1 References

2 Schemes vs Representable Functors (Thursday January 9th)

Last time: fix an \(S{\hbox{-}}\)scheme, i.e. a scheme over \(S\). Then there is a map \begin{align*} {\operatorname{Sch}}_{_{/S}} &\to {\operatorname{Fun}}( {\operatorname{Sch}}_{_{/S}}^{\operatorname{op}}, {\operatorname{Set}}) \\ x &\mapsto h_x(T) = \hom_{{\operatorname{Sch}}_{_{/S}} }(T, x) .\end{align*}

where \(T' \xrightarrow{f} T\) is given by

\begin{align*} h_x(f): h_x(T) &\to h_x(T') \\ \qty{ T \mapsto x } &\mapsto \text{triangles of the form} .\end{align*}

2.1 Representability

\begin{align*}\hom_{{\operatorname{Fun}}}(h_x, F) = F(x).\end{align*}

\begin{align*}\hom_{{\operatorname{Sch}}_{/S}}(x, y) \cong \hom_{{\operatorname{Fun}}}(h_x, h_y).\end{align*}

A moduli functor is a map

\begin{align*} F: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}&\to {\operatorname{Set}}\\ F(x) &= \text{ "Families of something over $x$" } \\ F(f) &= \text{"Pullback"} .\end{align*}

A moduli space for that “something” appearing above is an \(M \in \mathrm{Obj}({\operatorname{Sch}}_{/S})\) such that \(F \cong h_M\).

Now fix \(S = \operatorname{Spec}(k)\), and write \(h_m\) for the functor of points over \(M\). Then \begin{align*} h_m(\operatorname{Spec}(k)) = M(\operatorname{Spec}(k)) \cong \text{families over } \operatorname{Spec}k = F(\operatorname{Spec}k) .\end{align*}

\(h_M(M) \cong F(M)\) are families over \(M\), and \(\operatorname{id}_M \in \mathrm{Mor}_{{\operatorname{Sch}}_{/S}}(M, M) = \xi_{Univ}\) is the universal family.

Every family is uniquely the pullback of \(\xi_{\text{Univ}}\). This makes it much like a classifying space. For \(T\in {\operatorname{Sch}}_{/S}\), \begin{align*} h_M &\xrightarrow{\cong} F \\ f\in h_M(T) &\xrightarrow{\cong} F(T) \ni \xi = F(f)(\xi_{\text{Univ}}) .\end{align*}

where \(T\xrightarrow{f} M\) and \(f = h_M(f)(\operatorname{id}_M)\).

If \(M\) and \(M'\) both represent \(F\) then \(M \cong M'\) up to unique isomorphism.

which shows that \(f, g\) must be mutually inverse by using universal properties.

A length 2 subscheme of \({\mathbb{A}}^1_k\) (??) then \begin{align*} F(S) = \left\{{ V(x^2 + bx + c)}\right\} \subset {\mathbb{A}}^5 \end{align*} where \(b, c \in {\mathcal{O}}_s(s)\), which is functorially bijective with \(\left\{{b, c \in {\mathcal{O}}_s(s)}\right\}\) and \(F(f)\) is pullback. Then \(F\) is representable by \({\mathbb{A}}_k^2(b, c)\) and the universal object is given by \begin{align*} V(x^2 + bx + c) \subset {\mathbb{A}}^1(?) \times{\mathbb{A}}^2(b, c) \end{align*} where \(b, c \in k[b, c]\). Moreover, \(F'(S)\) is the set of effective Cartier divisors in \({\mathbb{A}}_5'\) which are length 2 for every geometric fiber. \(F''(S)\) is the set of subschemes of \({\mathbb{A}}_5'\) which are length 2 on all geometric fibers. In both cases, \(F(f)\) is always given by pullback.

Problem: \(F''\) is not a good moduli functor, as it is not representable. Consider \(\operatorname{Spec}k[\varepsilon]\), for which we have the following situation:

We think of \(T_p F^{', ''}\) as the tangent space at \(p\). If \(F\) is representable, then it is actually the Zariski tangent space.

Moreover, \(T_p M = ({\mathfrak{m}}_p / {\mathfrak{m}}_p^2)^\vee\), and in particular this is a \(k{\hbox{-}}\)vector space. To see the scaling structure, take \(\lambda \in k\).

\begin{align*} \lambda: k[\varepsilon] & \to k[\varepsilon] \\ \varepsilon &\mapsto \lambda \varepsilon \\ \\ \lambda^*: \operatorname{Spec}(k[\varepsilon]) &\to \operatorname{Spec}(k[\varepsilon]) \\ \\ \lambda: M(\operatorname{Spec}(k[\varepsilon])) &\to M(\operatorname{Spec}(k[\varepsilon])) .\end{align*}

Conclusion: If \(F\) is representable, for each \(p\in F(\operatorname{Spec}k)\) there exists a unique point of \(T_p F\) that are invariant under scaling.

