☆ A number of solutions are adaped from Howard Nuer's own solutions to his Technion 2019 Algebraic Geometry course. I will later go through and accredit the exercises where his approach is used.
Section 6.1: Introduction
There are no exercises in this section, just commentary.
Section 6.2: Morphisms of ringed spaces
Exercise 6.2.A:
Suppose \((X, \OO_X)\) and \((Y,\OO_Y)\) are ringed spaces, \(X=\bigcup_i U_i \) is an open cover of \(X\), and we have morphisms of ringed spaces \(\pi_i : U_i \to Y\) that "agree on the overlaps", i.e., \( \pi_i\vert_{U_i\cap U_j} = \pi_j\vert{U_i \cap U_j} \). Show that there is a unique morphism of ringed spaces \(\pi : X \to Y\) such that \(π\vert_{U_i} = \pi_i\). (Exercise 2.2.F essentially showed this for topological spaces.)
Proof:
For each \( i \) we get maps \( \OO_Y \to (\pi_i)_\ast \OO_{U_i} \). Since \( (\pi_i)_\ast \OO_{U_i} \vert_{U_i \cap U_j} = (\pi_j)_\ast \OO_{U_j} \vert_{U_i \cap U_j} \) by assumption, the gluability axiom for sheaves implies that there exists a sheaf \( \widetilde{\pi}_\ast \OO_X \) such that \( (\widetilde{\pi}_\ast \OO_X)\vert_{U_i} = (\pi_i)_\ast \OO_{U_i} \). By defining \( \pi : X \to Y \) by \( \pi_i \) on each open set \( i\) and endowing it with the morphism \( \OO_Y \to \pi_\ast \OO_X \) in the natural way, the identity axiom for sheaves tells us that this morphism must be the unique morphism extending that \( \pi_i \).
$$\tag*{$\blacksquare$}$$
Exercise 6.2.B:
Given a morphism of ringed spaces \(\pi: X \to Y\), show that sheaf pushforward induces a functor \(\Mod{\OO_X} \to \Mod{\OO_Y}\).
Proof:
This exercise is almost identical to Exercise 2.3.B with a marginal amount of extra data to show. Letting \( \FF \) be an \( \OO_X \)-module and \( V \subset Y \), the construction of the pushforward tells us \( \pi_\ast \FF(V) := \FF(\pi^{-1}(V)) \); in order to make this into an \( \OO_Y \)-module, consider some \( f \in \OO_Y(V) \) (again noting our open set \(V\) is arbitrary). Since \( \pi : X \to Y \) is a morphism of ringed spaces, it comes with the additional data of a map \( \OO_Y \to \pi_\ast \OO_X \) (which by abuse of notation we also call \( \pi_\ast \)). Then \( \pi_\ast f = f \circ \pi \in \OO_X(\pi^{-1}(V)) = \pi_\ast \OO_X(V) \), so we define our \( \OO_Y \)-action in the natural way as
$$
f \cdot \pi_\ast \FF (V) := \pi_\ast f \cdot \FF(\pi^{-1} (V))
$$
It easily follows that this action satisfies the axioms of an \( \OO_Y \) module since the induced map \( \OO_Y \to \pi_\ast \OO_X \) is a morphism of sheaves of rings, so that it preserves the identity, composition, and commutes with restrictions.
With this construction, showing that \( \pi_\ast \) maps \( \OO_X \)-linear morphisms to \( \OO_Y \)-linear morphisms is fairly straightforward. Indeed if \( \FF, \GG \) are \( \OO_X \)-modules, \( \phi : \FF \to \GG \) is \( \OO_X \)-linear, and \( V \subset Y \), the only sensible way to define \( \pi_\ast \phi(V) : \pi_\ast \FF(V) \to \pi_ast \GG(V) \) is via \( \pi_\ast \phi(V) := \phi (\pi^{-1} (V)) \). To see that this is now \( \OO_Y \)-linear (in the sense of our \(\OO_Y \)-action defined above), notice that for any \( s \in \OO_Y(V) \) and \( f \in \pi_\ast \FF(V) \), we have
$$
s \cdot \pi_\ast \phi(V) f = (s \circ \pi) \cdot \phi(\pi^{-1} V)f = \phi(\pi^{-1}(V)) \Big( (s \circ \pi) f \Big) = \pi_\ast \phi \big( \pi_\ast s \cdot f \big)
$$
$$\tag*{$\blacksquare$}$$
Exercise 6.2.C:
Given a morphism of ringed spaces \(\pi: X \to Y\) with \(\pi(p) = q\), show that there is a map of stalks \((\OO_Y)_q \to (\OO_X)_p \).
Proof:
This follows immediately from Exercise 2.3.A by considering the map \( \OO_Y \to \pi_\ast \OO_X \) (as part of the data of \(\pi \)) — it is worth noting that the proof only relies on the universal properties of the colimit.
$$\tag*{$\blacksquare$}$$
Exercise 6.2.D:
Suppose \(\pi^\sharp : B \to A\) is a morphism of rings. Define a morphism of ringed spaces \(\pi : \spec A \to \spec B\) as follows. The map of topological spaces was given in Exercise 3.4.H. To describe a morphism of sheaves \(\OO_{\spec B} \to \pi_\ast \OO_{\spec A}\) on \(\spec B\), it suffices to describe a morphism of sheaves on the distinguished base of \(\spec B\). On \(D(g) \subset \spec B\), we define
$$
\OO_{\spec B}(D(g)) \to \OO_{\spec A}( \pi^{-1} D(g) ) = \OO_{\spec A} (D(\pi^\sharp g) )
$$
by \(B_g \to A_{\pi^\sharp g}\). Verify that this makes sense (e.g., is independent of \(g\)), and that this describes a morphism of sheaves on the distinguished base. (This is the third in a series of exercises. We saw that a morphism of rings induces a map of sets in §3.2.9, a map of topological spaces in Exercise 3.4.H, and now a map of ringed spaces here.)
Proof:
To see that our map above is indepenent of \( g \), recall from Exercise 3.5.E that if \( D(g_1) = D(g_2) \) then \( g_2 \) is an invertible element of \( B_{g_1} \) and \( g_1 \) is an invertible element of \( B_{g_2} \). Thus, we get a map \( B_{g_1} \to B_{g_2} \) given by \( \frac{b}{g_1^r} \mapsto \frac{ g_1^{-r} b }{ 1 } \) and vice-versa. Since \( \pi^\sharp \) is a morphism of rings, it maps units to units so that our morphism \( B_g \to A_{\pi^\sharp g} \) does not depend on \(g\).
Both \( \OO_{\spec B} \) and \( \OO_{\spec A} \) were shown to be sheaves on a base in §4.1; thus it remains to show that \( B_g \to A_{\pi^\sharp g} \) commutes with restrictions. First notice that by Exercise 3.5.E \( D(g) \subset D(f) \) iff \( g^n = hf \) for some \( n \geq 0 \) and \( h \in B \) iff \( \pi^\sharp(g)^n = \pi^\sharp(h) \pi^\sharp(f) \) iff \( D(\pi^{\sharp} g ) \subset D(\pi^\sharp f) \). Using our choice of \(h\) above, our restriction map \( B_f \to B_g \) is simply given by
$$
\frac{b}{f^r} = \frac{ b \cdot h^r }{ f^r h^r } \longmapsto \frac{ b \cdot h^r }{ g^{nr} }
$$
But then our map \( B_g \to A_{\pi^\sharp g} \) must clearly commute with the restriction maps since
Recall (Exercise 3.4.K) that \(\spec k[y]_{(y)} \) has two points, \([(0)]\) and \([(y)]\), where the second point is closed, and the first is not. Describe a map of ringed spaces \(\spec k(x) \to \spec k[y]_{(y)} \) sending the unique point of \(\spec k(x)\) to the closed point \([(y)]\), where the pullback map on global sections sends \(k\) to \(k\) by the identity, and sends \(y\) to \(x\). Show that this map of ringed spaces is not of the form described in Key Exercise 6.2.D.
