Math 764  Algebraic Geometry II  Homeworks
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Homeworks (Spring 2017)
Here are homework problems for Math 764 from Spring 2017 (by Dima Arinkin). I tried to convert the homeworks into the wiki format with pandoc. This does not always work as expected; in case of doubt, check the pdf files.
 Homework 1 (PDF), due February 3rd.
 Homework 2 (PDF), due February 10th.
 Homework 3 (PDF), due February 17th.
 Homework 4 (PDF), due February 24th.
 Homework 5 (PDF), due March 3rd.
Homework 1
Due Friday, February 3rd
In all these problems, we fix a topological space [math]X[/math]; all sheaves and presheaves are sheaves on [math]X[/math].
 Example: Let [math]X[/math] be the unit circle, and let [math]{\mathcal{F}}[/math] be the sheaf of [math]C^\infty[/math]functions on [math]X[/math]. Find the (sheaf) image and the kernel of the morphism [math]\frac{d}{dt}:{\mathcal{F}}\to{\mathcal{F}}.[/math] Here [math]t\in{\mathbb{R}}/2\pi{\mathbb{Z}}[/math] is the polar coordinate on the circle.
 Sheaf operations: Let [math]{\mathcal{F}}[/math] and [math]{\mathcal{G}}[/math] be sheaves of sets. Recall that a morphism [math]\phi:{\mathcal{F}}\to {\mathcal{G}}[/math] is a (categorical) monomorphism if and only if for any sheaf [math]{\mathcal{F}}'[/math] and any two morphisms [math]\psi_1,\psi_2:{\mathcal{F}}'\to {\mathcal{F}}[/math], the equality [math]\phi\circ\psi_1=\phi\circ\psi_2[/math] implies [math]\psi_1=\psi_2[/math]. Show that [math]\phi[/math] is a monomorphism if and only if it induces injective maps on all stalks.
 Let [math]{\mathcal{F}}[/math] and [math]{\mathcal{G}}[/math] be sheaves of sets. Recall that a morphism [math]\phi:{\mathcal{F}}\to{\mathcal{G}}[/math] is a (categorical) epimorphism if and only if for any sheaf [math]{\mathcal{G}}'[/math] and any two morphisms [math]\psi_1,\psi_2:{\mathcal{G}}\to{\mathcal{G}}'[/math], the equality [math]\psi_1\circ\phi=\psi_2\circ\phi[/math] implies [math]\psi_1=\psi_2[/math]. Show that [math]\phi[/math] is a epimorphism if and only if it induces surjective maps on all stalks.
 Show that any morphism of sheaves can be written as a composition of an epimorphism and a monomorphism. (You should know what order of composition I mean here.)
 Let [math]{\mathcal{F}}[/math] be a sheaf, and let [math]{\mathcal{G}}\subset{\mathcal{F}}[/math] be a subpresheaf of [math]{\mathcal{F}}[/math] (thus, for every open set [math]U\subset X[/math], [math]{\mathcal{G}}(U)[/math] is a subset of [math]{\mathcal{F}}(U)[/math] and the restriction maps for [math]{\mathcal{F}}[/math] and [math]{\mathcal{G}}[/math] agree). Show that the sheafification [math]\tilde{\mathcal{G}}[/math] of [math]{\mathcal{G}}[/math] is naturally identified with a subsheaf of [math]{\mathcal{F}}[/math].
 Let [math]{\mathcal{F}}_i[/math] be a family of sheaves of abelian groups on [math]X[/math] indexed by a set [math]I[/math] (not necessarily finite). Show that the direct sum and direct product of this family exists in the category of sheaves of abelian groups. (E.g., a direct sum would be a sheaf of abelian groups [math]{\mathcal{F}}[/math] together with a universal family of homomorphisms [math]{\mathcal{F}}_i\to {\mathcal{F}}[/math].) Do these operations agree with (a) taking stalks at a point [math]x\in X[/math] (b) taking sections over an open subset [math]U\subset X[/math]?

Locally constant sheaves:
Definition. A sheaf [math]{\mathcal{F}}[/math] is constant over an open set [math]U\subset X[/math] if there is a subset [math]S\subset F(U)[/math] such that the map [math]{\mathcal{F}}(U)\to{\mathcal{F}}_x:s\mapsto s_x[/math] (the germ of [math]s[/math] at [math]x[/math]) gives a bijection between [math]S[/math] and [math]{\mathcal{F}}_x[/math] for all [math]x\in U[/math].
