Example 38.21.2. Let $X = S = \mathop{\mathrm{Spec}}(k[x, y])$ where $k$ is a field. Let $\mathcal{F} = \widetilde{M}$ where $M = k[x, x^{-1}, y]/(y)$. For a $k[x, y]$-algebra $A$ set $F_{flat}(A) = F_{flat}(\mathop{\mathrm{Spec}}(A))$. Then $F_{flat}(k[x, y]/(x, y)^ n) = \{ *\}$ for all $n$, while $F_{flat}(k[[x, y]]) = \emptyset$. This means that $F_{flat}$ isn't representable (even by an algebraic space, see Formal Spaces, Lemma 86.33.3). Thus the universal flattening does not exist in this case.

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