Proof:
Since we want the pullback map on global sections to send \( k \) to \( k \) and \( y \) to \(x\), if we assume to the contrary that our morphism did come from a morphism of rings, then we could define \( \phi^\sharp : B \to A \) where \( A = k(x) \) and \( B = k[y]_{(y)} \) using the above description. But then the induced map \( \phi : \spec A \to \spec B \) would send the unique point \( (0) \) to \( (\phi^\sharp)^{-1}(0) \). Since \( x \neq 0 \) in \( k(x) \), we cannot have that the map from the previous exercise sends \( (0) \) to \( (y) \) (i.e. the preimage of \((x)\)).
$$\tag*{$\blacksquare$}$$
Section 6.3: From locally ringed spaces to morphisms of schemes
Exercise 6.3.A:
Show that morphisms of locally ringed spaces glue (cf. Exercise 6.2.A). (Hint: your solution to Exercise 6.2.A may work without change.)
Proof:
The previous proof suffices.
$$\tag*{$\blacksquare$}$$
Exercise 6.3.B:
Show that \(\spec A\) is a locally ringed space. (Hint: Exercise 4.3.F.)
Show that the morphism of ringed spaces \( \pi : \spec A \to \spec B \) defined by a ring morphism \(\pi^\sharp : B \to A\) (Exercise 6.2.D) is a morphism of locally ringed spaces.
By Exercise 4.3.F we know that the stalk of \( \OO_{\spec A, [\pp]} \) is \( A_\pp \) and the stalk of \( \OO_{\spec B, [\pi^\sharp \pp] } \) is \( B_{\pi^\sharp \pp} \). By Exercise 6.2.D, we have that the map of stalks \( \pi^\sharp_\pp : B_{\pi^\sharp \pp} \to A_\pp \) sends the unique maximal ideal \( \pi^\sharp \pp \) to \( \pp \) by construction.
$$\tag*{$\blacksquare$}$$
Exercise 6.3.C:
Show that a morphism of schemes \(\pi: X \to Y\) is a morphism of ringed spaces that looks locally like morphisms of affine schemes. Precisely, if \(\spec A\) is an affine open subset of \(X\) and \(\spec B\) is an affine open subset of \(Y\), and \(\pi(\spec A) \subset \spec B\), then the induced morphism of ringed spaces is a morphism of affine schemes. (In case it helps, note: if \(W \subset X\) and \(Y \subset Z\) are both open embeddings of ringed spaces, then any morphism of ringed spaces \(X \to Y\) induces a morphism of ringed spaces \(W \to Z\), by composition \(W \to X\to Y \to Z\).) Show that it suffices to check on a set (\(\spec A_i , \spec B_i \)) where the \(\spec A_i\) form an open cover of \(X\) and the \(\spec B_i\) form an open cover of \(Y\).
Proof:
Recall that a scheme is simply a ringed space that is locally affine. By definition, every scheme comes equipped with an affine cover, say \( \spec A_i \) for \(X\) and \( \spec B_i \) for \(Y\) — since the question is referring to the local nature of our morphisms, it clearly suffices to check on our open cover (addressing the latter part first.)
By considering the restriction of our morphism \( \pi \) to some \( \spec A \) such that \( \pi(\spec A) \subset \spec B \) (as in the problem statement), then as \( \pi \) is a morphism of locally ringed spaces (as a morphism of schemes) the restriction \( \pi \vert_{\spec A} \) must also be a morphism of locally ringed spaces. By Key Proposition 6.3.2, it follos that this map is induced by some map \( \pi^\sharp \vert_{\spec A} : B \to A \) as in Exercise 6.2.D. But this is precisely what is needed to show that \( \pi\vert_{\spec A}\) is morphism of affine schemes.
$$\tag*{$\blacksquare$}$$
Exercise 6.3.D:
Show that the category of rings and the opposite category of affine schemes are equivalent (see §1.2.21 to read about equivalence of categories).
Proof:
The majority of the information for this proof we have had at our disposal since Chapter 4. Let \( \FF : \mathrm{Aff} \to \mathrm{Ring} \) denote the global section functor \( ( \spec A, \OO_{\spec A} ) \to \Gamma( D(1), \OO_{\spec A} ) = A \) which contravariantly transforms morphisms of affine schemes \( \phi^{\sharp} : (\spec A, \OO_{\spec A}) \to (\spec B, \OO_{\spec B}) \) by globalizing the induced map \( \OO_{\spec B}(D(1)) \to \phi_\ast \OO_{\spec B}(D(1)) \) to give us the ring morphism \( B \to A \).
As before, we define the functor \( \GG : \mathrm{Ring} \to \mathrm{Aff} \) to be the map of objects \( A \mapsto \spec A \). For morphisms \( \phi : B \to A \), we are able to describe the map \( \phi : (\spec A, \OO_{\spec A}) \to (\spec B, \OO_{\spec B}) \) as a morphism of sets, topological spaces, and now a morphism of ringed spaces by the construction in Exercise 6.2.D. Indeed this construction will help us show that the composition is naturally isomorphic to the identity.
First we show that \( \FF \circ \GG \) is naturally isomorphic to \( Id_{\mathrm{Ring}} \). Let \( \phi^\sharp : B \to A \) be a morphism of rings. Applying the functor \( \GG \), we obtain the morphism \( \phi : (\spec A, \OO_{\spec A}) \to (\spec B, \OO_{\spec B}) \) as a morphism of locally ringed spaces as in Exercise 6.2.D. By definition, we have that the induced map of sheaves is \( B_g = \Gamma(D(g), \OO_{\spec B}) \to \Gamma(D(\phi^\sharp g ), \OO_{\spec A}) = A_{\phi^\sharp g} \). Thus, by applying \( \FF \), we obtain the morphism \( B_1 = \Gamma(D(1), \OO_{\spec B}) \to \Gamma(D(1), \OO_{\spec A}) = A_1 \). Since \( B_1 \) is clearly isomorphic to \(B\) and \( A_1 \) clearly isomorphic to \(A\), we get that \( \FF \circ \GG\) is naturally isomorphic to \( Id_{\mathrm{Ring}} \).
Now to see that \( \GG \circ \FF \) is naturally isomorphic to \( Id_{\mathrm{Aff}} \), we refer to Key proposition 6.3.2 as this shows that any morphism \( \phi : (\spec A, \OO_{\spec A}) \to (\spec B, \OO_{\spec B}) \) must be induced by a ring morphism \( \phi^\sharp : B \to A \), which is precisely what we obtain by applying \( \GG \circ \FF \).
Since a contravariant functor \( \mathcal{D} : \mathcal{A} \to \mathcal{B} \) can be described as a covariant functor \( \mathcal{D}^\prime : \mathcal{A}^{op} \to \mathcal{B} \), it follows that \( \mathrm{Ring} \) is equivalent to \( \mathrm{Aff}^{op} \).
$$\tag*{$\blacksquare$}$$
Exercise 6.3.E:
(This exercise can give you some practice with understanding morphisms of schemes by cutting up into affine open sets.) Make
sense of the following sentence: "\(\A^{n+1}_k \backslash \{ 0 \} \to \P^n_k\) given by
$$
(x_0, x_1, \dots, x_n) \to [x_0, x_1, \dots, x_n]
$$
is a morphism of schemes." Caution: you can’t just say where points go; you have to say where functions go. So you may have to divide these up into affines, and describe the maps, and check that they glue. (Can you generalize to the case where \(k\) is replaced by a general ring \(B\)? See Exercise 6.3.M for an answer.)
Proof:
Let \( \pi : \A^{n+1}_k \backslash \{ 0 \} \to \P^n_k \) denote our morphism. Taking the standard open cover \( U_i = \{ x_i \neq 0 \} \subset \P^n_k \). From our work in section §4.5 we know that \( U_i = \spec k[x_{0/i}, \dots, x_{n/i}] / (x_{i/i} - 1) \) and \( \pi^{-1}(U) = \spec k[x_0, \dots, x_n]_{x_i} \). By letting \( t_j = x_{j/i} \), our morphism of sheaves \( k[t_0, \dots, t_{i-1}, t_{i+1}, \dots, t_n] \to k[x_0, \dots, x_n]_{x_i} \) is simply given by \( t_j \mapsto x_j / x_i \) as expected. It was shown in Exercise 4.5.L that these maps indeed glue; since our morphism \( \pi \) is globally defined, this gives us a morphism of schemes.
$$\tag*{$\blacksquare$}$$
Exercise 6.3.F:
Show that morphisms \( X \to \spec A \) are in natural bijection with ring morphisms \( A \to \Gamma(X, \OO_X) \). Hint: Show that this is true when \(X\) is affine. Use the fact that morphisms glue, Exercise 6.3.A. (This is even true in the category of locally ringed spaces. You are free to prove it in this generality, but it is easier in the category of schemes.)