[math]{\mathcal{F}}[/math] is locally constant (on [math]X[/math]) if every point of [math]X[/math] has a neighborhood on which [math]{\mathcal{F}}[/math] is constant.
Recall that a covering space [math]\pi:Y\to X[/math] is a continuous map of topological spaces such that every [math]x\in X[/math] has a neighborhood [math]U\ni x[/math] whose preimage [math]\pi^{1}(U)\subset U[/math] is homeomorphic to [math]U\times Z[/math] for some discrete topological space [math]Z[/math]. ([math]Z[/math] may depend on [math]x[/math]; also, the homeomorphism is required to respect the projection to [math]U[/math].)
Show that if [math]\pi:Y\to X[/math] is a covering space, its sheaf of sections [math]{\mathcal{F}}[/math] is locally constant. Moreover, prove that this correspondence is an equivalence between the category of covering spaces and the category of locally constant sheaves. (If [math]X[/math] is pathwise connected, both categories are equivalent to the category of sets with an action of the fundamental group of [math]X[/math].)
 Sheafification: (This problem may be hard, but it is still a good idea to try it) Prove or disprove the following statement (contained in the lecture notes). Let [math]{\mathcal{F}}[/math] be a presheaf on [math]X[/math], and let [math]\tilde{\mathcal{F}}[/math] be its sheafification. Then every section [math]s\in\tilde{\mathcal{F}}(U)[/math] can be represented as (the equivalence class of) the following gluing data: an open cover [math]U=\bigcup U_i[/math] and a family of sections [math]s_i\in{\mathcal{F}}(U_i)[/math] such that [math]s_i_{U_i\cap U_j}=s_j_{U_i\cap U_j}[/math].
Homework 2
Due Friday, February 10th
Extension of a sheaf by zero. Let [math]X[/math] be a topological space, let [math]U\subset X[/math] be an open subset, and let [math]{\mathcal{F}}[/math] be a sheaf of abelian groups on [math]U[/math].
The extension by zero [math]j_{!}{\mathcal{F}}[/math] of [math]{\mathcal{F}}[/math] (here [math]j[/math] is the embedding [math]U\hookrightarrow X[/math]) is the sheaf on [math]X[/math] that can be defined as the sheafification of the presheaf [math]{\mathcal{G}}[/math] such that [math]{\mathcal{G}}(V)=\begin{cases}{\mathcal{F}}(V),&V\subset U\\0,&V\not\subset U.\end{cases}[/math]
 Is the sheafication necessary in this definition? (Or maybe [math]{\mathcal{G}}[/math] is a sheaf automatically?)
 Describe the stalks of [math]j_!{\mathcal{F}}[/math] over all points of [math]X[/math] and the espace étalé of [math]j_!{\mathcal{F}}[/math].
 Verify that [math]j_![/math] is the left adjoint of the restriction functor from [math]X[/math] to [math]U[/math]: that is, for any sheaf [math]{\mathcal{G}}[/math] on [math]X[/math], there exists a natural isomorphism [math]{\mathop{\mathrm{Hom}}}({\mathcal{F}},{\mathcal{G}}_U)\simeq{\mathop{\mathrm{Hom}}}(j_!{\mathcal{F}},{\mathcal{G}}).[/math]
(The restriction [math]{\mathcal{G}}_U[/math] of a sheaf [math]{\mathcal{G}}[/math] from [math]X[/math] to an open set [math]U[/math] is defined by [math]{\mathcal{G}}_U(V)={\mathcal{G}}(V)[/math] for [math]V\subset U[/math].)
Side question (not part of the homework): What changes if we consider the version of extension by zero for sheaves of sets (‘the extension by empty set’)?
Examples of affine schemes.
 Let [math]R_\alpha[/math] be a finite collection of rings. Put [math]R=\prod_\alpha R_\alpha[/math]. Describe the topological space [math]{\mathop{\mathrm{Spec}}}(R)[/math] in terms of [math]{\mathop{\mathrm{Spec}}}(R_\alpha)[/math]’s. What changes if the collection is infinite?
