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 $\mathbf{P}^n\_{C\_R}$.



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> Is $\mathcal{X}$ necessarily flat over $C\_R$? What if $C\_R = \mathbf{P}^1\_R$?
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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 $2$.



Flat scheme-theoretic closure
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.


