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 immersion1 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
B = (Z \to T)\left((Z \to X_ T \to X)^{-1}(X \setminus U)\right)
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
[1] This is clear if X \to S is separated as in this case Morphisms, Lemma 29.41.7 tells us that the immersion \varphi : Z \to X_ T has closed image and hence is a closed immersion by Schemes, Lemma 26.10.4. We suggest the reader skip the rest of this footnote as we don't know of any instance where the assumptions on X \to S hold but X \to S is not separated. In the general case, let x \in X_ T be a point in the closure of \varphi (Z). We have to show that x \in \varphi (Z). Let t \in T be the image of x. By assumption on X \to S we can choose an affine open W \subset X_ T containing x and \varphi (Z_ t). Then \varphi ^{-1}(W) is an open containing the whole fibre Z_ t and since Z \to T is closed, we may after replacing T by an open neighbourhood of t assume that Z = \varphi ^{-1}(W). Then \varphi (Z) \subset W is closed by the separated case (as W \to T is separated) and we conclude x \in \varphi (Z).
Comments (0)