Proof:
The forward direction is obvious by considering the induced map of schemes \( \OO_{\spec A} \to \phi_\ast \OO_{\spec A} \) (as part of the data of a morphism \( \phi: X \to \spec A \)) and looking at the global sections \( A = \Gamma(\spec A, \OO_{\spec A} ) \to \Gamma(\phi^{-1} \spec A, \phi_\ast\OO_{\spec A}) = \Gamma(X, \OO_X) \).
For the reverse direction, let \( \{ U_i = \spec B_i \} \) be an affine cover of \(X\) and \( \phi : A \to \Gamma(X, \OO_X) \). For each \( i \) we get a restriction map \( \res{X}{U_i} : \Gamma(X, \OO_X) \to \Gamma(\spec B_i, \OO_X) = B_i \); taking compositions gives us a morphism \( A \to B_i \), which induces maps \( \spec B_i \to \spec A \). By Exercise 6.3.A these morphisms glue giving us the desired map \( X \to \spec A \).
$$\tag*{$\blacksquare$}$$
Exercise 6.3.G:
Show that this definition of \(A\)-scheme given in §6.3.7 agrees with the earlier definition of §5.3.6.
Proof:
Recall that in §5.3.6 an \(A\)-scheme was defined as a scheme in which the ring of sections over any open set is an \(A\)-algebra, and the restriction morphisms are maps of \(A\)-algebras. To see that this implies the definition given in §6.3.7, suppse \( X \) is an \(A\)-scheme in the former sense. Then on the open set \(X\), we must have that \( \Gamma(X, \OO_X) \) is an \(A\)-algebra so there is a ring morphism \( A \to \Gamma(X, \OO_X) \). By the previous exercise, this corresponds to a morphism of schemes \( X \to \spec A \) so that \( X \) is an \(A\)-scheme in the latter sense.
Conversely, if \( X \) is an \(A\)-algebra in the sense of §6.3.7 then there exists a morphism of schemes \( X \to \spec A \). Again using the previous exercise, this corresponds to a ring morphism \( A \to \Gamma(X, \OO_X) \). By composing with restrictions, this makes each ring of sections an \(A\)-algebra.
$$\tag*{$\blacksquare$}$$
Exercise 6.3.H:
If \(S_\bullet \) is a finitely generated graded \(A\)-algebra, describe a natural "structure morphism" \(\proj S_\bullet \to \spec A\).
Proof:
For a general finitely generated graded algebra \(S_\bullet\) there unfortunately isn't any nice way to describe \( \Gamma(\proj S_\bullet, \OO_{\proj S_\bullet}) \) globally without gluing. Proceeding in this way, for any \( f \in S_+ \) homoegeneous we know by the \( \proj \) construction that sets of the form \( \spec ((S_\bullet)_f)_0 \) form a base for the topology on \( \proj S_\bullet \). Now since \( S_\bullet \) is an \(A\)-algebra, we also know that \( ((S_\bullet)_f)_0 \) is an \(A\)-algebra, giving us a morphism \( A \to \Gamma(D(f), \OO_{\proj S_\bullet}) \). By Exercsie 4.5.K these \( \Gamma(D(f), \OO_{\proj S_\bullet}) \) glue together to give a well-defined ring homomorphism \( A \mapsto \Gamma(\proj S_\bullet, \OO_{\proj S_\bullet}) \). Therefore, Exercise 6.3.F tells us this corresponds to a morphism of schemes \( \proj S_\bullet \to \spec A \).
$$\tag*{$\blacksquare$}$$
Exercise 6.3.I:
Show that \(\spec \Z\) is the final object in the category of schemes. In other words, if \(X\) is any scheme, there exists a unique morphism to \(\spec \Z\). (Hence the category of schemes is isomorphic to the category of \(\Z\)-schemes.) If \(k\) is a field, show that \(\spec k\) is the final object in the category of \(k\)-schemes.
Proof:
Since \( \Z \) is initial in the category of rings and \( \Gamma(X, \OO_X) \) always carries a ring structure, there exists a unique morphism \( \Z \to \Gamma(X, \OO_X) \) for every scheme \(X\). By Exercise 6.3.F, this implies that there exists a unique morphism \( X \to \spec \Z \) for every scheme \(X\) such that \( \spec \Z \) is the initial object in \( \mathrm{Sch} \).
Similarly, let \( X \) be a \( k \)-scheme so that there is a canonical map \( X \to \spec k \). Since \( \spec k \) is a point, \( \spec k \) is clearly final in the category of sets since there is only one morphism \( X \to \spec k \) sending every point of \(X\) to the unique point \( [(0)] \). But then it must be final in the category \( \mathrm{Sch}_k \) as well by virtue of the forgetful functor.
$$\tag*{$\blacksquare$}$$
Exercise 6.3.J:
Suppose \(p\) is a point of a scheme \(X\). Describe a canonical (choice-free) morphism \(\spec \OO_{X,p} \to X\). (Hint: do this for affine \(X\) first. But then for general \(X\) be sure to show that your morphism is independent of choice.)
Define a canonical morphism \(\spec \kappa(p) \to X\). (This is often written \(p \to X\); one gives \(p\) the obvious interpretation as a scheme, \(\spec \kappa(p)\).)
Proof:
First let \( U = \spec A \) be an affine neighborhood of \( p \) so that there exists some prime ideal \( \pp \subset A \) with \( p = [\pp] \). Then \( \OO_{X, p} = A_\pp \) so that \( \spec \OO_{X, p} = \spec A_\pp \). Since we have a localization map \( A \to A_\pp \), there is a clear surjection \( \spec A_\pp \to \spec A \). Lastly, composing by inclusion gives us the desired map \( \spec \OO_{X,p} = \spec A_{\pp} \to U \hookrightarrow X \).
To show that the map is canonical, suppose \( V = \spec B \) is another affine neighborhood containing \( p \) so that there exists some prime ideal \( \qq \subset B \) with \( \OO_{X, p} = B_\qq = A_\pp \). Then on \( U \cap V \) there should exist some affine neighborhood \( W \subset U \cap V \) that is a distinguished open of the form \( D_A(f) \) in \(U\) and \( D_B(g) \) in \(V\) (i.e. \( f \in A \), \( g \in B \)). Since \( p = [\pp] \in D_A(f) \), \( f \notin \pp \) so \( f \) is invertible in \( A_\pp \) — by the universal property of localization we get a morphism \( A_f \to A_{\pp} \). Making a similar argument for \( B \) and noting that \( A_f \cong B_g \) and \( A_\pp = B_\qq \), the maps must clearly be the same. Applying our functor \( \spec \), we see that \( \spec \OO_{X, p} \to U\) and \( \spec \OO_{X, p} \to V \) must factor through \( W \subset U \cap V \) and thus our desired map is independent of choice of affine neighborhood.
Similar to above, if \( U = \spec A \) is an affine neighborhood of \( p \) then we can find some \( \pp \subset A \) such that \( p = [\pp] \). Thus, \( \kappa(p) = A_\pp / \pp A_\pp \) so that the surjection \( A_\pp \to A_\pp / \pp A_\pp \) yields a morphism of schemes \( \spec \kappa(p) \to \spec \OO_{X, p} \). Since the map \( \spec \OO_{X, p} \to X \) was indepenedent of affine chart, the composition \( \spec \kappa(p) \to X \) is as well.
$$\tag*{$\blacksquare$}$$
Exercise 6.3.K:
Suppose \(X\) is a scheme, and \((A, \mm)\) is a local ring. Suppose we have a scheme morphism \(\pi : \spec A \to X\) sending \([\mm]\) to \(\pp\). Show that any open set containing \(\pp\) contains the image of \(\pi\). Show that there is a bijection between \(\textrm{Mor}\,(\spec A, X)\) and
$$
\{ p \in X, \textrm{local}\ \textrm{homeomorphisms}\ \OO_{X, p} \to A \}.
$$
(Possible hint: Exercise 6.3.J(a))
Proof:
We wish to show (in a somewhat stronger sense) that any morphism from the spectrum of a local ring must factor through \( \spec \OO_{X, \pp} \). That is to say we obtain a diagram
where the vertical arrow \( \spec \OO_{X, p} \to X \) is the canonical morphism from the previous exercise.
