The Stacks project

Proof. Let $\{ T_ i \to T\} _{i \in I}$ be an fpqc covering of schemes over $S$. Set $X_ i = X_{T_ i} = X \times _ S T_ i$. Note that $\{ X_ i \to X_ T\} _{i \in I}$ is an fpqc covering of $X_ T$ (Topologies, Lemma 34.9.8) and that $X_{T_ i \times _ T T_{i'}} = X_ i \times _{X_ T} X_{i'}$. Suppose that $Z_ i \in \mathrm{Hilb}^ d_{X/S}(T_ i)$ is a collection of elements such that $Z_ i$ and $Z_{i'}$ map to the same element of $\mathrm{Hilb}^ d_{X/S}(T_ i \times _ T T_{i'})$. By effective descent for closed immersions (Descent, Lemma 35.37.2) there is a closed immersion $Z \to X_ T$ whose base change by $X_ i \to X_ T$ is equal to $Z_ i \to X_ i$. The morphism $Z \to T$ then has the property that its base change to $T_ i$ is the morphism $Z_ i \to T_ i$. Hence $Z \to T$ is finite locally free of degree $d$ by Descent, Lemma 35.23.30. $\square$

Proof. Let $T = \mathop{\mathrm{lim}}\nolimits T_ i$ be a limit of affine schemes over $S$. We have to show that $\mathrm{Hilb}^ d_{X/S}(T) = \mathop{\mathrm{colim}}\nolimits \mathrm{Hilb}^ d_{X/S}(T_ i)$. Observe that if $Z \to X_ T$ is an element of $\mathrm{Hilb}^ d_{X/S}(T)$, then $Z \to T$ is of finite presentation. Hence by Limits, Lemma 32.10.1 there exists an $i$, a scheme $Z_ i$ of finite presentation over $T_ i$, and a morphism $Z_ i \to X_{T_ i}$ over $T_ i$ whose base change to $T$ gives $Z \to X_ T$. We apply Limits, Lemma 32.8.5 to see that we may assume $Z_ i \to X_{T_ i}$ is a closed immersion after increasing $i$. We apply Limits, Lemma 32.8.8 to see that $Z_ i \to T_ i$ is finite locally free of degree $d$ after possibly increasing $i$. Then $Z_ i \in \mathrm{Hilb}^ d_{X/S}(T_ i)$ as desired. $\square$

Proof. Let $T$ be a scheme over $S$ and let $Z \in \mathrm{Hilb}^ d_{Y/S}(T)$. Claim: there is a closed subscheme $T_ X \subset T$ such that a morphism of schemes $T' \to T$ factors through $T_ X$ if and only if $Z_{T'} \to Y_{T'}$ factors through $X_{T'}$. Applying this to a scheme $T_{univ}$ representing $\mathrm{Hilb}^ d_{Y/S}$ and the universal object¹ $Z_{univ} \in \mathrm{Hilb}^ d_{Y/S}(T_{univ})$ we get a closed subscheme $T_{univ, X} \subset T_{univ}$ such that $Z_{univ, X} = Z_{univ} \times _{T_{univ}} T_{univ, X}$ is a closed subscheme of $X \times _ S T_{univ, X}$ and hence defines an element of $\mathrm{Hilb}^ d_{X/S}(T_{univ, X})$. A formal argument then shows that $T_{univ, X}$ is a scheme representing $\mathrm{Hilb}^ d_{X/S}$ with universal object $Z_{univ, X}$.

Proof of the claim. Consider $Z' = X_ T \times _{Y_ T} Z$. Given $T' \to T$ we see that $Z_{T'} \to Y_{T'}$ factors through $X_{T'}$ if and only if $Z'_{T'} \to Z_{T'}$ is an isomorphism. Thus the claim follows from the very general More on Flatness, Lemma 38.23.4. However, in this special case one can prove the statement directly as follows: first reduce to the case $T = \mathop{\mathrm{Spec}}(A)$ and $Z = \mathop{\mathrm{Spec}}(B)$. After shrinking $T$ further we may assume there is an isomorphism $\varphi : B \to A^{\oplus d}$ as $A$-modules. Then $Z' = \mathop{\mathrm{Spec}}(B/J)$ for some ideal $J \subset B$. Let $g_\beta \in J$ be a collection of generators and write $\varphi (g_\beta ) = (g_\beta ^1, \ldots , g_\beta ^ d)$. Then it is clear that $T_ X$ is given by $\mathop{\mathrm{Spec}}(A/(g_\beta ^ j))$. $\square$

Proof. In this proof all unadorned products are over $S$. Let $H = \underline{\mathrm{Hilb}}^ d_{X/S}$ and let $Z \in \mathrm{Hilb}^ d_{X/S}(H)$ be the universal object. Consider the two objects $Z_1, Z_2 \in \mathrm{Hilb}^ d_{X/S}(H \times H)$ we get by pulling back $Z$ by the two projections $H \times H \to H$. Then $Z_1 = Z \times H \subset X_{H \times H}$ and $Z_2 = H \times Z \subset X_{H \times H}$. Since $H$ represents the functor $\mathrm{Hilb}^ d_{X/S}$, the diagonal morphism $\Delta : H \to H \times H$ has the following universal property: A morphism of schemes $T \to H \times H$ factors through $\Delta$ if and only if $Z_{1, T} = Z_{2, T}$ as elements of $\mathrm{Hilb}^ d_{X/S}(T)$. Set $Z = Z_1 \times _{X_{H \times H}} Z_2$. Then we see that $T \to H \times H$ factors through $\Delta$ if and only if the morphisms $Z_ T \to Z_{1, T}$ and $Z_ T \to Z_{2, T}$ are isomorphisms. It follows from the very general More on Flatness, Lemma 38.23.4 that $\Delta$ is a closed immersion. In the proof of Lemma 44.2.3 the reader finds an alternative easier proof of the needed result in our special case. $\square$