 Recall that the image of a regular map of varieties is constructible (Chevalley’s Theorem); that is, it is a union of locally closed sets. Give an example of a map of rings [math]R\to S[/math] such that the image of a map [math]{\mathop{\mathrm{Spec}}}(S)\to{\mathop{\mathrm{Spec}}}(R)[/math] is
(a) An infinite intersection of open sets, but not constructible.
(b) An infinite union of closed sets, but not constructible. (This part may be very hard.)
Contraction of a subvariety.
Let [math]X[/math] be a variety (over an algebraically closed field [math]k[/math]) and let [math]Y\subset X[/math] be a closed subvariety. Our goal is to construct a [math]{k}[/math]ringed space [math]Z=(Z,{\mathcal{O}}_Z)=X/Y[/math] that is in some sense the result of ‘gluing’ together the points of [math]Y[/math]. While [math]Z[/math] can be described by a universal property, we prefer an explicit construction:
 The topological space [math]Z[/math] is the ‘quotientspace’ [math]X/Y[/math]: as a set, [math]Z=(XY)\sqcup \{z\}[/math]; a subset [math]U\subset Z[/math] is open if and only if [math]\pi^{1}(U)\subset X[/math] is open. Here the natural projection [math]\pi:X\to Z[/math] is identity on [math]XY[/math] and sends all of [math]Y[/math] to the ‘center’ [math]z\in Z[/math].
 The structure sheaf [math]{\mathcal{O}}_Z[/math] is defined as follows: for any open subset [math]U\subset Z[/math], [math]{\mathcal{O}}_Z(U)[/math] is the algebra of functions [math]g:U\to{k}[/math] such that the composition [math]g\circ\pi[/math] is a regular function [math]\pi^{1}(U)\to{k}[/math] that is constant along [math]Y[/math]. (The last condition is imposed only if [math]z\in U[/math], in which case [math]Y\subset\pi^{1}(U)[/math].)
In each of the following examples, determine whether the quotient [math]X/Y[/math] is an algebraic variety; if it is, describe it explicitly.
 [math]X={\mathbb{P}}^2[/math], [math]Y={\mathbb{P}}^1[/math] (embedded as a line in [math]X[/math]).
 [math]X=\{(s_0,s_1;t_0:t_1)\in{\mathbb{A}}^2\times{\mathbb{P}}^1:s_0t_1=s_1t_0\}[/math], [math]Y=\{(s_0,s_1;t_0:t_1)\in X:s_0=s_1=0\}[/math].
 [math]X={\mathbb{A}}^2[/math], [math]Y[/math] is a twopoint set (if you want a more challenging version, let [math]Y\subset{\mathbb{A}}^2[/math] be any finite set).
Homework 3
Due Friday, February 17th
 (Gluing morphisms of sheaves) Let [math]F[/math] and [math]G[/math] be two sheaves on the same space [math]X[/math]. For any open set [math]U\subset X[/math], consider the restriction sheaves [math]F_U[/math] and [math]G_U[/math], and let [math]Hom(F_U,G_U)[/math] be the set of sheaf morphisms between them.
Prove that the presheaf on [math]X[/math] given by the correspondence [math]U\mapsto Hom(F_U,G_U)[/math] is in fact a sheaf.
 (Gluing morphisms of ringed spaces) Let [math]X[/math] and [math]Y[/math] be ringed spaces. Denote by [math]\underline{Mor}(X,Y)[/math] the following presheaf on [math]X[/math]: its sections over an open subset [math]U\subset X[/math] are morphisms of ringed spaces [math]U\to Y[/math] where [math]U[/math] is considered as a ringed space. (And the notion of restriction is the natural one.) Show that [math]\underline{Mor}(X,Y)[/math] is in fact a sheaf.
 (Affinization of a scheme) Let [math]X[/math] be an arbitrary scheme. Prove that there exists an affine scheme [math]X_{aff}[/math] and a morphism [math]X\to X_{aff}[/math] that is universal in the following sense: any map form [math]X[/math] to an affine scheme factors through it.
 Let us consider direct and inverse limits of affine schemes. For simplicity, we will work with limits indexed by positive integers.