Now suppose that \( U = \spec R \subset X \) is an affine open neighborhood of \(p\) such that \( p = [\pp] \). There are a few helpful facts about specialization and generization that were not proved earlier since we did not have the machinery of Key Proposition 6.3.2. Most notably, we wish to show that \( \pi \) preserves any generizations of our closed point \( \mm \). This follows from Key proposition 6.3.2 since if \( \qq \subset \mm \) is a generalization of \( \mm \), then we know our morphism of schemes (when restricted to affine opens) is induced by some morphism of rings \( \widetilde{\pi} : B \to A \). On the level of spectra, this means that \( \pp \subset \mm \) implies \( \widetilde{\pi}^{-1} (\pp) \subset \widetilde{\pi} (\mm)\) so that generizations map to generizations. Lastly, any open affine containing our closed point \( p \in X \) must contain any generization of it. Since \(( A, \mm )\) is a local ring, every prime ideal \( \qq \in A \) is contained in \( \mm \) (by Zorn's lemma) so that every point of \( \spec A \) is a generization of \( [\mm] \in \spec A \). Therefore, any affine open set containing \( p \in X \) must contain the entire image of \( \pi \).
Applying Key Proposition 6.3.2 a second time, we know that our morphism of schemes \( \pi : \spec A \to \spec R \) is induced by a ring morphism \( \widetilde{\pi} : R \to A \). If we take \( \pp \) to be the preimage of the maximal ideal \( \mm \), then any \( f \notin \pp \) is sent to a unit in \( A \) (since \( (\widetilde{\pi}(f)) = A \) ) so that \( \widetilde{\pi} \) factors through \( A_\pp \). But \( A_\pp = \spec \OO_{X,p} \) so we're done.
$$\tag*{$\blacksquare$}$$
Exercise 6.3.L:
(easy) Show that a morphism of schemes \(X \to Y\) induces a map of \(Z\)-valued points \(X(Z) \to Y(Z)\).
Note that morphisms of schemes \(X \to Y\) are not determined by their "underlying" map of points. (What is an example?) Show that they are determined by their induced maps of \(Z\)-valued points, as \(Z\) varies over all schemes. (Hint: pick \(Z = X\). In the course of doing this exercise, you will largely prove Yoneda’s Lemma in the guise of Exercise 9.1.C.)
Proof:
Given any \( \{Z \to X\} \in X(Z) \), \( \{ Z \to X \to Y \} \in Y(Z) \). This is just a special case of talking about the covariant functor \( \textrm{Hom}_\mathcal{C}(Z, -) \) (see §1.2.14).
Clearly morphisms of schemes need not be determined by their maps of points, even on the level of affine schemes. For example, there is only one map \( \spec k \to \spec k \) in terms of points, though the map \( k \to k \) can have several choices via \( x \mapsto a\cdot x \).
The second part is indeed a repitition of Exercise 1.3.Y(b) (which is the contravariant version of Yoneda's lemma). Indeed by part (a) above we get that for any \( Z \in \mathrm{Sch} \) there is a map of sets \( i_Z : \textrm{Hom}_{\mathrm{Sch}}(Z, X) \to \textrm{Hom}_{\mathrm{Sch}}(Z, Y) \). By taking \( Z = X \) and considering the image of \( \textrm{Id}_X \) we obtain a unique map \( g : X \to Y \) such that for all \( Z \in \mathrm{Sch} \), \( i_Z \) is \( u \mapsto g \circ u \).
$$\tag*{$\blacksquare$}$$
Exercise 6.3.M<:
Suppose \(B\) is a ring. If \(X\) is a \(B\)-scheme, and \(f_0, \dots, f_n\) are \(n + 1\) functions on \(X\) with no common zeros, then show that \([f_0, \dots , f_n]\) gives a morphism of \(B\)-schemes \(X \to \P^n_B\).
Suppose \(g\) is a nowhere vanishing function on \(X\), and \(f_i\) are as in part (a). Show that the morphisms \([f_0,\dots,f_n]\) and \([gf_0,\dots,gf_n]\) to \(\P^n_B\) are the same.
Proof:
By assumption, for every point \( p \in X \) there must exist some \( f_j \) with \( f_j(p) \neq 0 \) — that is to say \( X = \bigcup_{i=0}^{n} X_{f_i} \). Let \( U_i = \spec (( B[x_0, \dots, x_n] )_{x_i})_0 \) denote the usual affine open cover of \( \P^n_B \). If \( V = \spec A \subset X\) is an affine neighborhood, then \( V \cap X_{f_i} = V_{f_i} = D(f_i) \) so that we get an obvious morphism of \( B \)-algebras \( B[x_0, \dots, x_n] \to A_{f_i}\) given by \( x_{j/i} \mapsto f_j / f_i \). Since this map is independent of choice of affine \( V \subset X \) and glues in the same way as \( \P^n_B \) we obtain a morphism \( X \to \P^n_B \).
On each affine distinguished open our map of \( B \)-algebras \( B[x_0, \dots, x_n] \to A_{gf_i} \) is given by \( x_{j/i} \mapsto gf_j / gf_i = f_j / f_i \) which is the same as in part (a).
$$\tag*{$\blacksquare$}$$
Exercise 6.3.N:
Recall the analytification construction of Exercise 5.3.G. For each morphism of reduced finite type \(\C\)-schemes \(\pi : X \to Y\) (over \(C\)), define a morphism of complex analytic prevarieties \(\pi_{an} : X_{an} \to Y_{an}\) (the analytification of \(\pi\)). Show that analytification gives a functor from the category of reduced finite type \(\C\)-schemes to the category of complex analytic prevarieties. (Remark: Two nonisomorphic varieties can have isomorphic analytifications. For example, Serre described two different algebraic structures on the complex manifold \(\C_\ast \times \C_\ast\), see [Ha2, p. 232] and [MO68421]; one is "the obvious one", and the other is a \(\P_1\)-bundle over an elliptic curve, with a section removed. For an example of a smooth complex surface with infinitely many algebraic structures, see §19.11.3. On the other hand, a compact complex variety can have only one algebraic structure (see [Se3, §19]).)
Section 6.4: Maps of graded rings and maps of projective schemes
Exercise 6.4.A:
Suppose that \(\phi: S_\bullet \to R_\bullet\) is a morphism of \((\Z_{\geq 0})\)-graded rings. (By map of graded rings, we mean a map of rings that preserves the grading as a map of "graded semigroups" (or "graded monoids"). In other words, there is a \(d > 0\) such that \(S_n\) maps to \(R_{dn}\) for all \(n\).) Show that this induces a morphism of schemes \( (\proj R_\bullet ) \backslash V (\phi(S_+ )) \to \proj S_\bullet \) . (Hint: Suppose \(f\) is a homogeneous element of \(S_+\). Define a map \(D(\phi(f)) \to D(f)\). Show that they glue together (as \(f\) runs over all homogeneous elements of \(S_+\)). Show that this defines a map from all of \(\proj R_\bullet \backslash V(\phi(S_+)) \) .) In particular, if
$$
V(\phi(S_+)) = \emptyset
$$
then we have a morphism \(\proj R_bullet \to \proj S_\bullet\). From your solution, it will be clear that if \(\phi\) is furthermore a morphism of \(A\)-algebras, then the induced morphism \(\proj R_\bullet \backslash V (\phi(S_\bullet )) \to \proj S_\bullet \) is a morphism of \(A\)-schemes.
Proof:
We first check this on an affine cover of our schemes \( \proj S_\bullet \) and \( \proj R_\bullet \). From §4.5 we know the affine cover of \( \proj S_\bullet \) is sets of the form \( \spec ((S_\bullet)_f)_0 \) as \( f \) runs over \( f \in S_\plus \). If we fix \( f \in S_+ \) with \( \deg f = n \), then by assumption we have that \( \deg \phi(f) = dn \) so that \( \phi(f) \in R_+\). By Key Proposition 6.3.2, our morphism \( D(\phi(f)) = \spec ((R_\bullet)_{\phi(f)})_0 \to \spec ((S_\bullet)_f)_0 = D(f) \) should be induced by a ring morphism \( \psi_f : ((S_\bullet)_f)_0 \to ((R_\bullet)_{\phi(f)})_0 \). The obvious choice is
This is clearly well defined since if \( \frac{g}{f^k} = \frac{g^\prime}{f^l} \), then as \( \deg f > 0 \) it must be the case that \( g - g^\prime = 0 \). Notice however that \( \deg g = dk \) and \( \deg g = dm \), so this is only possible when \( m = l \). But then \( \phi(g) - \phi(g^\prime) = 0 \) and \( \psi( \frac{g}{f^k} ) = \psi( \frac{g^\prime}{f^l} ) \). Thus, for each \( f \in S_+ \) we obtain a morphism \( \spec ((R_\bullet)_{\phi(f)})_0 \to \spec ((S_\bullet)_f)_0\).
