Hartshorne Notes/Solutions

Christopher Eur

This document was created to help the author study the book, and as a result is likely filledwith abundance of inelegance if not inaccuracies. Please use with caution. The solutions writtenup here are: exercises that are really propositions, exercises that serve as good examples, exercisesthat correct wrong intuitions, and exercises that just subjectively seem interesting. Some solutionshere and they come not from the author but from various sources which the author has failedto keep track of, and hence the author does not claim credit to any of the work in this docu-ment. Any and all errors are due to the author. Comments and errata are welcome; please emailchrisweur@gmail.com.

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Contents

I Varieties 4I.1 Affine varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4I.2 Projective varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4I.3 Morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5I.4 Rational maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6I.5 Nonsingular varieties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7I.6 Nonsingular curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9I.7 Intersections in Projective Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

II Schemes 12II.1 Sheaves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

II.1.1 A bit of category nonsense... . . . . . . . . . . . . . . . . . . . . . . . . . . . 12II.1.2 Back to exercises... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

II.2 Schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14II.3 First properties of schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17II.4 Separated and proper morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19II.5 Sheaves of modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22II.6 Divisors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24II.7 Projective morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27II.8 Differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

IIICohomology 32III.1 Derived Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32III.2 Cohomology of Sheaves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32III.3 Cohomology of Noetherian Affine Scheme . . . . . . . . . . . . . . . . . . . . . . . . 33III.4 Cech Cohomology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34III.5 The Cohomology of Projective Space . . . . . . . . . . . . . . . . . . . . . . . . . . . 35III.6 Ext Groups and Sheaves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38III.7 The Serre Duality Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38III.8 Higher Direct Images of Sheaves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38III.9 Flat Morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

IV Curves 40IV.1 Riemann-Roch Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40IV.2 Hurwitz’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42IV.3 Embeddings in Projective Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44IV.4 Elliptic Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

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IV.5 The Canonical Embedding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47IV.6 Classification of Curves in P3

k . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

V Surfaces 50V.1 Geometry on a Surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

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Chapter I

Varieties

I.1 Affine varieties

I.1.1(c). Let F := ax2 + bxy + cy2 + dx + ey + f define the (irreducible) quadric curve C in A2.We can tell that C ' A1 or C ' A1 \ 0 by inspecting a, b, c as follows (note that we heavily usek = k). Noting that (by)2 − 4acy2 = (b2 − 4ac)y2, let D = b2 − 4ac. If D 6= 0, then C ' A1 \ 0since there is a linear change of coordinates so that F becomes XY = 1. If D = 0, then eithera = b = c = 0 or at least two among three are 0, or all nonzero, so that change of coordinates thengives X = F or X2 +DX + F = Y .

I.1.10(c,e). (c): Consider Speck[x](x) and its generic points as the open subset. (e): Nagata hasan example of Noetherian ring that is infinite dimensional.

I.1.11. First, some easy general observations:

Lemma 1: If A is a domain, then ker : Bϕ→ A for any ring B is prime; i.e. im(ϕ#) is irreducible.

Lemma 2: If ϕ : B → A is finite, then B/ kerϕ → A is finite, so dimA = dim im(ϕ#) ≤ dimB.

The two lemmas immediately imply that the variety C ⊂ A3 defined by x = t3, y = t4, z = t5 isa variety of dimension 1, and thus p := I(C) is prime of height 2 (∵ I.1.8A). Lastly, to see that itrequires three generators, consider solving for the kernel K of [t3 t4 t5] in k(t), and note that thereare three k[t]-linearly independent elements in K ∩ k[t].

I.1.12. x2 + y2(y − 1)2 = 0

I.2 Projective varieties

I.2.6. Let Yi := Y ∩ Ui 6= ∅, so that A(Yi)[x±i ] ' S(Y )xi (as graded rings even). Thus, since

Y, Yi are irreducible, dimS(Y ) = dimA(Yi)[x±i ] = tr degk S(Y )xi = tr degkA(Yi) + 1 = dimYi + 1.

