Hilbert Functors and Projective Hilbert Schemes — Examples

By the finite-scheme supplier in [F1], every invertible twist restricted to $Z$ has $d$-dimensional sections and zero higher cohomology. Thus its Euler characteristic is the constant $d$ for every integer twist. This includes nilpotent structure and residue-field degrees, since length is the full dimension of the finite coordinate algebra over $k$.

The Hilbert polynomial is consequently $P(r)=d$. For a flat finitely presented family of total fibre length $d$, the classifying map lands in the corresponding open and closed stratum by [F1]. The finite-scheme supplier also gives $d=0$ for the empty subscheme.

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.

The family has no points on $Y=0$: there its equation becomes $X^2=0$ but $X$ is invertible. On $Y\mathrm{\ne }0$ its algebra is $A[x]\mathrm{/}(x^2-t)$, free over $A$ with basis $1,x$ by division by the monic polynomial. It is finite flat of rank two, and its closed immersion is finitely presented. Every fibre has length two and therefore constant Hilbert polynomial $2$ by [F1]. At $t=0$ it is a double point. In characteristic different from two, nonzero fibres are either two distinct rational points or a quadratic field point. In characteristic two a nonzero fibre can also be a double point: at $t=1$, $x^2-1=(x-1{)}^2$. Total length is always two.

For an arbitrary $A$-algebra $B$, its algebra pulls back to $B[x]\mathrm{/}(x^2-t_B)$, still free on $1,x$. Thus the family remains finitely presented and flat after any base change. By [F1], the corresponding map to the Hilbert scheme is the composite of $\mathrm{Spec}B\to \mathrm{Spec}A$ with the original classifying map, and its scheme theoretic pulled-back family is exactly this algebra. In particular the assertion applies when $B$ has nilpotents.

For every $d\ge 1$, the subscheme on the affine chart given by $x^d=0$, viewed as a closed subscheme of ${\mathbb{P}}_k^1$ supported at $[0:1]$, is finitely presented and has length $d$. Being over a field it is flat. Thus the polynomial-$d$ stratum is nonempty by [F2]. These are distinct strata for distinct $d$.

All strata are open and closed by [F1], and they form an open cover of the full Hilbert scheme. No finite subfamily of this cover contains the nonempty strata for all $d$. This cover has no finite subcover, proving failure of quasi-compactness and therefore of finite type or properness over $k$.