By Exercise 3.5.D we know that if \( f, h \in S_+ \) then \( D(f) \cap D(h) = D(fh) \). Thus, \( \phi^\sharp ( D(f) \cap D(h) ) = \phi^\sharp(D(fh)) = D( \phi(fh) ) = D(\phi(f)) \cap D(\phi(h)) \). Since our morphisms \( \psi_f \) and \( \psi_h\) clearly restrict to \( \psi_{fh} \) by construction, this clearly glues to a morphism of schemes. Now since \( \phi \) maps \( S_n \) to \( R_{dn} \) for \( d>0 \) and we locally define our maps \( \psi_f \) for \( f \in S_+ \), we obtain a morphism of schemes \( \phi^\sharp : \proj R_\bullet \backslash V( \phi(S_+) ) \to \proj S_\bullet \).
$$\tag*{$\blacksquare$}$$
Exercise 6.4.B:
Show that if \(\phi: S_\bullet \to R_\bullet\) satisfies \( \sqrt{(\phi(S_+))} = R_+\), then hypothesis (6.4.0.1) is satisfied. (Hint: Exercise 4.5.I.) This algebraic formulation of the more geometric hypothesis can sometimes be easier to verify.
Proof:
By Exercise 4.5.I we know that \( V(I) = \emptyset \) is equivalent to \( \sqrt{I} \supset R_+ \) so the claim follows.
$$\tag*{$\blacksquare$}$$
Exercise 6.4.C:
This exercise shows that different maps of graded rings can give the same map of schemes. Let \(R_\bullet = k[x, y, z]/(xz, yz, z^2 )\) and \(S_\bullet = k[a, b, c]/(ac, bc, c^2)\), where every variable has degree 1. Show that \(\proj R_\bullet \cong \proj S_\bullet \cong \P^1_k\). Show that the maps \(S_\bullet \to R_\bullet\) given by \((a,b,c) \mapsto (x,y,z)\) and \((a, b, c) \mapsto (x, y, 0) \) give the same (iso)morphism \(\proj R_\bullet \to \proj S_\bullet\) . (The real reason is that all of these constructions are insensitive to what happens in a finite number of degrees. This will be made precise in a number of ways later, most immediately in Exercise 6.4.F.)
Proof:
Similar to the approach of hilbert polynomials, we look at the generators of the graded ring \( R_\bullet \) (and thus \(S_\bullet\)). In degree 1 we have all the linear terms \( \{x, y, z\} \); however, in degree 2 we notice that the only generators are \( \{ a^2, b^2, ab \} \). In degree 3, this pattern continues as \( \{ a^3, b^3, a^2b, ab^2 \} \).
Following our construction above, let \( \phi_1\) denote the first morphism \( S_\bullet \to R_\bullet \) and \( \phi_2 \) the second. Then under the first morphism, one of our local maps \( \psi_a : D(a) \to D(x) \) sends \( \frac{g}{a^r} \mapsto \frac{\phi_1(g)}{x^r} \). When \( r > 1 \), \( g \) must be a polynomial strictly in \(a, b\) so that we may consider the resulting fraction as a polynomial in \( \frac{b}{a} \) (by homogeneity). Now in degree 1, \( \phi_1(g) \) may be any linear term \( a, b, c \). However, \( ( \frac{c}{a} ) \) cant be prime since \( c^2 = 0 \in (a) \) so that the map on the spectra \( \spec ((R_\bullet)_x)_0 \to \spec ((S_\bullet)_a)_0 \) doesnt see this nilpotent. On the structure sheaf, we notice that \( \Gamma(D(c), \OO_{\proj S_\bullet}) = ( \C[a, b, c] / (ac, bc, c^2) )_c \) --- since \( c \) is a nilpotent this is the zero ring. Thus, it doesnt matter if \( \phi \) maps \( c \) to \(z\) (another nilpotent) or \( 0 \) as our morphism of sheaves should map the zero ring to the zero ring on the affine set \( D(c) \).
$$\tag*{$\blacksquare$}$$
Exercise 6.4.D:
Show that the map of graded rings \(S_{n\bullet} \hookrightarrow S_\bullet\) induces an isomorphism \(\proj S_\bullet \to \proj S_{n\bullet}\) . (Hint: if \(f \in S_+\) is homogeneous of degree divisible by \(n\), identify \(D(f)\) on \(\proj S_\bullet\) with \(D(f)\) on \(\proj S_{n\bullet}\). Why do such distinguished open sets cover \(\proj S_\bullet\)?)
Proof:
Answering the last part first, notice that by Exercise 3.5.D \( D(f^n) = D(f) \cap D(f^{n-1}) = D(f) \cap D(f) \cap \dots \) so that \( D(f^n) = D(f) \). Thus we may cover \( \proj S_\bullet \) by homogeneous functions with degree divisible by \( n \) for the same reason that we may cover \( \proj S_\bullet \) by homogeneous functions of positive degree.
Following the hint, let \( f \in S_+ \) be homogeneous of degree \( \deg f = kn \) and consider \( ((S_\bullet)_f)_0 = \{ \frac{g}{f^r} \mid r \geq 1, g \in S_{rkn} \} \). Since \( f \in (S_{n})_k\) and \( (S_n)_{rk} := S_{rkn} \), the "identity" morphism of rings \( ((S_\bullet)_f)_0 \to ((S_{n\bullet})_f)_0 \) given by \( \frac{g}{f^r} \mapsto \frac{g}{f^r} \) is inverse to the map \(S_{n\bullet} \hookrightarrow S_\bullet \) from the problem statement. By Exercise 4.3.A, we get an isomorphism \( D_{S_{n\bullet}}(f) = \spec ((S_{n\bullet})_f)_0 \cong \spec ((S_\bullet)_f)_0 \). Since these distinguished opens cover \( \proj S_\bullet \) by the previous paragraph, we obtain an isomorphism of schemes.
$$\tag*{$\blacksquare$}$$
Exercise 6.4.E:
If \(S_\bullet\) is generated in degree \(1\), show that \(S_{n\bullet}\) is also generated in degree 1. (You may want to consider the case of the polynomial ring first.)
Proof:
Since \( S_\bullet \) generated in degree 1, there is a surjective graded morphism of \( S_0 \)-algebras \( S_0[X_1, \dots, X_n] \twoheadrightarrow S_\bullet \). Since the map \( S_\bullet \to S_{n\bullet} \) is obviously a surjection, we get a morphism \( S_0[X_1, \dots, X_n] \twoheadrightarrow S_{n\bullet} \).
$$\tag*{$\blacksquare$}$$
Exercise 6.4.F:
Show that if \(R_\bullet\) and \(S_\bullet\) are the same finitely generated graded rings except in a finite number of nonzero degrees (make this precise!), then \( \proj R_\bullet \cong \proj S_\bullet \).
Proof:
The interpretation here is that \( R_m = S_m \) except for finitely many \( m \in \mathbb{N} \). To see that the \( \proj R_\bullet \cong \proj S_\bullet \), let \( n > 0 \) be some integer such that \( R_m = S_m \) for all \( m \geq n \). Then by Exercise 6.4.D we have that the map of graded rings \( S_{n \bullet} = R_{n \bullet} \hookrightarrow S_\bullet \) induces an isomorphism \( \proj S_\bullet \cong \proj R_\bullet \).
$$\tag*{$\blacksquare$}$$
Exercise 6.4.G:
Suppose \(S_\bullet\) is generated over \(S_0\) by \(f_1, \dots, f_n\). Find a \(d\) such that
\(S_{d•}\) is finitely generated in "new" degree 1 (= "old" degree d). (This is surprisingly
tricky, so here is a hint. Suppose there are generators \(x_1, \dots , x_n\) of degrees \(d_1, \dots ,
d_n\) respectively. Show that any monomial \(x_1^{a_1} \dots x_n^{a_n}\) of degree at least \(nd_1 \dots d_n\)
has \(a_i \geq ( \prod_j d_j)/d_i\) for some \(i\). Show that the \((nd_1 \dots d_n)\)th Veronese subring is generated by elements in "new" degree 1.)