If \(F, F', G \in {\operatorname{Fun}}( ({\operatorname{Sch}}_{/S})^{\operatorname{op}}, {\operatorname{Set}})\), there exists a fiber product

where \begin{align*} (F \times_G F')(T) = F(T) \times_{G(T)} F'(T) .\end{align*}

This works with the functor of points over a fiber product of schemes \(X \times_T Y\) for \(X, Y \to T\), where \begin{align*} h_{X \times_T Y}= h_X \times_{h_t} h_Y .\end{align*}

If \(F, F', G\) are representable, then so is the fiber product \(F \times_G F'\).

For any functor \begin{align*} F: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}\to {\operatorname{Set}} ,\end{align*} for any \(T \xrightarrow{f} S\) there is an induced functor \begin{align*} F_T: ({\operatorname{Sch}}_{/T}) &\to {\operatorname{Set}}\\ x &\mapsto F(x) .\end{align*}

\(F\) is representable by \(M_{/S}\) implies that \(F_T\) is representable by \(M_T = M \times_S T / T\).

2.2 Projective Space

Consider \({\mathbb{P}}^n_{\mathbb{Z}}\), i.e. “rank 1 quotient of an \(n+1\) dimensional free module.”

\({\mathbb{P}}^n_{/{\mathbb{Z}}}\) represents the following functor \begin{align*} F: {\operatorname{Sch}}^{\operatorname{op}}&\to {\operatorname{Set}}\\ S &\mapsto \left\{{ {\mathcal{O}}_S^{n+1} \to L \to 0 }\right\} / \sim .\end{align*}

where \(\sim\) identifies diagrams of the following form:

and \(F(f)\) is given by pullbacks.

\({\mathbb{P}}^n_{/S}\) represents the following functor: \begin{align*} F_S: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}&\to {\operatorname{Set}}\\ T &\mapsto F_S(T) = \left\{{ {\mathcal{O}}_T^{n+1} \to L \to 0}\right\} / \sim .\end{align*}

This gives us a cleaner way of gluing affine data into a scheme.

2.2.1 Proof of Proposition

Note that \({\mathcal{O}}^{n+1} \to L \to 0\) is the same as giving \(n+1\) sections \(s_1, \cdots s_n\) of \(L\), where surjectivity ensures that they are not the zero section. So \begin{align*} F_i(S) = \left\{{{\mathcal{O}}_s^{n+1} \to L \to 0}\right\}/\sim ,\end{align*} with the additional condition that \(s_i \neq 0\) at any point. There is a natural transformation \(F_i \to F\) by forgetting the latter condition, and is in fact a subfunctor.1

It is enough to show that each \(F_i\) and each \(F_{ij}\) are representable, since we have natural transformations:

and each \(F_{ij} \to F_i\) is an open embedding on the level of their representing schemes.

For \(n=1\), we can glue along open subschemes

For \(n=2\), we get overlaps of the following form:

This claim implies that we can glue together \(F_i\) to get a scheme \(M\). We want to show that \(M\) represents \(F\). \(F(s)\) (LHS) is equivalent to an open cover \(U_i\) of \(S\) and sections of \(F_i(U_i)\) satisfying the gluing (RHS). Going from LHS to RHS isn’t difficult, since for \({\mathcal{O}}_s^{n+1} \to L \to 0\), \(U_i\) is the locus where \(s_i \neq 0\) and by surjectivity, this gives a cover of \(S\). The RHS to LHS comes from gluing.

We have \begin{align*} F_i(S) = \left\{{{\mathcal{O}}_S^{n+1} \to L \cong {\mathcal{O}}_s \to 0, s_i \neq 0}\right\} ,\end{align*} but there are no conditions on the sections other than \(s_i\).
So specifying \(F_i(S)\) is equivalent to specifying \(n-1\) functions \(f_1 \cdots \widehat{f}_i \cdots f_n \in {\mathcal{O}}_S(s)\) with \(f_k \neq 0\). We know this is representable by \({\mathbb{A}}^n\). We also know \(F_{ij}\) is obviously the same set of sequences, where now \(s_j \neq 0\) as well, so we need to specify \(f_0 \cdots \widehat{f}_i \cdots f_j \cdots f_n\) with \(f_j \neq 0\). This is representable by \({\mathbb{A}}^{n-1} \times{\mathbb{G}}_m\), i.e. \(\operatorname{Spec}k[x_1, \cdots, \widehat{x}_i, \cdots, x_n, x_j^{-1}]\). Moreover, \(F_{ij} \hookrightarrow F_i\) is open.