(a) Let [math]R_i[/math] be a collection of rings ([math]i\gt 0[/math]) together with homomorphisms [math]R_i\to R_{i+1}[/math]. Consider the direct limit [math]R:=\lim\limits_{\longrightarrow} R_i[/math]. Show that in the category of schemes, [math]{\mathop{\mathrm{Spec}}}(R)=\lim\limits_{\longleftarrow}{\mathop{\mathrm{Spec}}}R_i.[/math]
(b) Let [math]R_i[/math] be a collection of rings ([math]i\gt 0[/math]) together with homomorphisms [math]R_{i+1}\to R_i[/math]. Consider the inverse limit [math]R:=\lim\limits_{\longleftarrow} R_i[/math]. Show that generally speaking, in the category of schemes, [math]{\mathop{\mathrm{Spec}}}(R)\neq\lim\limits_{\longrightarrow}{\mathop{\mathrm{Spec}}}R_i.[/math]
 Here is an example of the situation from 4(b). Let [math]k[/math] be a field, and let [math]R_i=k[t]/(t^i)[/math], so that [math]\lim\limits_{\longleftarrow} R_i=k[[t]][/math]. Describe the direct limit [math]\lim\limits_{\longrightarrow}{\mathop{\mathrm{Spec}}}R_i[/math] in the category of ringed spaces. Is the direct limit a scheme?
 Let [math]S[/math] be a finite partially ordered set. Consider the following topology on [math]S[/math]: a subset [math]U\subset S[/math] is open if and only if whenever [math]x\in U[/math] and [math]y\gt x[/math], it must be that [math]y\in U[/math].
Construct a ring [math]R[/math] such that [math]\mathop{\mathrm{Spec}}(R)[/math] is homeomorphic to [math]S[/math].
 Show that any quasicompact scheme has closed points. (It is not true that any scheme has closed points!)
 Give an example of a scheme that has no open connected subsets. In particular, such a scheme is not locally connected. Of course, my convention here is that the empty set is not connected...
Homework 4
Due Friday, February 24th
 Show that the following two definitions of quasiseparatedness of a scheme [math]S[/math] are equivalent:
 The intersection of any two quasicompact open subsets of [math]S[/math] is quasicompact;
 There is a cover of [math]S[/math] by affine open subsets whose (pairwise) intersections are quasicompact.
 In class, we gave the following definition: a scheme [math]S[/math] is integral if it is irreducible and reduced. Show that this is equivalent to the definition from Vakil’s notes: a scheme is integral if for any nonempty open [math]U\subset S[/math], [math]O_S(U)[/math] is a domain.
 Let us call a scheme [math]X[/math] locally irreducible if every point has an irreducible neighborhood. (Since a nonempty open subset of an irreducible space is irreducible, this implies that all smaller neighborhoods of this point are irreducible as well.) Prove or disprove the following claim: a scheme is irreducible if and only if it is connected and locally irreducible.
 Show that a locally Noetherian scheme is quasiseparated.
 Show that the following two definitions of a Noetherian scheme [math]X[/math] are equivalent:
 [math]X[/math] is a finite union of open affine sets, each of which is the spectrum of a Noetherian ring;
 [math]X[/math] is quasicompact and locally Noetherian.
 Show that any Noetherian scheme [math]X[/math] is a disjoint union of finitely many connected open subsets (the connected components of [math]X[/math].) (A problem from the last homework shows that things might go wrong if we do not assume that [math]X[/math] is Noetherian.)
 A locally closed subscheme [math]X\subset Y[/math] is defined as a closed subscheme of an open subscheme of [math]Y[/math]. Accordingly, a locally closed embedding is a composition of a closed embedding followed by an open embedding (in this order). In principle, one can try to reverse the order, and consider open subschemes of closed subschemes of [math]Y[/math]. Does this yield an equivalent definition?
Remark. The difficulty of such questions (and, sometimes, the answer to them) depends on the class of schemes one works with: often, very mild assumptions (such as, say, quasicompactness) would make the question easy. A complete answer to this problem would include both the mild assumptions that would make the two versions equivalent, and a description of what happens for general schemes.
Homework 5
Due Friday, March 3rd
 Fix a field [math]k[/math], and put [math]X={\mathop{Spec}}k[x][/math] and [math]Y={\mathop{Spec}}k[y][/math]. Consider the morphism [math]f:X\to Y[/math] given by [math]y=x^2[/math]. Describe the fiber product [math]X\times_YX[/math] as explicitly as possible. (The answer may depend on [math]k[/math].)