Now, noting that dimY = supi dimYi, we have that dimS(Y ) = dimY + 1, and in fact, we have astronger statement that dimS(Y ) = dimYi + 1 for any Yi 6= ∅.

Remark: The statement is true even for Y a projective algebraic set (not just a projectivevariety). However, the stronger statement is no longer true (as tr deg depends on having integraldomains).

I.2.7(b). Let Yi := Y ∩ Ui be one where dim is maximized so that dimY = dimYi. ThendimYi = dim(Yi ⊂ Ui) = dim(Y ∩ Ui) = dimY (last equality uses strong form of Ex.I.2.6).

I.2.8. One direction is easy; for the other direction: Geometric soln: If dim(Y ⊂ Pn) = n−1, then(WLOG) dimY0 = n−1 so that Y0 ' Speck[x1, . . . , xn]/f0 for f0 irreducible. Now, Y = V (fh0 ) can

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be checked on all affine patches. Algebraic soln: Let’s use that S is a UFD. Let p be the prime suchthat ProjS/p = Y , and so there is no (homogeneous) prime between (0) ( p. Take any nonzerohomogeneous element f ∈ p, and factor it (note that each factor is homogeneous too), so that atleast one irreducible factor g is in p. Now, (0) ( (g) ⊂ p with (g) prime so that (g) = p.

I.2.12. (a): Pullback of homogeneous prime is homogeneous prime. (b): Let’s re-index the y’s by

I = I ∈ (([n+1]

d

)). Let (y•) ∈ Z(a), so that yI 6= 0 for some I ∈ I . WLOG assume 0 ∈ I. Then

we claim thatxix0

=yI\0∪i

yIworks (and this solution is unique up to scaling). To see this, note that

for any J ∈ I , we have

∏`∈J x`

xd0=

∏`∈J yI\0∪`

ydI=

yJy0d

(where last equality follows from (y•) ∈ Z(a),

and letting J = I we have y0d 6= 0). (c): With part (b), we have that D(yid)’s (i = 0, . . . , n) cover

Z. On D(y0d), part (b) solution above shows thatk[y/y

0d’s]

a ' k[x1/x0, . . . , xn/x0]. (d): The mapis [s, t] 7→ [s3, s2t, st2, t3] so that on an affine patch Us it is [t, t2, t3], which is the twisted cubic.

I.3 Morphisms

I.3.1(c,e). (c): Let the conic be V (f) ⊂ P2. Then fz (dehomogenize at Uz) is either xy−1 or x2−y.In either case, we conclude f = xy − z2 (up to coordinate change). Hence, V (f) = ν2(P1) (thedegree 2 Veronese embedding of P1 → P2). (e): Then A(Y ) ' O(Y ) ' k, and hence dimY = 0.

I.3.3. (a): A morphism ϕ : X → Y induces ϕ∗ : O(Y )→ O(X). Define a map ϕ∗p : OY,ϕ(p) → OX,p

by [(U, f)] 7→ [(ϕ−1(U), f ϕ)]; that this is well-defined and is a local homomorphism of rings areeasy to check. (b): If ϕ is an isomorphism, its inverse morphism ψ gives natural inverse to ϕ∗p. Forthe converse, use the fact that isomorphism is local on the target, by taking an affine cover of Yand noting that Op ' Ap for affines. (c): If ϕ(X) = Y , then ϕ(ϕ−1(U)) = U for any U ⊂ Y , sothat ϕ∗ is injective.

I.3.5. Let H = V (f) ⊂ Pn with deg f = d. Then under νd : Pn → PN (N =(n+dn

)− 1), the image

of H is now a hyperplane. Thus, since νd is isomorphism onto its image, we have that νd(Pn)\νd(H)is affine, and thus Pn \H is affine.

I.3.7(b). If Y ⊂ Pn \H, then Y is affine and thus is a point (contradiction to dimY ≥ 1).