What is the compatibility we are using to glue? For any subset \(I \subset \left\{{0, \cdots, n}\right\}\), we can define \begin{align*} F_I = \left\{{{\mathcal{O}}_s^{n+1} \to L \to 0, s_i\neq 0 \text{ for } i\in I}\right\} = {\prod{i\in I}}_F F_i ,\end{align*} and \(F_I \to F_J\) when \(I \supset J\).

3 Functors as Spaces (Tuesday January 14th)

Last time: representability of functors, and specifically projective space \({\mathbb{P}}_{/{\mathbb{Z}}}^n\) constructed via a functor of points, i.e.

\begin{align*} h_{{\mathbb{P}}^n_{/{\mathbb{Z}}} }: {\operatorname{Sch}}^{\operatorname{op}}&\to {\operatorname{Set}}\\ s &\mapsto {\mathbb{P}}^n_{/{\mathbb{Z}}}(s) = \left\{{ {\mathcal{O}}_s^{n+1} \to L \to 0}\right\} .\end{align*}

for \(L\) a line bundle, up to isomorphisms of diagrams:

That is, line bundles with \(n+1\) sections that globally generate it, up to isomorphism. The point was that for \(F_i \subset {\mathbb{P}}_{/{\mathbb{Z}}}^n\) where \begin{align*} F_i(s) = \left\{{{\mathcal{O}}_s^{n+1} \to L \to 0 {~\mathrel{\Big|}~}s_i \text{ is invertible}}\right\} \end{align*} are representable and can be glued together, and projective space represents this functor.

Because projective space represents this functor, there is a universal object:

and other functors are pullbacks of the universal one. (Moduli Space)

Show that \({\mathbb{P}}_{/{\mathbb{Z}}}^n\) is proper over \(\operatorname{Spec}{\mathbb{Z}}\). Use the evaluative criterion, i.e. there is a unique lift

3.1 Generalizing Open Covers

For a category \(C\), we say a diagram \(X \to Y \rightrightarrows Z\) is an equalizer iff it is universal with respect to the following property:

where \(X\) is the universal object.

For sets, \(X = \left\{{y {~\mathrel{\Big|}~}f(y) = g(y)}\right\}\) for \(Y \xrightarrow{f, g} Z\).

A coequalizer is the dual notion,

Take \(C = {\operatorname{Sch}}_{/S}\), \(X_{/S}\) a scheme, and \(X_\alpha \subset X\) an open cover. We can take two fiber products, \(X_{\alpha \beta}, X_{\beta, \alpha}\):

These are canonically isomorphic.

In \({\operatorname{Sch}}_{/S}\), we have

where \begin{align*} f_{\alpha\beta}: X_{\alpha\beta} &\to X_\alpha \\ g_{\alpha\beta}: X_{\alpha\beta} &\to X_\beta ;\end{align*}

form a coequalizer. Conversely, we can glue schemes. Given \(X_\alpha \to X_{\alpha\beta}\) (schemes over open subschemes), we need to check triple intersections:

Then \(\varphi_{\alpha\beta}: X_{\alpha\beta}\xrightarrow{\cong} X_{\beta\alpha}\) must satisfy the cocycle condition:

Maps \(\varphi_{\alpha\beta}: X_{\alpha\beta}\xrightarrow{\cong} X_{\beta\alpha}\) satisfy the cocycle condition iff

  1. \begin{align*}\varphi_{\alpha\beta}^{-1}\qty{ X_{\beta\alpha} \cap X_{\beta\gamma} } = X_{\alpha\beta} \cap X_{\alpha \gamma},\end{align*} noting that the intersection is exactly the fiber product \(X_{\beta\alpha} \times_{X_\beta} X_{\beta \gamma}\).

  2. The following diagram commutes:

Then there exists a scheme \(X_{/S}\) such that \({\coprod}_{\alpha\beta} X_{\alpha\beta} \rightrightarrows {\coprod}X_\alpha \to X\) is a coequalizer; this is the gluing.

Subfunctors satisfy a patching property because morphisms to schemes are locally determined. Thus representable functors (e.g. functors of points) have to be (Zariski) sheaves.

A functor \(F: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}\to {\operatorname{Set}}\) is a Zariski sheaf iff for any scheme \(T_{/S}\) and any open cover \(T_\alpha\), the following is an equalizer: \begin{align*} F(T) \to \prod F(T_\alpha) \rightrightarrows \prod_{\alpha\beta} F(T_{\alpha\beta}) \end{align*} where the maps are given by restrictions.