Proof:
Following the hint, let \( x_1, \dots, x_n \) be as described above and let \( x_1^{a_1} \dots x_n^{a_n} \) be a monomial of degree at least \( n d_1 \dots d_n \) — notice we want all \( d_i \geq 1 \) since anything in degree \(0\) may be pulled out as a constant. Moreover we want all \( a_i \geq 0 \); since the exact degree of this polynomial is \( a_1d_1 + \dots + a_n d_n \) and \( n \geq 1 \) we need at least one of the \( a_i > 0 \). Let \( a_jd_j = \max_{1 \leq i \leq n}\{ a_id_i \} \) so that
$$
n a_jd_j \geq a_1 d_1 + \dots + a_n d_n \geq n d_1 \dots d_n
$$
Next we wish to use induction to show that \( S_{nd_1\dots d_n \bullet} \) is generated in degree \(1\) --- that is to say, we wish to show that any monomial in degree \( mnd_1\dots d_n \) can be written as a monomial in elements of degree \( nd_1\dots d_n \). The base case of \( m = 1 \) is trivial, and if we suppose we have shown this to hold for some \( m_0 \geq 1 \) and let \( f_1^{a_1} \dots f_n^{a_n} \) be a monomial of degree \( (m_0+1)nd_0\dots d_n \). By the result proved above, we may factor out some \( f_i^{d_1\dots d_n / d_i} \) to obtain a monomial of degree \( ((m_0 + 1)n - 1)d_1\dots d_n \geq nd_1 \dots d_n \). Repeating this process \( n - 1 \) times gives a monomial of degree \( m_0nd_1\dots d_n \) (times the \( n \) factors of degree \( d_1 \dots d_n \) that we pulled out, thus making a factor of degree \( nd_1 \dots d_n \) ). By inductive hypothesis, the resulting monomial of degree \( m_0nd_1 \dots d_n \) may be written as a monomial in degree \( n d_0 \dots d_n \) as desired.
$$\tag*{$\blacksquare$}$$
Exercise 6.4.H:
Suppose \(S_\bullet\) is a finitely generated ring. Show that \(S_{n\bullet}\) is a finitely generated graded ring. (Possible approach: use the previous exercise, or something similar, to show there is some \(N\) such that \(S_{nN\bullet}\) is generated in degree 1, so the graded ring \(S_{nN\bullet}\) is finitely generated. Then show that for each \(0 < j < N\), \(S_{nN \bullet +nj}\) is a finitely generated module over \(S_{nN\bullet}\).)
Proof:
The hint seems a bit excessive; by the previous paragraph, Exercise 6.4.G and Exercise 6.4.D we may assume without loss of generality that \( S_\bullet \) is generated in degree 1. By Exercise 4.5.D \( S_\bullet \) is a finitely generated graded ring iff \( S_\bullet \) is a finitely generated \(S_0\)-algebra. Since \( S_\bullet \) finitely generated, we have a surjective algebra morphism \( S_0[X_0, \dots, X_n] \twoheadrightarrow S_\bullet \) for some \( n \). Similar to Exercise 6.4.E, the surjection \( S_\bullet \twoheadrightarrow S_{n\bullet} \) shows \( S_{n\bullet} \) is finitely generated in degree 1.
$$\tag*{$\blacksquare$}$$
Section 6.5: Rational maps from reduced schemes
Exercise 6.5.A:
Show that a rational map \(\pi: X \dashrightarrow Y\) of irreducible schemes is dominant if and only if \(\pi\) sends the generic point of \(X\) to the generic point of \(Y\).
Proof:
For the forward direction, let \( \eta_X, \eta_Y \) denote the generic points of \(X\) and \(Y\), respectively, and suppose \( \pi : X \dashrightarrow Y \) is dominant. Let \( f : U \to Y \) be a representative for some dense open subset \(U \subset X\); since \( f \) is continuous, \( f(U) = f( \overline{ \{\eta_X\} }) \subset \overline{ f(\eta_X) } \). Since the latter is closed, we must also have \( Y = \overline{f(U)} \subset \overline{f(\eta_X)} \) so that \( f(\eta_X) \) is a generic point of \(Y\). Since we are assuming all schemes are reduced, generic points are unique so that \( f(\eta_X) = \eta_Y \).
Conversely, if \( f \) maps \( \eta_X \) to \( \eta_Y \), then \( \{ \eta_Y \} \subset f(U) \) so that \( Y = \overline{ \{ \eta_Y \} } \subset \overline{ f(U) }\).
$$\tag*{$\blacksquare$}$$
Exercise 6.5.B:
Show that dominant rational maps of integral schemes give morphisms of function fields in the opposite direction.
Proof:
Let \( X \) and \( Y \) be integral schemes with generic points \( \eta_X \) and \( \eta_Y \) (respectively), \( \pi : X \dashrightarrow Y \) be dominant map and \( f : U \to Y \) a representative. As a morphism of schemes, \( f \) comes with the additional data of a morphism of sheaves \( f_\ast : \OO_Y \to f_\ast \OO_Y \), which by Exercise 2.2.I indues a map of stalks \( \OO_{Y, q} \to \OO_{X, p} \) for every set of points with \( f(p) = q \). By the previous exercise and Exercise 5.2.H we obtain a map \( K(Y) = \OO_{Y, \eta_Y} \to \OO_{X, \eta_X} = K(X) \).
$$\tag*{$\blacksquare$}$$
Exercise 6.5.C:
Let \(K\) be a finitely generated field extension of \(k\). (Recall that a field extension \(K\) over \(k\) is finitely generated if there is a finite "generating set" \(x_1, \dots , x_n\) in \(K\) such that every element of \(K\) can be written as a rational function in \(x_1 , \dots , x_n\) with coefficients in \(k\).) Show that there exists an irreducible affine \(k\)-variety with function field \(K\). (Hint: Consider the map \(k[t_1, \dots , t_n] \to K\) given by \(t_i \mapsto x_i\), and show that the kernel is a prime ideal \(\pp\), and that \(k[t_1 , \dots , t_n ]/\pp\) has fraction field \(K\). Interpreted geometrically: consider the map \(\spec K \to \spec k[t_1 , \dots , t_n ]\) given by the ring map \(t_i \mapsto x_i\), and take the closure of the one-point image.)
Proof:
Since the image of \( k[t_1, \dots, t_n] \) is a subring of a field, it is an integral domain. By the first isomorphism theorem this implies that \( \pp\) must be prime. Then \( K( k[t_1, \dots, t_n] / \pp ) \hookrightarrow K\) is a subfield; however, since the \( x_i \) generated \( K \) and the \( \overline{t_i} \) (i.e. images in the fraction field of quotient) generate K( k[t_1, \dots, t_n] / \pp ), the above map is a surjection and thus an isomorphism.
$$\tag*{$\blacksquare$}$$
Exercise 6.5.D:
Describe equivalences of categories among the following
The category with objects "integral \(k\)-varieties", and morphisms "dominant rational maps defined over \(k\)";
the category with objects "integral affine \(k\)-varieties",and morphisms "dominant rational maps defined over \(k\)"; and
the opposite ("arrows-reversed") category with objects finitely generated field extensions of \(k\)", and morphisms "inclusions extending the identity on \(k\)"
Proof:
There is an obvious forgetful functor going from (b) to (a). To go from \( (a) \) to \( (c) \) we simply apply Exercise 6.5.B. This clearly preserves the identity, since if we consider the identity map \( \textrm{id}_X : X \dashrightarrow X \) (as a dominant rational map possibly restricted to some dense open subset \(U\)) and \( \eta_X \) the generic point, then \( (\textrm{id}_X)_\ast : \OO_U \to \OO_U \) is necessarily the identity so that the induced map of stalks at the generic point \( K(U) \to K(U) \) is also the identity. In a similar fashion, if \( X, Y \) and \(Z\) are integral \(k\)-schemes with dominant morphisms \( \phi : X \dashrightarrow Y \) and \( \psi : Y \dashrightarrow Z \), then by taking representatives \( \overline{\phi} \) and \( \overline{\psi} \) we may assume WLOG that the domain of \( \overline{\psi} \) is \( \overline{\phi}(U) \) (where \( U \) dense in \( X \)). For the sake of notation, let us also denote the image of \( \overline{\psi} \) as \(W\). Letting \( \eta_X, \eta_Y \) and \( \eta_Z \) denote the respective generic points, we must have that \( \overline{\psi} \circ \overline{\phi} \) sends \( \eta_X \) to \( \eta_Z \) so that \( \psi \circ \phi \) is dominant by Exercise 6.5.A. Since the pullbacks commute, we get that the map \( (\overline{\psi} \circ \overline{\phi})_\ast : \OO_W \to \OO_U \) is the same as the composition of \( \overline{\phi}_\ast \) with \( \overline{\psi}_\ast \). Taking stalks at the generic points shows that the maps of function fields commute. Notice that since our \(k \)-varieties are by definition finite-type, we must have that the function field is a finite extension.