 (The Frobenius morphism.) Let [math]X[/math] be a scheme of characteristic [math]p[/math]: by definition, this means that [math]p=0[/math] in the structure sheaf of [math]X[/math]. Define the (absolute) Frobenius morphism [math]Fr_X:X\to X[/math] as follows: it is the identity map on the underlying set, and the pullback [math]Fr_X^*(f)[/math] equals [math]f^p[/math] for any (local) function [math]f\in{\mathcal{O}}_X[/math].
Verify that this defines an affine morphism of schemes. Assuming [math]X[/math] is a scheme locally of finite type over a perfect field, verify that [math]Fr_X[/math] is a morphism of finite type (it is in fact finite, if you know what it means).
 (The relative Frobenius morphism.) Let [math]X\to Y[/math] be a morphism of schemes of characteristic [math]p[/math]. Put [math]\overline X:=X\times_{Y,Fr_Y}Y,[/math] where the notation means that [math]Y[/math] is considered as a [math]Y[/math]scheme via the Frobenius map.
 Show that the Frobenius morphism [math]Fr_X[/math] naturally factors as the composition [math]X\to\overline{X}\to X[/math], where the first map [math]X\to\overline{X}[/math] is naturally a morphism of schemes over [math]Y[/math] (while the second map, generally speaking, is not). The map [math]X\to\overline{X}[/math] is called the relative Frobenius morphism.
 Suppose [math]Y={\mathop{Spec}}(\overline{\mathbb{F}}_p)[/math], and [math]X[/math] is an affine variety (that is, an affine reduced scheme of finite type) over [math]\overline{\mathbb{F}}_p[/math]. Describe [math]\overline X[/math] and the relative Frobenius [math]X\to\overline{X}[/math] explicitly in coordinates.
 Let [math]X[/math] be a scheme over [math]\mathbb{F}_p[/math]. In this case, the absolute Frobenius [math]Fr_X:X\to X[/math] is a morphism of schemes over [math]\mathbb{F}_p[/math] (and it coincides with the relative Frobenius of [math]X[/math] over [math]\mathbb{F}_p[/math].
Consider the extension of scalars [math]X'=X_{\overline{\mathbb{F}}_p}=X\otimes_{\mathbb{F}_p}\overline{\mathbb{F}}_p=X\times_{{\mathop{Spec}}(\mathbb{F}_p)}{\mathop{Spec}}(\overline{\mathbb{F}}_p).[/math] Then [math]Fr_X[/math] naturally extends to a morphism of [math]\overline{\mathbb{F}}_p[/math]schemes [math]X'\to X'[/math]. Compare the map [math]X'\to X'[/math] with the relative Frobenius of [math]X'[/math] over [math]\overline{\mathbb{F}}_p[/math].
 A morphism of schemes is surjective if it is surjective as a morphism of sets. Show that surjectivity is preserved by base changes. That is, if [math]f:X\to Z[/math] is surjective and [math]g:Y\to Z[/math] is arbitrary, then [math]X\times_ZY\to Y[/math] is surjective.
 (Normalization) A scheme is normal if all of its local rings are integrally closed domains. Let [math]X[/math] be an integral scheme. Show that there exists a normal integral scheme [math]\tilde{X}[/math] together with a morphism [math]\tilde{X}\to X[/math] that is universal in the following sense: any dominant morphism [math]Y\to X[/math] from a normal integral scheme to [math]X[/math] factors through [math]\tilde{X}[/math]. (Just like in the case of varieties, a morphism is dominant if its image is dense.)
 Let [math]X[/math] be a scheme of finite type over a field [math]k[/math]. For every field extension [math]K\supset k[/math], put [math]X_K:=X\otimes_kK={\mathop{Spec}}(K)\times_{{\mathop{Spec}}(k)}X.[/math] </p>
Show that [math]X[/math] is geometrically irreducible (that is, the morphism [math]X\to{\mathop{Spec}}(k)[/math] has geometrically irreducible fibers) if and only if [math]X_K[/math] is irreducible for all finite extensions [math]K\supset k[/math].