Any representable functor is a Zariski sheaf precisely because the gluing is a coequalizer. Thus if you take the cover \begin{align*} {\coprod}_{\alpha\beta} T_{\alpha\beta} \to {\coprod}_{\alpha}T_\alpha \to T ,\end{align*} since giving a local map to \(X\) that agrees on intersections if enough to specify a map from \(T\to X\).

Thus any functor represented by a scheme automatically satisfies the sheaf axioms.

Suppose we have a morphism \(F' \to F\) in the category \({\operatorname{Fun}}({\operatorname{Sch}}_{/S}, {\operatorname{Set}})\).

This says that we can test if pullbacks are contained in a subfunctors by checking factorization. This is the same as asking if the subfunctor \(F'\), which maps to \(F\) (noting a section is the same as a map to the functor of points), and since \(T\to F\) and \(F' \to F\), we can form the fiber product \(F' \times_F T\):

and \(F' \times_F T \cong U\). Note: this is almost tautological! Thus \(F' \to F\) is open/closed/locally closed iff \(F' \times_F T\) is representable and \(g\) is open/closed/locally closed. I.e. base change is representable.

  1. If \(F' \to F\) is open/closed/locally closed and \(F\) is representable, then \(F'\) is representable as an open/closed/locally closed subscheme

  2. If \(F\) is representable, then open/etc subschemes yield open/etc subfunctors

Treat functors as spaces.

We have a definition of open, so now we’ll define coverings.

A collection of open subfunctors \(F_\alpha \subset F\) is an open cover iff for any \(T_{/S}\) and any section \(\xi \in F(T)\), i.e. \(\xi: T\to F\), the \(T_\alpha\) in the following diagram are an open cover of \(T\):

Given \begin{align*} F(s) = \left\{{{\mathcal{O}}_s^{n+1} \to L \to 0}\right\} \end{align*} and \(F_i(s)\) given by those where \(s_i \neq 0\) everywhere, the \(F_i \to F\) are an open cover. Because the sections generate everything, taking the \(T_i\) yields an open cover.

3.2 Results About Zariski Sheaves

A Zariski sheaf \(F: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}\to {\operatorname{Set}}\) with a representable open cover is representable.

Let \(F_\alpha \subset F\) be an open cover, say each \(F_\alpha\) is representable by \(x_\alpha\). Form the fiber product \(F_{\alpha\beta} = F_\alpha \times_F F_\beta\). Then \(x_\beta\) yields a section (plus some openness condition?), so \(F_{\alpha\beta} = x_{\alpha\beta}\) representable. Because \(F_\alpha \subset F\), the \(F_{\alpha\beta} \to F_\alpha\) have the correct gluing maps.

This follows from Yoneda (schemes embed into functors), and we get maps \(x_{\alpha\beta} \to x_\alpha\) satisfying the gluing conditions. Call the gluing scheme \(x\); we’ll show that \(x\) represents \(F\). First produce a map \(x\to F\) from the sheaf axioms. We have a map \(\xi \in \prod_\alpha F(x_\alpha)\), and because we can pullback, we get a unique element \(\xi \in F(X)\) coming from the diagram \begin{align*} F(x) \to \prod F(x_\alpha) \rightrightarrows \prod_{\alpha\beta} F(x_{\alpha\beta}) .\end{align*}

If \(E \to F\) is a map of functors and \(E, F\) are Zariski sheaves, where there are open covers \(E_\alpha \to E, F_\alpha \to F\) with commutative diagrams

(i.e. these are isomorphisms locally), then the map is an isomorphism.

With the following diagram, we’re done by the lemma:

For \(S\) and \(E\) a locally free coherent \({\mathcal{O}}_s\) module, \begin{align*} {\mathbb{P}}E(T) = \left\{{f^* E \to L \to 0}\right\} / \sim \end{align*} is a generalization of projectivization, then \(S\) admits a cover \(U_i\) trivializing \(E\). Then the restriction \(F_i \to {\mathbb{P}}E\) were \(F_i(T)\) is the above set if \(f\) factors through \(U_i\) and empty otherwise. On \(U_i\), \(E \cong {\mathcal{O}}_{U_i}^{n_i}\), so \(F_i\) is representable by \({\mathbb{P}}_{U_i}^{n_i - 1}\) by the proposition. Note that this is clearly a sheaf.

For \(E\) locally free over \(S\) of rank \(n\), take \(r<n\) and consider the functor \begin{align*} {\operatorname{Gr}}(k, E)(T) = \left\{{f^*E \to Q \to 0}\right\} /\sim \end{align*} (a Grassmannian) where \(Q\) is locally free of rank \(k\).