Next, to go from (c) to (b) we simply apply the previous exercise. Similarly, to go from (c) to (a) one applies Proposition 6.5.7. It is fairly easy to check that (a) and (c) are equivalent via Exercise 6.5.B and Exercise 6.5.D. We obtain a map (b) to (c) in the same way as (a) to (c), giving us the second equivalence (which we will not prove.)
$$\tag*{$\blacksquare$}$$
Exercise 6.5.E:
Use the above to find a "formula" yielding all Pythagorean triples.
Proof:
A Pythagorean triple is a triple of integers \( (x, y, z) \) so that \( x^2 + y^2 = z^2 \); since \( (0, 0, 0) \) is trivial, we consider \( z \neq 0 \). Dividing by \(z\), this gives rational solutions to \( (x/z)^2 + (y/z)^2 = 1 \) — by the previous paragraph we know that we must have \( x / z = (m^2 - 1) / (m^2 + 1) \) and \( y / z = (-2m) / (m^2 + 1) \) for some \( m \in \Q \). Assuming that \( \gcd(x, y, z) = 1 \), we get \( m \in \Z \) and \( x = m^2 - 1, y = -2m, z = m^2 + 1 \)
$$\tag*{$\blacksquare$}$$
Exercise 6.5.F:
Show that the conic \(x^2 + y^2 = z^2\) in \(\P^2_k\) is isomorphic to \(P^1_k\) for any field \(k\) of characteristic not 2. (Aside: What happens in characteristic 2?)
Proof:
On any affine open set, say \( \spec (( k[x_0, x_1] )_{x_0})_0 \subset \P^1_k \), we wish to map \( x_{1/0} \) to the previous pythagorean triple \( [(x_{1/0})^2 - 1, -2x_{1/0}, (x_{1/0})^2 + 1 ] \). In order to make this agree on both (usual) affine opens, we effectively multiply through by \(x_0^2 \) to obtain the map \( [x_0, x_1] \mapsto [x_1^2 - x_0^2, -2x_0x_1, x_1^2 + x_0^2] \). However, in characteristic 2 we have \( z^2 = (x - y)^2 \) which is the double line along \( x = y \) --- this is certainly non-normal.
$$\tag*{$\blacksquare$}$$
Exercise 6.5.G:
Find all rational solutions to \(y^2 = x^3+x^2\), by finding a birational map to \(A^1_\Q\), mimicking what worked with the conic. Hint: what point should you project from? (In Exercise 19.10.F, we will see that these points form a group, and that this is a degenerate elliptic curve.)
Proof:
To find a \( \Q \)-algebra morphism \( \Q[t] \to \Q[x, y] / (y^2 - x^2 (x+1)) \), we project from the origin \( x = y = 0 \) in order to look for solutions to \( y = tx \) lying ond the nodal cubic. Then \( y^2 - x^2 (x + 1) \) implies \( x^2 (t^2 - x - 1) \) so that \( x = t^2 - 1 \) and \( y = t^3 - t \). Thus, we get a ring of \( \Q \)-algebras \( \phi : \Q[t] \to \Q[x, y] / ( y^2 - x^2(x + 1) ) \) via \( t \mapsto (t^2 - 1, t^3 - t) \) which in turn gives us a morphism \( \spec \Q[x, y] / (y^2 - x^3 - x^2) \to \A^1_\Q\). In the reverse direction, we only get a dominant rational map \( \Q[x, y] / (y^2 - x^2(x + 1)) \to \Q[t]\) via \( t = y/x \) away from the origin.
$$\tag*{$\blacksquare$}$$
Exercise 6.5.H:
Use a similar idea to find a birational map from the quadric surface \(Q = \{x^2 +y^2 = w^2 +z^2\} \subset \P^3_\Q\) to \(\P^2_\Q\). Use this to find all rational points on \(Q\). (This illustrates a good way of solving Diophantine equations. You will find a dense open subset of \(Q\) that is isomorphic to a dense open subset of \(\P^2\), where you can easily find all the rational points. There will be a closed subset of \(Q\) where the rational map is not defined, or not an isomorphism, but you can deal with this subset in an ad hoc fashion.)
Proof:
It is easier (for me) to deal with the change of coordinates to \( xw = yz \) (which works as \(\textrm{char} \Q \neq 2 \) since we may take \( (x - w) (x + w) = (z - y)(z + y) \) and relabel). In the forward direction we may project from \( [0 , 0 , 0 , 1] \) to \( \P^2 \) via \( [x , y , z , w] \mapsto [x , y , z ] \). In the reverse direction, as \( w = yz / x \) we may multiply through by \(x\) to obtain the map \( [x^2, xy, xz, yz] \) which is well-defined for \( x \neq 0 \). Thus, we get a morphism on the open affine set \( \spec ((\Q[x,y,z])_x )_0 \) which is isomorphic to \( \A^2_\Q \). Since \( \A^2_\Q \) is dense in \( \P^2_\Q \) we obtain the desired birational map.
$$\tag*{$\blacksquare$}$$
Exercise 6.5.I:
Consider the rational map \(\P^2_k \dashrightarrow \P^2_k\), given by \([x,y,z] \mapsto [1/x,1/y,1/z]\). What is the the domain of definition? (It is bigger than the locus where \(xyz \neq 0\)!) You will observe that you can extend it over "codimension 1 sets" (ignoring the fact that we don’t yet know what codimension means). This again foreshadows the Curve-to-Projective Extension Theorem 16.5.1.
Proof:
Multiplying through by \( xyz \), we may rewrite the cremona transformation as \( [x, y, z] \mapsto [yz, xz, xy] \). Thus, we get that \( yz = xz = xy = 0 \) if and only if two of the coordinates \( x, y, z \) are zero. Thus, the Cremona transformation is defined on \( \P^2 \backslash \{ [1, 0, 0], [0, 1, 0], [0, 0, 1] \} \).
$$\tag*{$\blacksquare$}$$
Section 6.6: Representable functors and group schemes
Exercise 6.6.A:
Suppose \(X\) is a \(\C\)-scheme. Verify that there is a natural bijection between maps \(X \to \A^1_\C\) in the category of \(C\)-schemes and functions on \(X\). (Here the base ring \(\C\) can be replaced by any ring \(A\).)
Proof:
Working in the generality of any ring \(A\), suppose for the forward direction that \( \phi^\sharp : X \to \spec A[t] \) is a morphism of \( A \)-schemes (that is, yields a commuting diagram)
from the data of this morphism, we also obtain a commuting diagram with reversed arrows corresponding to the structure morphism \( \phi : \OO_{\spec A[t]} \to \phi^\sharp_\ast \OO_X \) along with the structure morphism making these two sheaves into \( A \)-algebras on each open set. By considering the globalization functor \( \Gamma(-) \), we obtain a morphism of \(A\)-algebras \( A[t] \to \Gamma(X, \OO_X) \). Then similar to the paragraph above, looking at the image of \(t \in A[t]\) gives us a function \( f \in \Gamma(X, \OO_X) \).
In the reverse direction, we proceed in a similar way to Exercise 6.3.F: let \( f \in \Gamma(X, \OO_X) \) be a function on \(X\) and \( U_i = \spec B_i \) an affine open cover of \(X\). For each \( i \), let \( f_i = \res{X}{U_i}(f) \). If we let \( s_\ast : \OO_{\spec A} \to s_\ast \OO_X \) denote the structure morphism, for each \( U_i \) we should get some structure morphism \( s_i : A \to B_i = \Gamma(U_i, \OO_X) \). Since \( A \hookrightarrow A[t] \) is the natural inclusion, we define an \( A \)-algebra morphism \( A[t] \to B_i\) via \( 1 \mapsto s_i(1) \) and \( t \mapsto f_i \). Thus we obtain morphisms \( \spec A[t] \to U_i \) for each \( i \) which glue by construction, giving us the desired morphism \( \A_A^1 \to X \).
$$\tag*{$\blacksquare$}$$
Exercise 6.6.B:
Interpret rational functions on an integral scheme (Exercise 5.5.Q, see also Definition 5.5.6) as rational maps to \(\A^1_\Z\).