I.3.14(a). Fix a representative (p0, . . . , pn+1) ∈ kn+1 for the point P . WLOG let Pn = V (x0) ⊂

Pn+1. Then for Q = [q0, . . . , qn+1] ∈ Pn+1 \ P , we have ϕ(Q) = [p0q1 − p1q0, . . . p0qn+1 − pn+1q0],which is morphism by the Lemma below.

Lemma. Let ϕ : Y → Pn be a map and k[x0, . . . , xn] the coordinate ring of Pn. Then ϕ is amorphism iff xi

xj ϕ : ϕ−1(Uj) → k is a morphism for every 0 ≤ i, j ≤ n. In particular, a map

ϕ : Y → Pn given by n+ 1 homogeneous polynomials in yi’s (homogeneous coordinates for Y ) is amorphism. Proof: That a map is a morphism is a local property, so take distinguished affine coversof Pn and use Lemma I.3.6.

I.3.15(a,d). (a): Let X × Y = Z1 ∪ Z2. Define Xi := x ∈ X | x× Y ⊂ Zi. X = X1 ∪X2 sincex× Y is irreducible (hence for each x ∈ X we have x× Y ⊂ Z1 or x× Y ⊂ Z2). Moreover, X1 andX2 are closed since ideal membership test is an algebraic condition on the coefficients. But thenX = X1 (WLOG) so that X×Y ⊂ Z1. (d): Take transcendental bases (x1, . . . , xd) and (y1, . . . , ye)for K(X) and K(Y ). As K(X × Y ) = K(X)K(Y ), we have that (x, y) is transcendental basis forK(X × Y ). (Use Vakil’s equivalence relation definitions for transcendental basis).

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I.3.16. (a/b): Writing the coordinates of PN as

x0y0 · · · x0yn...

. . ....

xny0 · · · xnyn

, it is clear what the equations

in I(X) and I(Y ) become in these coordinates. (c): work via distinguished affine patches by notingthat Pn × Pm ⊃ Ui × Uj ' Upij ∩ σ ⊂ PN where σ is the image of the Segre embedding.

I.3.17. (a)/(b): UFD is integrally closed, and localization preserves integral closure. (c): (y/x)2−x = 0. (d): Suppose Am is normal for all m. If f ∈ K(A) integral over A, then f is integral overAm for all m, and hence f ∈ Am for all m, and thus f ∈ A. (e): If A′ ⊃ A is integral closure of A,and A → B where B is integrally closed in K(B), then A′ ⊂ B.

I.3.18(b). That the twisted quartic is normal is easy (on affine patches, it is actually isomorphic

to A1). However, it is not projective normal since t2u2 = z2

w /∈ S(Y ) and ( z2

w )2 − xw = 0.

I.3.20 (a): WLOG we can assume Y is affine. Let A = A(Y ), and m correspond to P . Then Am

is Noetherian normal domain. Now, for any prime p ⊂ m of height 1, that dimY ≥ 2 guaranteesthat p 6= m and moreover there exists m 6= m′ ⊃ p. Thus,

Am =⋂

p⊂m, ht p=1

Ap

by algebraic Hartog’s lemma. Now, if f ∈ O(Y \ P ), then f ∈⋂

m′ 6=mAm′ . Moreover, f ∈ Am byour observation above, and hence f ∈

⋂m′ Am′ = A = O(Y ).

I.4 Rational maps

I.4.4(c) Yx ' V (y2 = z + 1) and hence Y consists of points [1, t, t2 − 1] and [0, 0, 1], [0, 1, 0]. Thus,the image of the projection from [0, 0, 1] to V (z) is [1, t, 0] ∪ [0, 1, 0] ' P1. The inverse map[s, t] 7→ [s2, st, t2 − s2] is clearly a morphism. So, Y \ P is in fact isomorphic to P1.