  1. Show that this is representable

  2. For the Plucker embedding \begin{align*} {\operatorname{Gr}}(k, E) \to {\mathbb{P}}\wedge^k E ,\end{align*} a section over \(T\) is given by \(f^*E \to Q \to 0\) corresponding to \begin{align*} \wedge^k f^*E \to \wedge^k Q \to 0 ,\end{align*} noting that the left-most term is \(f^* \wedge^k E\). Show that this is a closed subfunctor.

That it’s a functor is clear, that it’s closed is not.

Take \(S = \operatorname{Spec}k\), then \(E\) is a \(k{\hbox{-}}\)vector space \(V\), then sections of the Grassmannian are quotients of \(V \otimes{\mathcal{O}}\) that are free of rank \(n\). Take the subfunctor \(G_w \subset {\operatorname{Gr}}(k, V)\) where \begin{align*} G_w(T) = \left\{{{\mathcal{O}}_T \otimes V \to Q \to 0}\right\} \text{ with } Q \cong {\mathcal{O}}_t\otimes W \subset {\mathcal{O}}_t \otimes V .\end{align*} If we have a splitting \(V = W \oplus U\), then \(G_W = {\mathbb{A}}(\hom(U, W))\). If you show it’s closed, it follows that it’s proper by the exercise at the beginning.

Thursday: Define the Hilbert functor, show it’s representable. The Hilbert scheme functor gives e.g. for \({\mathbb{P}}^n\) of all flat families of subschemes.

4 Thursday January 16th

4.1 Subfunctors

A functor \(F' \subset F: ({\operatorname{Sch}}_{/S})^{\operatorname{op}}\to {\operatorname{Set}}\) is open iff for all \(T \xrightarrow{\xi} F\) where \(T = h_T\) and \(\xi \in F(T)\).

We can take fiber products:

So we can think of “inclusion in \(F\)” as being an open condition: for all \(T_{/S}\) and \(\xi \in F(T)\), there exists an open \(U \subset T\) such that for all covers \(f: T' \to T\), we have \begin{align*} F(f)(\xi) = f^*(\xi) \in F'(T') \end{align*} iff \(f\) factors through \(U\).

Suppose \(U \subset T\) in \({\operatorname{Sch}}/T\), we then have \begin{align*} h_{U/T}(T') = \begin{cases} \emptyset & T' \to T \text{ doesn't factor } \\ {\{\operatorname{pt}\}}& \text{otherwise} \end{cases} .\end{align*}

which follows because the literal statement is \(h_{U/T}(T') = \hom_T(T', U)\). By the definition of the fiber product, \begin{align*} (F' \times_F T)(T') = \left\{{ (a,b) \in F'(T) \times T(T) {~\mathrel{\Big|}~}\xi(b) = \iota(a) \text{ in } F(T)}\right\} ,\end{align*} where \(F' \xrightarrow{\iota} F\) and \(T \xrightarrow{\xi} F\). So note that the RHS diagram here is exactly given by pullbacks, since we identify sections of \(F/T'\) as sections of \(F\) over \(T/T'\) (?).

We can thus identify \begin{align*} (F' \times_F T)(T') = h_{U_{/S}}(T') ,\end{align*} and so for \(U \subset T\) in \({\operatorname{Sch}}_{/S}\) we have \(h_{U_{/S}} \subset h_{T_{/S}}\) is the functor of maps that factor through \(U\). We just identify \(h_{U_{/S}}(T') = \hom_S(T', U)\) and \(h_{T_{/S}}(T') = \hom_S(T', T)\).

\({\mathbb{G}}_m, {\mathbb{G}}_a\). The scheme/functor \({\mathbb{G}}_a\) represents giving a global function, \({\mathbb{G}}_m\) represents giving an invertible function.

where \(T' = \left\{{f\neq 0}\right\}\) and \({\mathcal{O}}_T(T)\) are global functions.

4.2 Actual Geometry: Hilbert Schemes

The best moduli space!

Unless otherwise stated, assume all schemes are Noetherian.

We want to parameterize families of subschemes over a fixed object. Fix \(k\) a field, \(X_{/k}\) a scheme; we’ll parameterize subschemes of \(X\).

The Hilbert functor is given by \begin{align*} \operatorname{Hilb}_{X_{/S}}: ({\operatorname{Sch}}_{/S})^{op} \to {\operatorname{Set}} \end{align*} which sends \(T\) to closed subschemes \(Z \subset X \times_S T \to T\) which are flat over \(T\).

