Proposition 44.2.6. Let $X \to S$ be a morphism of schemes. Let $d \geq 0$. Assume for all $(s, x_1, \ldots , x_ d)$ where $s \in S$ and $x_1, \ldots , x_ d \in X_ s$ there exists an affine open $U \subset X$ with $x_1, \ldots , x_ d \in U$. Then $\mathrm{Hilb}^ d_{X/S}$ is representable by a scheme.

**Proof.**
Either using relative glueing (Constructions, Section 27.2) or using the functorial point of view (Schemes, Lemma 26.15.4) we reduce to the case where $S$ is affine. Details omitted.

Assume $S$ is affine. For $U \subset X$ affine open, denote $F_ U \subset \mathrm{Hilb}^ d_{X/S}$ the subfunctor such that for a scheme $T/S$ an element $Z \in \mathrm{Hilb}^ d_{X/S}(T)$ is in $F_ U(T)$ if and only if $Z \subset U_ T$. We will use Schemes, Lemma 26.15.4 and the subfunctors $F_ U$ to conclude.

Condition (1) is Lemma 44.2.1.

Condition (2)(a) follows from the fact that $F_ U = \mathrm{Hilb}^ d_{U/S}$ and that this is representable by Lemma 44.2.5. Namely, if $Z \in F_ U(T)$, then $Z$ can be viewed as a closed subscheme of $U_ T$ which is finite locally free of degree $d$ over $T$ and hence $Z \in \mathrm{Hilb}^ d_{U/S}(T)$. Conversely, if $Z \in \mathrm{Hilb}^ d_{U/S}(T)$ then $Z \to U_ T \to X_ T$ is a closed immersion^{1} and we may view $Z$ as an element of $F_ U(T)$.

Let $Z \in \mathrm{Hilb}^ d_{X/S}(T)$ for some scheme $T$ over $S$. Let

This is a closed subset of $T$ and it is clear that over the open $T_{Z, U} = T \setminus B$ the restriction $Z_{t'}$ maps into $U_{T'}$. On the other hand, for any $b \in B$ the fibre $Z_ b$ does not map into $U$. Thus we see that given a morphism $T' \to T$ we have $Z_{T'} \in F_ U(T')$ $\Leftrightarrow $ $T' \to T$ factors through the open $T_{Z, U}$. This proves condition (2)(b).

Condition (2)(c) follows from our assumption on $X/S$. All we have to do is show the following: If $T$ is the spectrum of a field and $Z \subset X_ T$ is a closed subscheme, finite flat of degree $d$ over $T$, then $Z \to X_ T \to X$ factors through an affine open $U$ of $X$. This is clear because $Z$ will have at most $d$ points and these will all map into the fibre of $X$ over the image point of $T \to S$. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)