I.4.5. Birational is easy (they both contain A2 as an open set). For non-isomorphic, note that anytwo curves in P2 intersect, whereas P1 × P1 has a family of lines whose members don’t intersect.

I.4.6. (a,b): For a1a2a3 6= 0, it is easy to see that ϕ is an involution.(c): The map ϕ is defined on P2 \ S where S = [1 : 0 : 0], [0 : 1 : 0], [0 : 0 : 1]. To see that

the map ϕ does not extend to a larger open subset U of P2, note that ClP2 = ClU = ClP2 \ S aspoints are of codimension 2 in P2. So, it ϕ extends, then the linear system on U is the same asthat of the original, i.e. kxy, yz, zx, which has three base-points (hence a contradiction).

I.4.7. WLOG, we can assume that X,Y are affine varieties. Now, taking Frac of an isomorphismθ : OY,q

∼→ OX,p gives θ : K(Y )∼→ K(X), which gives a rational map X → Y by functions

θ(yi) = θ(yi) ∈ OX,p. So the rational map is defined on an open set containing p. To see thatp 7→ q, use the following lemma:

Lemma. Suppose ϕ : X → Y is a morphism, and p, q are points of X,Y , and there is a mapof local rings OY,q → OX,p that extends ϕ∗ : O(Y ) → O(X). Then ϕ(p) = q. Proof: Supposeϕ(p) = q′ 6= q. If f ∈ O(Y ) vanishes on q then ϕ∗(f) must vanish on p. Now, restricting to U ⊂ Yaffine if necessary (so that q, q′ correspond to two different maximal ideals) we can find a functionf ∈ OY,q such that f(q) 6= 0 but f(q′) = 0 (f is a unit). Then ϕ∗(f)(p) = f(q′) = 0 so that ϕ∗(f)is not a unit in OX,p, which is a contradiction.

I.4.10. Let (x, y, t : u) be the coordinates for A2 × P1, and X := V (xu− ty) be the blow-up. Thecurve C ′ = C \ (0, 0) is equal to C \ V (x) or C \ V (y). In the ring k[x, y]/(x3 − y2), note that

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inverting x makes y invertible and vice versa, so the in either case the coordinate ring k[C ′] isk[x±, y±]/(x3 − y2). We now compute the fiber

ϕ−1(C ′) −−−−→ X ⊂ A2 × P1y yC ′ −−−−→ A2

in two charts: (i) where t 6= 0 i.e. X ∩ (A2 × Ut) and (ii) u 6= 0 i.e. X ∩ (A2 × Uu).(i) On Ut, we have k[x±, y±, ut ]/(x3− y2, xut − y) = k[x±, y±, ut ]/(x− (ut )2, y− (ut )3) so that the

closure of ϕ−1(C ′) inside A2 × Ut is the twisted cubic V (x− (u/t)2, y − (u/t)3).(ii) On the other patch Uu, we get that the closure in A2 ∩Uu is V (x( tu)2− 1, y( tu)− 1), so that

the closure does not contain a fiber over (0, 0). In fact, all the points of C belong to Ut, and themap C → C is given by k[x, y]/(x3 − y2) → k[x, y, ut ]/(x− (ut )2, y − (ut )3) ' k[ut ]. In Ex.I.3.2. weshowed that this map is homeomorphism but not an isomorphism.

I.5 Nonsingular varieties

I.5.1. They all have singularity at the origin (0,0) only. By inspecting the tangent directions at(0,0) (cf. Ex.I.5.3), we can see that (a) tacnode, (b) node, (c) cusp, (d) triple point.

I.5.3. (a): If µp(Y ) = 1 then [df ]p has rank 1, so Y is smooth at p. If µp(Y ) > 1, then [df ]p is anull matrix, so Y is singular at p. (b): node 2, triple point 3, cusp 2, tacnode 2.

Lemma (few things on length). The length of a module M over a Noetherian local ring (A,m,k)can be computed in many ways. One can find a composition series for M . Or equivalently, itis dimk(grmM) = dimkM/mM + dimkmM/m2M + · · · (if M is Artinian, then mNM = 0 forN >> 0). When M = A with A a finite type k-algebra, we are thus counting the number of(linearly independent) monomials.

Let f = xm and let n be the least such that the coefficient of yn is nonzezo in g. ThenO0/(f, g) = mn. Proof: The monomials are exactly xαyβ : 0 ≤ α ≤ m− 1, 0 ≤ β ≤ n− 1.

I.5.4. (a): Since f 6= g ⊂ p are irreducible, (f, g) is a regular sequence on Op, so that dim Op/(f, g) =0 by Krull’s height theorem. Thus, Op/(f, g) is Artinian and hence has finite length. Inequality Idont’ really know....

(b): WLOG let P = (0, 0), and let L : ax+ by = 0. If a = 0, then (L · Y )p = µp(Y ) iff fr (thenonzero homogeneous part of f of least degree) has nonzero xr coefficient. If a 6= 0 (so WLOGa = 1), then do change of coordinates x′ = x+by, y′ = y. Then the equality holds iff f ′r has nonzeroy′r coefficient (the cases when the coefficient is zero is a polynomial equation in b, so there are onlyfinite many such).

(c): Let Y = V (f) and L : V (z) where z 6 | f . Thus, g = f(x, y, 0) is a polynomial in x, y ofdegree d. Now, note that for p = (px, py, 0) ∈ Y ∩ L ∩ Ux, we have Op/(f, z) ' (k[ yx ]/fx0 )p wherefx0 is f dehomogenized and then z set to 0. Note that fx0 is gx, the dehomogenization of g. Since pis a point on Y ∩ L, it corresponds to a root of fx0 so that `(Op/(f, z)) = power of ( yx −

pypx

) in thelinear factorization of fx0 , i.e. the power of xpy − ypx in the linear factorization of g. From this itis clear that, (L · Y ) = d.

I.5.5. xd + yd + zd works as long as d 6= p. For d = p, consider f = xyp−1 + yzp−1 + zxp−1. Then[df ] = [yp−1 + (p− 1)zxp−2, zp−1 + (p− 1)xyp−2, xp−1 + (p− 1)yzp−2]. Setting [df ] = 0, and notingthat char = p, we have that the singular points are given by f = 0 and xyp−1 = yzp−1 = zxp−1. Theonly solution is x = y = z = 0 if p 6= 3. Actually, the better thing to do is this: f = xp+yp+zp+xyz.

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I.5.6(b,d). (b): WLOG P = (0, 0), so let f = xy + g define the nodal curve where deg g ≥ 3.Consider the pullback of V (f) in the blow-up X in two patches t 6= 0 and u 6= 0. In t 6= 0, wehave f = 0 and xu = y, so that plugging in we have x2u + g(x, xu) = 0. But x2|g(x, xu) sincedeg g > 2 so that we have x2(u + g(x, xu)/x2) = 0. The x2 = 0 gives the exceptional divisorE, and u + g(x, xu)/x2 = 0 gives the pullback of our curve. Here, if x = 0 then u = 0 (sincedeg g > 2), so on this affine patch the blow-up curve meets E at u = 0 (i.e. once). Moreover, thepoint (x, u) = (0, 0) is clearly smooth since ∂u(u+ g/x2) = 1. The same holds on the patch u 6= 0.

(d): Blow-up once and we get x2 − u3, the cuspidal cubic.

Lemma. Dehomogenization commutes with derivation. Precisely, if f(x0, . . . , xn) is a homoge-neous polynomial in k[x], then ∂f

∂xi(x0, . . . , xj = 1, . . . , xn) = ∂

∂xif(x0, . . . , xj = 1, . . . , xn) for i 6= j.

Proof: Both operations are k-linear, so suffices to check on monomials, for which the it is obvious.