Here flatness will replace the Cartier condition:

For \(X \xrightarrow{f} Y\) and \({\mathbb{F}}\) a coherent sheaf on \(X\), \(f\) is flat over \(Y\) iff for all \(x\in X\) the stalk \(F_x\) is a flat \({\mathcal{O}}_{y, f(x)}{\hbox{-}}\)module.

Note that \(f\) is flat if \({\mathcal{O}}_x\) is. Flatness corresponds to varying continuously. Note that everything works out if we only play with finite covers.

If \(X_{/k}\) is projective, so \(X \subset {\mathbb{P}}^n_k\), we have line bundles \({\mathcal{O}}_x(1) = {\mathcal{O}}(1)\). For any sheaf \(F\) over \(X\), there is a Hilbert polynomial \(P_F(n) = \chi(F(n)) \in {\mathbb{Z}}[n]\), i.e. we twist by \({\mathcal{O}}(1)\) \(n\) times. The cohomology of \(F\) isn’t changed by the pushforward into \({\mathbb{P}}_n\) since it’s a closed embedding, and so \begin{align*} \chi(X, F) = \chi({\mathbb{P}}^n, i_* F) = \sum (-1)^i \dim_k H^i({\mathbb{P}}^n, i_* F(n)) .\end{align*}

For \(n \gg 0\), \(\dim_k H^0 = \dim M_n\), the \(n\)th graded piece of \(M\), which is a graded module over the homogeneous coordinate ring whose \(i_*F = \tilde M\).

In general, for \(L\) ample of \(X\) and \(F\) coherent on \(X\), we can define a Hilbert polynomial, \begin{align*} P_F(n) = \chi(F\otimes L^n) .\end{align*}

This is an invariant of a polarized projective variety, and in particular subschemes. Over irreducible bases, flatness corresponds to this invariant being constant.

For \(f:X\to S\) projective, i.e. there is a factorization:

If \(S\) is reduced, irreducible, locally Noetherian, then \(f\) is flat \(\iff\) \(P_{{\mathcal{O}}_{x_s}}\) is constant for all \(s\in S\).

To be more precise, look at the base change to \(X_1\), and the pullback of the fiber? \({\mathcal{O}}\mathrel{\Big|}_{x_i}\)? Note that we’re not using the word “integral” here! \(S\) is flat \(\iff\) the Hilbert polynomial over the fibers are constant.

The zero-dimensional subschemes \(Z \in {\mathbb{P}}^n_k\), then \(P_Z\) is the length of \(Z\), i.e. \(\dim_k({\mathcal{O}}_Z)\), and \begin{align*} P_Z(n) = \chi({\mathcal{O}}_Z \otimes{\mathcal{O}}(n)) = \chi({\mathcal{O}}_Z) = \dim_k H^0(Z; {\mathcal{O}}_Z) = \dim_k {\mathcal{O}}_Z(Z) .\end{align*}

For two closed points in \({\mathbb{P}}^2\), \(P_Z = 2\). Consider the affine chart \({\mathbb{A}}^2 \subset {\mathbb{P}}^2\), which is given by \begin{align*} \operatorname{Spec}k[x, y]/(y, x^2) \cong k[x]/(x^2) \end{align*} and \(P_Z = 2\). I.e. in flat families, it has to record how the tangent directions come together.

Consider the flat family \(xy = 1\) (flat because it’s an open embedding) over \(k[x]\), here we have points running off to infinity.

A sheaf \(F\) is flat iff \(P_{F_S}\) is constant.

4.3 Proof That Flat Sheaves Have Constant Hilbert Polynomials

Assume \(S = \operatorname{Spec}A\) for \(A\) a local Noetherian domain.

For \(F\) a coherent sheaf on \(X_{/A}\) is flat, we can take the cohomology via global sections \(H^0(X; F(n))\). This is an \(A{\hbox{-}}\)module, and is a free \(A{\hbox{-}}\)module for \(n\gg 0\).

Assumed \(X\) was projective, so just take \(X = {\mathbb{P}}_A^n\) and let \(F\) be the pushforward. There is a correspondence sending \(F\) to its ring of homogeneous sections constructed by taking the sheaf associated to the graded module

\begin{align*} \sum_{n\gg0} H^0( {\mathbb{P}}_A^m; F(n) ) = \bigoplus_{n \gg 0} H^0({\mathbb{P}}_A^m; F(n)) \end{align*} and taking the associated sheaf (\(Y \mapsto \tilde Y\), as per Hartshorne’s notation) which is free, and thus \(F\) is free.2

Conversely, take an affine cover \(U_i\) of \(X\). We can compute the cohomology using Čech cohomology, i.e. taking the Čech resolution. We can also assume \(H^i({\mathbb{P}}^m; F(n)) = 0\) for \(n \gg 0\), and the Čech complex vanishes in high enough degree. But then there is an exact sequence \begin{align*} 0 \to H^0({\mathbb{P}}^m; F(n)) \to \mathcal C^0( \underline{U}; F(n) ) \to \cdots \to C^m( \underline{U}; F(n) ) \to 0 .\end{align*} Assuming \(F\) is flat, and using the fact that flatness is a 2 out of 3 property, the images of these maps are all flat by induction from the right. Finally, local Noetherian and finitely generated flat implies free.

