}^n_{C_R}$$.

Is $$\mathcal{X}$$ necessarily flat over $$C_R$$? What if $$C_R = \mathbf{P}^1_R$$?

I'd expect "no" because the scheme-theoretic closure is flat when $$C_K$$ is replaced by the open complement of a point in a regular integral scheme of dimension $$1$$, whereas $$C_R$$ is of dimension $$2$$.

• Suppose $\dim \mathcal{X} = \dim C_R = 2$. Even then it's not something you can guarantee, if $\mathcal{X}$ happens to have a blowup in it (ie, all fibers are not the same dimensional) then it won't be flat. Or if $\mathcal{X}$ is even finite over $C_R$ but $\mathcal{X}$ is not S2, it's not flat. Commented May 6, 2023 at 19:46

Explicitly, lets let $$R = \mathbb{C}[x]_{(x)}$$ so $$K = \mathbb{C}(x)$$. You can then let $$C_K = {\bf P}^1_K$$ and $$C_R = {\bf P}^1_R$$. $$C_R$$ has a chart that looks like $$\mathbb{C}[x]_(x)[y]$$. This is just a localization of $$\mathbb{C}[x,y]$$. Blowup $$(x,y)$$ in the latter, and localize (equivalently, blowup $$(x,y)$$ in the appropriate chart of $$C_R$$). You get $$\mathcal{X} \to C_R$$ which is not flat.
Now, $$\mathcal{X} \subseteq \mathbb{P}^1_{C_R}$$, so base change to $$C_K$$ and you get $$X \subseteq \mathbb{P}^1_{C_K}$$. It's obviously flat over $$C_K$$ (in fact $$X \to C_K$$ is an isomorphism). When you close it up, you just recover $$\mathcal{X}$$ though (and you can't recover anything else, since you have the generic point of $$\mathcal{X}$$ in $$X$$).
This has nothing to do with $$\mathbb{C}$$, you can do the same with any DVR.