Constant fibre polynomial does not give flatness over a nonreduced base

The scheme $Z$ is supported in $Y\mathrm{\ne }0$ and there has algebra $A[x]\mathrm{/}(x,ϵ)=k$. Its inclusion has finite presentation. The base has one geometric fibre, and after any extension of its residue field that fibre is one reduced point. Its Hilbert polynomial is therefore the constant polynomial $1$.

Nevertheless $k$ is not flat over $A$: tensoring the inclusion $(ϵ)\hookrightarrow A$ with $k$ gives the zero map from the nonzero module $(ϵ){\otimes }_{A}k\cong k$ to $k$. Tensoring has destroyed injectivity. Thus $Z$ is excluded from the Hilbert functor by [F1]. Its flattening locus for polynomial $1$ is the closed subscheme $ϵ=0$, since after a map $A\to B$ the quotient $B\mathrm{/}ϵB$ is locally free of rank one exactly when $ϵ$ maps to zero; a surjection $B\to B\mathrm{/}ϵB$ of locally free rank-one modules must be an isomorphism. The stratum and the base have the same underlying point but different scheme structures.