Proof:
By Exercise 5.5.Q rational functions on an integral scheme \(X\) are just elements of \( K(X) = \OO_{X, \eta} \). Fixing some \( f \in K(X) \), by Proposition 6.5.7 it suffices to provide a map \( \Q(t) = K(\Z[t]) \to K(X) \). The obvious choice here is \( t \mapsto f \).
$$\tag*{$\blacksquare$}$$
Exercise 6.6.C:
Show that if a contravariant functor \(F\) is represented by \(Y\) and by \(Z\), then we have a unique isomorphism \(Y \to Z\) induced by the natural isomorphism of functors \(h_Y \to h_Z\). Hint: this is a version of the universal property arguments of §1.3: once again, we are recognizing an object (up to unique isomorphism) by maps to that object. This exercise is essentially Exercise 1.3.Z(b). (This extends readily to Yoneda’s Lemma in this setting, Exercise 9.1.C. You are welcome to try that now.)
Proof:
We will simply refer to the proof of Exercise 9.1.C, as we only use universal properties.
$$\tag*{$\blacksquare$}$$
Exercise 6.6.D:
Suppose \(F\) is the contravariant functor \(\Sch \to \Set\) defined by \(F(X) = \{\textrm{Grothendieck, A.}\}\) for all schemes \(X\). Show that \(F\) is representable. (What is it representable by?)
Proof:
We interpret \(F\) as sending every scheme \(X\) to the same point. Since (the) singleton set is terminal in the category of sets, we wish to find some \( Y \) such that \( \hom (-, Y) \) is a singleton in the category of schemes for every source \(X\) — that is to say, final in the category of schemes. By Exercise 6.3.I we know this is \( Y = \spec \Z \).
$$\tag*{$\blacksquare$}$$
Exercise 6.6.E:
In this exercise, \( \Z \) may be replaced by any ring.
(Affine \(n\)-space represents the functor of \(n\) functions) Show that the contravariant functor from \((\Z)\)-schemes to \(\Set\)
$$
X \mapsto \{ (f_1, \dots, f_n) : f_i \in \Gamma(X, \OO_X) \}
$$
is represented by \( \A^n_\Z \). Show that \( \A^1_\Z \times_\Z \A^1_\Z \cong \A^2_\Z \), i.e., that \( \A^2_\Z \) satisfies the universal property of \( \A^1 \times \A^1 \). (You will undoubtedly be able to immediately show that \( \prod \A^{m_i}_\Z \cong \A^{\sum m_i}_\Z \))
(The functor of invertible functions is representable.) Show that the contravariant functor from \( \Z \)-schemes to \( \Set \) taking \( X \) to invertible functions on \(X\) is representable by \( \spec \Z[t, t^{-1}] \).
Proof:
Recall by Essential Exercise 6.3.F that morphisms \( X \to \spec A \) are in natural bijection with ring morphisms \( A \mapsto \Gamma(X, \OO_X) \). The ring of interest here is clearly \( A = \Z[x_1, \dots, x_n] \); thus, for any \(\Z\)-scheme \(X\), the set \( \Hom (X, \A^n_\Z) \) is in natural bijection with ring morphisms \( \Z[x_1, \dots, x_n] \to \Gamma(X, \OO_X) \). As \( X \) is a \(Z\)-scheme, the image of \( 1 \in \Z\) is already determined by the structure morphism. Thus, the data of such a morphism is determined by \( x_i \mapsto f_i \) for some \( f_i \in \Gamma(X, \OO_X) \) giving us an obvious natural isomorphism between \( F \) and \( h_{\A^n_\Z} \).
To see the second result, we don't exactly need to employ the previous functor. Though not explained until later, a fiber product of affine schemes
Since \( \Z[x_1] \otimes \Z[x_2] \cong \Z[x_1, x_2]\) we get the desired result.
Since morphisms \( X \to \A^1_\Z\) are in bijection with ring maps \( \Z[t] \to \Gamma(X, \OO_X) \) by Exercise 6.3.F, we simply localize at the generic point to get a morphism \( \Z[t, t^{-1}] \to \Gamma(X, \OO_X)^\times \) which again is in bijection with morphisms to \( \spec \Z[t, t^{-1}] \).
$$\tag*{$\blacksquare$}$$
Exercise 6.6.F:
Fix a ring \(A\). Consider the functor \(H\) from the category of locally ringed spaces to Sets given by \(H(X) = \{A \to \Gamma(X,\OO_X)\} \). Show that this functor is representable (by \(\spec A\)). This gives another (admittedly odd) motivation for the definition of \(\spec A\), closely related to that of §6.3.6.
Proof:
This is the same as Exercise 6.3.F with the changed hypothesis to locally ringed spaces (so that there need not exist a cover by affines.) Instead, one uses a map of stalks to construct a topological map.
For the forward direction, if \( \phi : X \to \spec A \) is a morphism of locally ringed spaces, we also get a stalk map \( \phi^\sharp : \OO_{\spec A} \to \OO_X \), which at each point \( x \) gives us a stalk map \( \phi^\sharp_x : A_{\phi(x)} \to \OO_{X, x} \). As a morphism of local rings, we know that \( \phi(x) \) must in fact be the inverse image of the maximal ideal \( \mm_x \subset \OO_{X, x} \) under the map \( A \to \OO_X \to \OO_{X, x} \) which itself must be maximal — therefore, the map \( \{ X \to \spec A \} \mapsto \{ A \to \Gamma(X, \OO_X) \} \) must be injective as it is determined at its stalks.
To see that it is surjective, suppose \( \phi^\sharp : A \to \Gamma(X, \OO_X) \) is a ring map. For every \( x \in X \), let \( \psi(x) \subset A \) denote the inverse image of \( \mm_x \subseteq \OO_{X, x} \) (which we only know is a prime ideal) under the map \( A \to \OO_X \to \OO_{X, x} \) determined by restriction morphisms. This gives us a map of sets \( \phi : X \to \spec A \); to show this extends over open sets, for any \( f \in A \) one has\( (\phi^{-1}(D(f)) = X_{\phi^{\sharp}(f)} \), where the latter set is the open set of all \( x \in X \) such that \( \phi^\sharp(f) \) is not in \( \mm_x \). Since each \( \phi^{\sharp}(f) \vert_{X_{\phi^\sharp(f)}} \) is a unit, there is a unique map \( A_f \to \OO_X( X_{\phi^\sharp(f)}) \) compatible with restriction, giving us our morphism of sheaves \( \OO_{\spec A} \to \phi_\ast \OO_X \)
$$\tag*{$\blacksquare$}$$
Section 6.7: The Grassmannian (initial construction)
Exercise 6.7.A:
Show that any two bases are related by an invertible \(n \times n\) matrix over \(A\) — a matrix with entries in \(A\) whose determinant is an invertible element of A.
Proof:
This is a standard linear algebra exercise: let \( \{v_1, \dots, v_n\} \) and \( \{w_1, \dots, w_n\} \) be two bases for \( A^{\oplus n} \). Then for each \( 1 \leq i \leq n \), \( v_i \) may be uniquely represented as a linear combination \( v_i = \sum_{j=1}^n c_{ij} w_j \). Similarly, for any \( 1 \leq i \leq n \), \( w_i \) may be uniquely represented as a linear combination \( w_i = \sum_{j=1}^n d_{ij} v_j \). Thus, \( C = [c_{ij}] \) is the desired matrix with inverse \( D = [d_{ij}] \).
$$\tag*{$\blacksquare$}$$
Exercise 6.7.B:
Given two bases \(v\) and \(w\), explain how to glue \(U_v\) to \(U_w\) along appropriate open sets. You may find it convenient to work with coordinates \(a_{ji}\) where \(i\) runs from \(1\) to \(n\), not just \(k+1\) to \(n\), but imposing \(a_{ji} = \delta_{ji}\) (i.e., \(1\) when \(i = j\) and \(0\) otherwise) when \(i \leq k\). This convention is analogous to coordinates \(x_{i/j}\) on the patches of projective space (§4.4.9). Hint: the relevant open subset of Uv will be where a certain determinant doesn’t vanish.
Proof:
Let \( C = [c_{ij}]\) denote our change of basis matrix from the problem above. We wish to show how to transform a \(k \)-plane on \( U_v \) of the form
Following the hint, we assume \( c_{ij} = \delta_{ij} \) for \( 1 \leq i \leq k \); then applying our change of basis matrix to the first equation above gives