By the lemma, we want to show \(H^0({\mathbb{P}}^m; F(n))\) is free for \(n\gg 0\) iff the Hilbert polynomials on the fibers \(P_{F_S}\) are all constant.

It suffices to show that for each point \(s\in \operatorname{Spec}A\), we have \begin{align*} H^0(X_s; F_S(n)) = H^0(X; F(n)) \otimes k(S) \end{align*} for \(k(S)\) the residue field, for \(n\gg 0\).

\(P_{F_S}\) measures the rank of the LHS.

\(\implies\): The dimension of RHS is constant, whereas the LHS equals \(P_{F_S}(n)\).

\(\impliedby\): If the dimension of the RHS is constant, so the LHS is free.

For a f.g. module over a local ring, testing if localization at closed point and generic point have the same rank. For \(M\) a finitely generated module over \(A\), we find that \begin{align*} 0 \to A^n \to M \to Q \end{align*} is surjective after tensoring with \(\mathrm{Frac}(A)\), and tensoring with \(k(S)\) for a closed point, if \(\dim A^n = \dim M\) then \(Q = 0\).

By localizing, we can assume \(s\) is a closed point. Since \(A\) is Noetherian, its ideal is f.g. and we have \begin{align*} A^m \to A \to k(S) \to 0 .\end{align*} We can tensor with \(F\) (viewed as restricting to fiber) to obtain \begin{align*} F(n)^m \to F(n) \to F_S(n) \to 0 .\end{align*} Because \(F\) is flat, this is still exact. We can take \(H^*(x, {\,\cdot\,})\), and for \(n\gg 0\) only \(H^0\) survives. This is the same as tensoring with \(H^0(x, F(n))\).

Given a polynomial \(P \in {\mathbb{Z}}[n]\) for \(X_{/S}\) projective, we define a subfunctor by picking only those with Hilbert polynomial \(p\) fiberwise as \(\operatorname{Hilb}^P_{X_{/S}} \subset \operatorname{Hilb}_{X_{/S}}\). This is given by \(Z \subset X \times_S T\) with \(P_{Z} = P\).

If \(S\) is Noetherian and \(X_{/S}\) projective, then \(\operatorname{Hilb}_{X_{/S}}^P\) is representable by a projective \(S{\hbox{-}}\)scheme.

See cycle spaces in analytic geometry.

5 Hilbert Polynomials (Thursday January 23)

Some facts about the Hilbert polynomial:

  1. For a subscheme \(Z \subset {\mathbb{P}}_k^n\) with \(\deg P_z = \dim Z = n\), then \begin{align*} p_z(t) = \deg z \frac{t^n}{n!} + O(t^{n-1}) .\end{align*}

  2. We have \(p_z(t) = \chi({\mathcal{O}}_z(t))\), consider the sequence \begin{align*} 0 \to I_z(t) \to {\mathcal{O}}_{{\mathbb{P}}^n}^{(t)} \to {\mathcal{O}}_z^{(t)} \to 0 ,\end{align*} then \(\chi(I_z(t)) = \dim H^0( {\mathbb{P}}^n, J_z(t) )\) for \(t \gg 0\), and \(p_z(0)\) is the Euler characteristic of \({\mathcal{O}}_Z\).

Keywords to look up here: Serre vanishing, Riemann-Roch, ideal sheaf.

Then \begin{align*} p_C(t) = (\deg C)t + \chi({\mathcal{O}}_{{\mathbb{P}}^1}) = 3t + 1 .\end{align*}

