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Let $X$ be a fixed curve (e.g. a Noetherian, projective scheme of dimension 1, of finite type over an algebraically closed field $k$) and let $S$ be an arbitrary parameter scheme over $k$. Let $D \subset X \times S$ be a flat family over $S$ of subschemes of $X$, of relative dimension 0 and degree $d$, with ideal sheaf $\mathcal I$.

Consider now the clolsed subscheme $D' \subset X \times S$ defined by the ideal $\mathcal I^n$. In my mind, $D' = nD$ in some geometric sense. Is $D'$ again flat over $S$?

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    $\begingroup$ No, that is not true. Let $X$ be $\text{Spec}\ k[x,y]/\langle xy \rangle.$ Let $S$ be $\text{Spec}\ k[\epsilon]/\langle \epsilon^2 \rangle.$ Let $I$ be $\langle x,y-\epsilon\rangle.$ Then $\mathcal{O}_{X\times S}/\mathcal{I}^2$ is a direct sum of two copies of $\mathcal{O}_S$ (generated by $1$ and $y$) plus one summand $\mathcal{O}_S/\epsilon \mathcal{O}_S$ (generated by $x$). It is not $\mathcal{O}_S$-flat. $\endgroup$
    – Jason Starr
    Commented Oct 5, 2018 at 12:22

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I am just posting my comment as an answer. No, that is not true. Let $X$ be $\text{Spec}\ k[t,u]/\langle tu\rangle $. Let $S$ be $\text{Spec}\ k[ϵ]/\langle ϵ^2 \rangle.$ Let $I$ be $\langle t,u−ϵ\rangle.$ Then $\mathcal{O}_{X×S}/I^2$ is a direct sum of two copies of $\mathcal{O}_S$ (generated by $1$ and $u$) plus one summand $\mathcal{O}_S/ϵ\mathcal{O}_S$ (generated by $t$). It is not $\mathcal{O}_S$-flat.

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