Proof. Say $S = \mathop{\mathrm{Spec}}(R)$. Then we can choose a closed immersion of $X$ into the spectrum of $R[x_ i; i \in I]$ for some set $I$ (of sufficiently large cardinality. Hence by Lemma 44.2.3 we may assume that $X = \mathop{\mathrm{Spec}}(A)$ where $A = R[x_ i; i \in I]$. We will use Schemes, Lemma 26.15.4 to prove the lemma in this case.

Condition (1) of the lemma follows from Lemma 44.2.1.

For every subset $W \subset A$ of cardinality $d$ we will construct a subfunctor $F_ W$ of $\mathrm{Hilb}^ d_{X/S}$. (It would be enough to consider the case where $W$ consists of a collection of monomials in the $x_ i$ but we do not need this.) Namely, we will say that $Z \in \mathrm{Hilb}^ d_{X/S}(T)$ is in $F_ W(T)$ if and only if the $\mathcal{O}_ T$-linear map

is surjective (equivalently an isomorphism). Here for $f \in A$ and $Z \in \mathrm{Hilb}^ d_{X/S}(T)$ we denote $f|_ Z$ the pullback of $f$ by the morphism $Z \to X_ T \to X$.

Openness, i.e., condition (2)(b) of the lemma. This follows from Algebra, Lemma 10.79.4.

Covering, i.e., condition (2)(c) of the lemma. Since

is surjective and since $(Z \to T)_*\mathcal{O}_ Z$ is finite locally free of rank $d$, for every point $t \in T$ we can find a finite subset $W \subset A$ of cardinality $d$ whose images form a basis of the $d$-dimensional $\kappa (t)$-vector space $((Z \to T)_*\mathcal{O}_ Z)_ t \otimes _{\mathcal{O}_{T, t}} \kappa (t)$. By Nakayama's lemma there is an open neighbourhood $V \subset T$ of $t$ such that $Z_ V \in F_ W(V)$.

Representable, i.e., condition (2)(a) of the lemma. Let $W \subset A$ have cardinality $d$. We claim that $F_ W$ is representable by an affine scheme over $R$. We will construct this affine scheme here, but we encourage the reader to think it through for themselves. Choose a numbering $f_1, \ldots , f_ d$ of the elements of $W$. We will construct a universal element $Z_{univ} = \mathop{\mathrm{Spec}}(B_{univ})$ of $F_ W$ over $T_{univ} = \mathop{\mathrm{Spec}}(R_{univ})$ which will be the spectrum of

where the $e_ l$ will be the images of the $f_ l$ and where the closed immersion $Z_{univ} \to X_{T_{univ}}$ is given by the ring map

mapping $1 \otimes 1$ to $\sum b^ le_ l$ and $x_ i$ to $\sum b_ i^ le_ l$. In fact, we claim that $F_ W$ is represented by the spectrum of the ring

where the ideal $\mathfrak a_{univ}$ is generated by the following elements:

1. multiplication on $B_{univ}$ is commutative, i.e., $c_{lk}^ m - c_{kl}^ m \in \mathfrak a_{univ}$,

2. multiplication on $B_{univ}$ is associative, i.e., $c_{lk}^ m c_{m n}^ p - c_{lq}^ p c_{kn}^ q \in \mathfrak a_{univ}$,

3. $\sum b^ le_ l$ is a multiplicative $1$ in $B_{univ}$, in other words, we should have $(\sum b^ le_ l)e_ k = e_ k$ for all $k$, which means $\sum b^ lc_{lk}^ m - \delta _{km} \in \mathfrak a_{univ}$ (Kronecker delta).

After dividing out by the ideal $\mathfrak a'_{univ}$ of the elements listed sofar we obtain a well defined ring map

sending $1 \otimes 1$ to $\sum b^ le_ l$ and $x_ i \otimes 1$ to $\sum b_ i^ le_ l$. We need to add some more elements to our ideal because we need

1. $f_ l$ to map to $e_ l$ in $B_{univ}$. Write $\Psi (f_ l) - e_ l = \sum h_ l^ me_ m$ with $h_ l^ m \in R[c_{kl}^ m, b^ l, b_ i^ l]/\mathfrak a'_{univ}$ then we need to set $h_ l^ m$ equal to zero.

Thus setting $\mathfrak a_{univ} \subset R[c_{kl}^ m, b^ l, b_ i^ l]$ equal to $\mathfrak a'_{univ} +$ ideal generated by lifts of $h_ l^ m$ to $R[c_{kl}^ m, b^ l, b_ i^ l]$, then it is clear that $F_ W$ is represented by $\mathop{\mathrm{Spec}}(R_{univ})$. $\square$

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 immersion² 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 $T \to S$. $\square$