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 $2$.
Best Answer
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.