5.0.1 Hypersurfaces

Recall that length 2 subschemes of \({\mathbb{P}}^1\) are the same as specifying quadratics that cut them out, each such \(Z \subset {\mathbb{P}}^1\) satisfies \(Z = V(f)\) where \(\deg f = d\) and \(f\) is homogeneous. So we’ll be looking at \({\mathbb{P}}H^0({\mathbb{P}}^n_k, {\mathcal{O}}(d))^\vee\), and the guess would be that this is \(\operatorname{Hilb}_{{\mathbb{P}}^n_k}\) Resolve the structure sheaf

\begin{align*} 0 \to {\mathcal{O}}_{{\mathbb{P}}^n}(-d) \to {\mathcal{O}}_{{\mathbb{P}}^n}(t) \to {\mathcal{O}}_D(t) \to 0 .\end{align*} so we can twist to obtain \begin{align*} 0 \to {\mathcal{O}}_{{\mathbb{P}}^n}(t-d) \to {\mathcal{O}}_{{\mathbb{P}}^n}(t) \to {\mathcal{O}}_D(t) \to 0 .\end{align*} Then \begin{align*} \chi({\mathcal{O}}_D(t)) = \chi({\mathcal{O}}_{{\mathbb{P}}^n}(t)) - \chi({\mathcal{O}}_{{\mathbb{P}}^n}(t-d)) ,\end{align*} which is \begin{align*} {n+t \choose n} - {n+t-d \choose n} = \frac{dt^{n-1}}{(n-1)!} + O(t^{n-2}) .\end{align*}

Anything with the Hilbert polynomial of a degree \(d\) hypersurface is in fact a degree \(d\) hypersurface.

We want to write a morphism of functors \begin{align*} \operatorname{Hilb}_{{\mathbb{P}}^n_k}^{P_{n, d}} \to {\mathbb{P}}H^0 ({\mathbb{P}}^n, {\mathcal{O}}(d) )^\vee .\end{align*} which sends flat families to families of equations cutting them out. Want \begin{align*} Z \subset {\mathbb{P}}^n \times S \to {\mathcal{O}}_s \otimes H^0( {\mathbb{P}}^n, {\mathcal{O}}(d) )^\vee\to L \to 0 .\end{align*} This happens iff \begin{align*} 0 \to L^\vee\to {\mathcal{O}}_s \otimes H^0({\mathbb{P}}^n, {\mathcal{O}}(d)) \end{align*} with torsion-free quotient. Note that we use \(L^\vee\) instead of \({\mathcal{O}}_s\) because of scaling. We have \begin{align*} 0 \to I_z &\to {\mathcal{O}}_{{\mathbb{P}}^n \times S} \to {\mathcal{O}}_z \to 0 \\ 0 \to I_z(d) &\to {\mathcal{O}}_{{\mathbb{P}}^n \times S}(d) \to {\mathcal{O}}_z(d) \to 0 \quad\text{by twisting} .\end{align*} We then consider \(\pi_s: {\mathbb{P}}^n \times S \to S\), and apply the pushforward to the above sequence. Notie that it is not right-exact:

This equality follows from flatness, cohomology, and base change. In particular, we need the following:

The scheme-theoretic fibers, given by \(H^0({\mathbb{P}}^n, I_z(d))\) and \(H^0({\mathbb{P}}^n, {\mathcal{O}}_z(d))\), are all the same dimension.


  1. Cohomology and base change, i.e. for \(X \xrightarrow{f} Y\) a map of Noetherian schemes (or just finite-type) and \(F\) a sheaf on \(X\) which is flat over \(Y\), there is a natural map (not usually an isomorphism) \begin{align*} R^i f_* f \otimes k(y) \to H^i(x_y, {\left.{{F}} \right|_{{x_y}} } ) ,\end{align*} but is an isomorphism if \(\dim H^i(x_y, {\left.{{F}} \right|_{{x_y}} } )\) is constant, in which case \(R^i f_* f\) is locally free.

  2. If \(Z \subset {\mathbb{P}}^n_k\) is a degree \(d\) hypersurface, then independently we know \begin{align*} \dim H^0({\mathbb{P}}^n, I_z(d)) = 1 \text{ and } \dim H^0({\mathbb{P}}^n, {\mathcal{O}}_z(d)) = {d+n \choose n} - 1 .\end{align*}

To get a map going backwards, we take the universal degree 2 polynomial and form \begin{align*} V(a_{00} x_0^2 + a_{11} x_1^2 + a_{12}x_2^2 + a_{01}x_0 x_1 + a_{02} x_0 x_2 + a_{12} x_1 x_2) \subset {\mathbb{P}}^2 \times{\mathbb{P}}^5 .\end{align*}

5.0.2 Example: Twisted Cubics

Consider a map \({\mathbb{P}}^1 \to {\mathbb{P}}^3\) obtained by taking a basis of a homogeneous cubic polynomial. The canonical example is \begin{align*} (x, y) \to (x^3, x^2y, xy^2, y^3) .\end{align*} Then \begin{align*} P_C(t) = 3t + 1 \end{align*} and \(\operatorname{Hilb}_{{\mathbb{P}}_k^3}^{3t+1}\) has a component with generic point a twisted cubic, and another component with points a curve disjoint union a point, and the overlap are nodal curves with a “fat” 3-dimensional point: