Exercise 3 (by Zi-Li).
Define α : HomA(M, Γ(X,F )) → HomX( fM ,F ) by:
Given ϕ : M → Γ(X,F ), define ψ(D(f)) :M (D(f )) = M ⊗f AAf →F (D(f)), m⊗
a 7→ aϕ(m). Glue ψ(D(f )) to get ψ : fM →F .
The inverse of α is taking global section, hence,e and Γ are adjoint pair.
Exercise 6 (by Tzu-Yang Tsai).
(a) (⊆) If p ∈Suppm, i.e. mp 6= 0, if p /∈ V (Annm),
∃r ∈ Annm \ p s.t. rm = 0 ⇒ m1 = 01 →←
(⊇) If p ∈ V (Annm), that is, p ⊇ Annm ⇒ @r ∈ M \ p s.t. rm = 0 ⇒ mp 6= 0
(b) (⊆) If p ∈SuppF , i.e. Fp = Mp 6= 0, by finitely generated, we may have M = A < m1, . . . , mn > for some {mi}ni=1 ⇒ p ⊇ ∩ni=1Annmi ⇒ p ∈ V (AnnM )
(⊇) If p ∈ V (AnnM ), then p ∈ Annmi∀i = 1 ∼ n ⇒ Mp =Fp 6= 0 ⇒ p ∈ SuppF
(c) Since F is coherent, F |U = ˜M for some A-module M , where U is an open affine subset = Spec A. Then SuppF |U = V (AnnM ) is closed, thus SuppF = SU ⊆affineX is closed.
(d) Recall that 0 → HZ0 → F → j∗(F |U) → 0 is an exact sequence, where U = X \ Z, j : U ,→ X is the inclusion map (Ex 1.20). Since j is open immersion, j∗ is quasi-compact and separated, and F |U is quasi-coherent, these imply j∗(F |U) is quasi-coherent. Combining that F is also quasi-coherent,HZ0is quasi-coherent. By definition ΓZ(F ) = {p ∈ X|Suppp ⊆ Z}
= {p ∈ X|p ∈ V (Anna)}, since A is Noetherian ⇒ a is finitely generated.
Therefore ΓZ(F ) = {m ∈ M|anmfor somen ∈ N} ∼= Γa(M )
⇒Γa(M ) ∼˜ =ΓZ(˜F ) = HZ0
(e) The quasi-coherent case has been proved in (d).
For the coherent case, ΓZ(F ) is finitely generated, thus = HZ0 =ΓZ(˜F ) is coherent.
Exercise 15 (by Shi-Xin).
(a) Let X = Spec A be an affine scheme where A is noetherian ring and F be a quasi-coherent sheaf on X. Then F = ˜M for some A-module M . Since M = limα∈AMα where Mα|α ∈ A are all finitely generated A-submodule of M , for any f ∈ A, we must have Mf = limα(Mα)f. It follows that F (D(f )) = limαM (D(f )). Moreover, because D(f ) form a basis, F (U ) =˜ limαM (U ) for any open set U . Thus F = lim˜ αFα where each Fα := ˜Mα is a coherent sheaf.
(b) Let i : U → X be the inclusion map. Since X is noetherian, U is also noetherian, and hence by Proposition 2.5.8, i∗(F ) is quasi-coherent. Then by (a), we can write i∗(F ) = limαFα. Therefore for any affine open subset V ⊂ U , we must have F |V = limα(Fα)|V. We might assume F |V ∼= ˜M , Fα|V ∼= ˜Mα where M, Mα are finitely generated. Then we have M =∼= Mα. Since M is finitely generated, there must be some Mα containing all generators of M . Moreover, U can be covered by finitely many affine open subsets, so we can choose F0 := Fα for some α such that F0|U ∼= F .
(c) Let ρ be the natural map G → i∗(G|U). Since ρ−1(i∗F )|U ∼= F and it is quasi-coherent, there is a coherent subsheaf F0 of ρ−1(i∗F ) such that FU0 ∼= F . Moreover, F0 ⊂ ρ−1(i∗F ) ⊂ ρ−1(i∗G|U) ⊂ G
(d) Let X be a noetherian scheme covered by affine open subsetsSn
i=1Ui. We have proved the desired result when n = 1. It suffices to show that we can extend over one of them at a time. We might assume n ≥ 2 and suppose that we have a coherent subsheaf F10 ⊂ G|U1 on U1 such that F10|U ∩U1 ∼= F |U ∩U1. Then we might apply (c) to F10 on U1, so we obtain a coherent sheaf F20 ⊂ G|U1 on U1∪ U2 such that F20|U ∩(U1∪U2)∼= F |U ∩(U1∪U2). Thus by induction we prove desired result. Moreover, taking G = i∗F shows that we can extend a coherent sheaf from an open subset U to X.
(e) Clearly, F ⊇ S Fα for all coherent subsheaves Fα of F . Conversely, if s is a section in F (U ) where U is an open set of X, let G be the subsheaf of FU generated by s. Then by (d) we can extend G to a coherent subsheaf G0 of F such that s ∈ G0(U ). Thus F ⊂S Fα.
6 Divisors
Exercise 1 (by Wei-Ping).
X × Pn is regular and codimension one since locally it is X × An. Irreducible since it is union of two affine pieces with nonempty intersection, and reduced is again local condition, hence X × Pn is integral. Also it is separated since X × Pn→ X is projective and X is separated.
Now we have Z → Cl(X × Pn) → Cl(X × An) → 0, where first map is 1 7→ X × H, H = {x0 = 0}. Since Cl(X ×An) ' Cl(X), it suffices to show the sequence is exact, then taking closure results a converse map of second map so sequence splits. Let f be element in function field, if (f ) = m · (X × H), then vX×H(f ) = m, f = hgxm0 . If m 6= 0 then h or g involves other divisors, a contradiction. Hence m = 0 and first map is injective, done.
Exercise 2 (by Wei-Ping).
(a) To prove well-defined, let η be generic point of divisor Yi, then consider two choices of covering, say {(Ui, fi)}, {(Vj, gj)}. Now choose η ∈ Ui, η ∈ Vj, then point of a divisor Z on X, say η, we consider an open neighborhood W of η disjoint with all divisors Yi occurring in (f ) such that η /∈ Yi∩ X. For every covering from Yi, choose one open set such that η is in it, say (Uiki, fiki), then consider product h =Q
ifikvYi(f )
i on their intersection. Then hf is unit on (T
iUiki) ∩ W since valuation is the same on this open set and note that the coefficient is same as those in the previous sum. Therefore vY
ij(f ) is same as coefficient of divisor Yij in (f ) · X, as we desired.
Combining (a) and the fact that any divisor on Pn can be change to some multiple of any hyperplane, in particular, not containing X, we get a homo-morphism Cl Pn→ X.
where aniis image under quotient and localization. Therefore ni = i(X, H, Yi), and by Bezout thm we get
deg(D · X) = (deg D) · (deg X)
(d) K(X) = S(X)((0)) so we can find some f ∈ K∗ such that f restricting on X is f , so by (b) get (f ) · X = (f ) Hence for any principal divisor D on X, deg(D · X) = deg((f ) · X) = (deg(f )) · (deg X) = 0. The degree function defines a homomorphism to Z and we get commutative diagram
Cl Pn Cl X
Z Z
deg ' deg
·(deg X)
Exercise 4 (by Shuang-Yen).
Since Quot(A) = k(x1, . . . , xn)[z]/(z2 − f ), [Quot(A) : k(x1, . . . , xn)] = 2. For α = g + hz ∈ Quot(K), where g, h ∈ k(x1, . . . , xn), the minimal polynomial of α over k(x1, . . . , xn) is
(X2− 2gX + (g2− h2f ), if h 6= 0
X − g, if h = 0
When h = 0, α is integral over k[x1, . . . , xn] if and only if α = g ∈ k[x1, . . . , xn] since k[x1, . . . , xn] is a UFD.
When h 6= 0, α is integral over k[x1, . . . , xn] if and only if −2g, g2 − h2f ∈ k[x1, . . . , xn]. Since the characteristic of k is not 2, it’s equivalent ti g, h2f ∈ k[x1, . . . , xn]. Write h = a/b with a, b ∈ k[x1, . . . , xn] and a, b are coprime to each other, then h2f = a2f /b2, but f is square-free, so h2f ∈ k[x1, . . . , xn] if and only if b ∈ k× if and only if h ∈ k[x1, . . . , xn]. Hence, α is integral over k[x1, . . . , xn] if and only if g, h ∈ k[x1, . . . , xn] if and only if α ∈ A, so A is integrally closed since z is also integral over k[x1, . . . , xn].
Exercise 6 (by Chun-Yi).
(a) (⇒) If P, Q, R are collinear. Let L be the line attaching P, Q, R. Since degX
= 3, L ∩ X has only three points P, Q, R. Since L ∼ [z = 0], P + Q + R ∼ 3P0
⇒ P − P0+ Q − Q0+ R − R0 ∼ 0 ⇒ P + Q + R = 0 in the group law of X.
(⇐) If P, Q, R = 0. Let L be the line passing through P, Q. By Bezout’s thm, L ∩ X has three points, say P, Q, S, then P + Q + T = 0. By (⇒), since the inverse of P + Q is unique, T = R ⇒ R ∈ L ⇒ P, Q, R are collinear.
(b) (⇒) Let L be the tangent line passing through P . By Bezout’s thm, L ∩ X has three points, say P, P, S, then by (a), P + P + T = 0. Since the inverse is unique and P + P = 0 ⇒ T = P0.
(⇐) Let L be the tangent line at P , passing through P0, then by Bezout’s thm, L intersect X at P with multiplicity 2. By (a), P + P + P0 = 0 ⇒ P + P = 0 ⇒ P has order 2.
(c) (⇒) Let L be the tangent line at P , then L ∩ X = {P, P S} for some S ∈ X, and P + P + S = 0. Since 3P = 0, S = P ⇒ L intersects X at P with multiplicity 3.
(⇐) If P is an inflection point, since L ∩ X has only three points, L ∩ X = {P, P, P }, thus P + P + P = 0, again by (a).
(d) By Mordell-Weil theorem, the points of X with coordinates in Q form a subgroup of X. If z = 0, the only rational point in (0,1,0). If z 6= 0, it suffices to find rational points of y2 = x3− x, which is only (1,0),(0,0),(-1,0)
⇒ the subgroup is Z/2Z × Z/2Z.
Exercise 7 (by Yi-Tsung).
In example 6.11.4, we have seen that there is 1-1 correspondence between the set of nonsingular closed points of X and the kernel CaCl0X of the degree map.
It suffices to show that the set of nonsingular closed points of X endowed the group structure is isomorphic to Gm. The set of nonsingular closed points of X is just X\ {(0, 0, 1)}, say Z = {(0, 0, 1)}. Consider φ : X\Z → Gm, (x, y, z) 7→
y − x
y + x. Consider the coordinates change: (x, y, z) = (4x0 − 4y0, −4x0 − 4y0, z0), then X\Z = {x0y0z0 = (x0 − y0)3} \ {(0, 0, 1)}, and φ(x0, y0, z0) = x0
y0. Now setting y0 = 1, then X\Z = {x0z0 = (x0 − 1)3} in A2k and φ0(x0, z0) := φ(x0, 1, z0) = x0, and clearly its inverse map Gm → X\Z is defined by t 7→
t,(t − 1)3 t
. Thus φ is bijective as sets. To show that φ is a group homomorphism, since for p ∈ X\Z, we have φ(−P ) = φ(P )−1, hence it suffices to show that for P, Q, R ∈ X\Z collinear, we have φ(P )φ(Q)φ(R) = 1, and it is enough to prove the same thing for φ0, i.e. to prove φ0(P )φ0(Q)φ0(R) = 1. Let L : z0 = ax0 + b be the line passing through P, Q, R, then φ0(P ), φ0(Q), φ0(R) are roots of x0(ax0 + b) = (x0 − 1)3, thus we see that φ0(P )φ0(Q)φ0(R) = 1, yielding that φ is a group homomophism, and hence an isomorphism. Therefore we see that CaCl0X ∼= {nonsingular closed points of X} ∼= Gm as groups.
Exercise 8 (by Chi-Kang).
(a) We need to show that f∗(L ⊗OY M ) ∼= (f∗L ) ⊗OX (f∗M ). Note that f∗(L ⊗OY M )(U) = f−1(L ⊗OY M ) ⊗f−1OY OX(U )
= lim
V ⊃f (U )
(L (V ) ⊗OY(V )M (V )) ⊗OY(V )OX(U )
= lim
V ⊃f (U )(L (V ) ⊗OY(V )OY(V )) ⊗OY(V )(M (V ) ⊗OY(V )OY(V )) ⊗OY(V )OX(U )
= lim
V ⊃f (U )(L (V ) ⊗OY(V )OX(U )) ⊗OY(V )(M (V ) ⊗OY(V )OX(U ))
= (f∗L (U)) ⊗OX(U )(f∗M (U)) so we are done.
(f∗(L ⊗ M ) ∼= (f∗L ) ⊗ (f∗M ) holds for any sheaves, not need invertible).
(b) We need to show that f∗L (D) ∼=L (f∗(D)). Since f is finite and sheaf isomorphism can be check locally, we may assume X = Spec B, Y = Spec A with B is a finite A module, f is induced by the map φ : A → B, and both A, B are integral domain of dimension 1.
Note that L (D)(U) := {s ∈ K(X) = K(U)|(div(s) + D)|U ≥ 0} by construction.
Now since X, Y are non-singular affine, every divisor is principal, so there is s ∈ K(Y ) ∼= Q(A) s,t, div(s) = D, henceL (D) =As, and sof L (f∗D) = ^Bφ(s).
Hence we have
f∗L (D)P ∼=L (D)f (P )⊗OY,f (P ) OX,P
∼= Asφ−1P ⊗A
φ−1P BP ∼= B(φ(s))P ∼=L (f∗D)P.
(c) We need to show f∗L (D) = L (D.X) for D ∈ Div(Pn) and D.X defined in 6.2. Since for U ⊂ Pn we haveL (D)(U) := {s ∈ K(X) = K(U)|(div(s) + D)|U ≥ 0}, we have
f∗L (D)(V ) = (f−1L (D) ⊗f−1OPn OX)(V )
= lim
U ⊃V L (D)(U) ⊗OPn(U )OX(V )
= {s ∈ K(X) = K(U )|(div(s) + D)|V ≥ 0} ⊗O
Pn|V OX(V )
= {s ∈ K(X) = K(U )|(div(s) + D.X)|V ≥ 0} ⊗OX(V )OX(V )
∼=L (D.X)(V ).
So we are done.
Exercise 10 (by Tzu-Yang Chou).
(a) A1k= Spec k[x], so given anyF ∈ Proj(A1k), we haveF ' ˜M for some finite k[x] module M . Since k[x] is a PID, M is finitely presented. Taking tilde functor we have 0 −→OXm −→OXn −→F −→ 0.
Define φ : K(X) −→ Z by F 7−→ n − m. Note that this is in fact mapping F ti the rank of the free part of M. Also, φ is epic since Oxn 7−→ n. φ is mono, sinceF 7−→ 0 ⇔ n = m ⇔ ∃0 −→ OXm −→OXn −→F −→ 0, that is [F ] = 0 in K(X). Hence φ is the desired isomorphism.
(b) The rank function define a map K(X) −→ Z since any short exact sequence of sheaf gives the stalk sequence at the generic point. Surjectivity is similar as in (a).
(c) The exactness at K(X) −→ K(X \ Y ) −→ 0 follows from Ex(II.5.15) For the exactness at K(Y ) −→ K(X) −→ K(X \ Y ), it’s clear that the composition is zero, so it remains to show that ifF |X\Y = 0, then ∃OY module G such that [i∗G ] = [F ] where i is the inclusion Y ,→ X.
LetF be a coherent sheaf on X only supported on Y , we’ll elaborate a finite filtration 0 =Fn ⊆ · · · ⊆F0 = F such that Fi/Fi+1 is anOY module. We amd thus the filtration is locally finite. X is a Noetherian scheme and in particular quasi-compact, again there exists a uniform n such that the filtration terminates atFn.
Exercise 11 (by Tzu-Yang Chou).
(a) Consider the exact sequence 0 −→ ID −→ OX −→ OD −→ 0, then we see that OD is isomorphic to the direct sum of the skyscraper sheaf of coker(ID,P −→ OX,P) at P , where P is a point whose coefficient in D is
may write both of them as differences of effective divisors, so we can assume they are effective. Now, ψ(D) = [OD] = [OX] − [ID] = [OX] − [L (−D)] = [OX] − [L (−D0)] = [OX] − [ID0] = [OD0] = ψ(D0).
(b) Find a locally free sheaf E00 with epimorphism E00 −→F0, where F0 is the extension of F to ¯X. Restricting on X and taking the kernel, we obtain 0 −→K −→ E00|X −→F −→ 0. We claim that K is locally free of finite rank. At P , KP ⊆ (E00|X)P = OX,Pm for some m. Since X is nonsingular, OX,P is PID ⇒KP is free of finite rank. This proves the existence of the resolution.
Independence of the choice: Conised two locally free resolutions of F , 0 −→ E1 −→E0 −→F −→ 0 and 0 −→ E10 −→ E00 −→ F −→ 0 with the maps named after f1, f0, f10, f00 respectively. We construct a third locally free resolution which maps surjectively to both of them. Let G0 := {(u.u0) ∈ E0⊕E00|f0(u) = f00(u0)} and letE ”0 be a locally free sheaf with a epimorphism E ”0 −→ G0. Let f ”0 : E ”0 −→ G0 −→ F . Then we have two surjections E ”0 −→E0 and E ”0 −→E00 by natural projections. Next, let H0,H00,H ”0
be kernels of f0, f00, f ”0 respectively. LetG1 := {(u, u0) ∈E1⊕E10|∃u” ∈H ”0
such that f1(u) = p(u”), f10(u”) = p0(u”)} where p, p0 are the epimorphisms from H ”0 to H0 and H00 respectively. We also let E ”1 be a locally free sheaf with an epimorphism E ”1 −→G . Then as above, we obtain that E ”1
surjects to allE1,E10 and E ”0. In fact, this works for locally free resolutions of any length inductively.
Now consider two resolutions with an epic chain map from one to another.
Taking kernels in each part and computing determinant we obtain the trivial sheaf, that is these two resolutions have the same determinant (and hecne by symmetry, any two resolutions have the same determinant.) Also, for a short exact sequence 0 → F0 −→ F −→ F ” −→ 0, we have det F = detF0⊗ detF ”. This gives a map from K(X) to Pic X.
To show that det(ψ(D)) = L (D), first consider the effective case. Since 0 −→ ID −→ OX −→ OD −→ 0 is a locally free resolution, we have det(ψ(D)) =OX⊗ID−1 =ID−1 =L (−D)−1 =L (D). For general D, write D = A − B with A, B effective and the same argument works.
(c) Given a coherent sheaf F on X, let L ∈ Pic(X) be an ample invertible sheaf, ∃n ∈ N such that F ⊗ L⊗n is globally generated, say by s1, ..., sm. Hence ∃ an epimorphism OXm −→F ⊗ L⊗n. At generic point ξ, this gives an epimorphism K(X)m −→Fξ, so ∃0 ≤ r ≤ m such that K(X)r −→Fξ. Call this isomorphism φ, and then there’s a dense open set U ⊆ X such that φ : OUr −→ F ⊗ L⊗n|U is an isomorphism. Now consider 0 −→
ker φ −→ OXr −→ F ⊗ L⊗n −→ cok φ −→ 0. Note that (ker φ)P is a submodule of OX,Pr and hence is free. Also, (ker φ)P = 0 on U and
hence ker φ = 0 on X by Ex(II.5.7)(a). Tensoring (L∗)⊗n, we obtain that 0 −→ (L∗)⊗n−→F −→ T −→ 0. It remains to check that T is a torsion sheaf: looking at the stalk sequence at ξ and we have a exact sequence of vector spaces, so Tξ= 0.
Now [F ] − r[OX] ∈ Imψ: [F ] = r[L (D)] + [T ] by above. Hence [F ] − r[OX] = [T ] + r([L (D)] − [OX]). The latter term lies in Imψ by part (a), so it suffices to check that the class of a torsion sheaf lies in the image of ψ.
But sinceTξ = 0, SuppT ( X is closed and hence is finitely many points;
thus T is a direct sum of skyscraper sheaves at each P ∈ SuppT , which in K(X) is equal to a multiple of [k(P )], which lies in the image of ψ.
(d) Combining all above, we obtain a split exact sequence 0 −→ Pic(X) −→
K(X) −→ Z −→ 0.
Exercise 12 (by Tzu-Yang Chou).
Define degree of F by the degree of the determinant of F , then all properties hold by Ex(II.6.11). For the uniqueness, we induct on n := rkF . When n = 0 we use (2); when n = 1 we use (1); when n > 1 we use (3) and the induction hypothesis applies.