I.5.7. (a): Let X = V (f). That X \ 0 is nonsigular follows easily from the lemma above, andmoreover, that X is in fact singular at (0, 0, 0) follows from deg f > 1.

(b): X is possibly singular at points on X ∩ E (where E is the exceptional divisor). Now, let(x, y, z, t : u : v) be coordinates on A3 × P2 and consider X ∩ Ut. The equations for Bl0(A3) isxu = ty, xv = tz, yv = uz, so that on this patch the equations we are considering is f(x, xu, xv).Now, since f is homogeneous of degree > 1, we have that f(x, xu, xv) = xdf(1, u, v). Now xd = 0gives the exceptional divisor E, and f(1, u, v) gives the strict transform. X ∩ Ut is nonsingular onpoints (x, u, v) = (0, u, v) satisfying f(1, u, v) = 0, since X is singular only at (0, 0, 0). Likewise forpatches Uu and Uv.

(c): The three affine patches are f(1, u, v), f(t, 1, v), f(t, u, 1).

I.5.8. That the rank of [J(f)] is independent of the homogeneous coordinates follows from that fi’s

are homogeneous (hence so are ∂fi∂xj

’s). By the lemma above, passing to an affine patch (WLOG)

Ux0 is the same as making the first column of [J(f)] into a zero column. This does not change therank however by the Euler’s formula.

I.5.9. Suppose f = gh is reducible (so g, h are not units). Then Z(g) ∩ Z(h) 6= ∅ by Ex.I.3.7, soconsider p in the intersection. Then Op/f is not an integral domain. However, V (f) is a nonsingularcurve, and hence Op/f must be a domain (see lemma below).

Lemma. A (Noetherian) regular local ring (A,m,k) is an integral domain. Proof: Induct onthe dimension of A. If dimA = 0, then A ' k so A is a domain. Now, if dimA = n, take aregular sequence (x1, . . . , xn) (exists by regular-ness). Now, consider A′ = A/x1. By Krull’s heighttheorem, dimA′ = n− 1, and moreover m/m2 ' (m/m2)/(k · x1), so that A′ is in fact regular localring as well. Thus, A′ is a domain. Now, suppose a, b ∈ A nonzero such that ab = 0. By Krull’sintersection theorem, we can write a = a′xm1 and b = b′xn1 where a′, b′ /∈ (x1). Since x1 is a nonzero

divisor, we thus have a′b′ = 0. However, this implies that a′, b′ 6= 0 but a′b

′= 0.

I.5.10(b,c). (b): A morphism ϕ : X → Y mapping p 7→ q induces a local ring map ϕ∗p : OY,q →OX,p. Let n,m be maximal ideals corresponding to p, q. Then we have a map of OY,q-modulesn → m, from which we get a map n/n2 → m/m2 of OY,q-modules and hence OY,q/n ' k-vectorspaces. Taking the dual, we get (m/m2)∨ → (n/n2)∨. More concretely, taking an affine open charts,suppose X,Y are affine so that the ring map is k[y1, . . . , ym]/I(Y )→ k[x1, . . . , xn]/I(X) given byyi 7→ fi(x). Actually, while this can be done, it will have to wait a bit until we hit differentials...(c): The map before taking duals is ψ : (x)/(x2) → (y)/(y2) which comes from restriction ofk[x] → k[x, y]/(x− y2). This is a zero map, and so is the dual thus.

I.5.11. The two equations are x2−xz = yw and yz = (x+z)w. Eliminating w, we get y2z−x3+xz2,the equation for the elliptic curve E ⊂ P2. For any points (x : y : z) ∈ E satisfying such that y 6= 0

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or x + z 6= 0, there exists a unique point (x, y, z, w) projecting down to (x, y, z) via ϕ. However,the fiber of ϕ over [1, 0,−1] (i.e. y = 0 and x + z = 0) is empty (the first equation then reads2 = 0). Thus, ϕ : Y \P → E \ [1, 0,−1] is a bijection. Moreover, we know check that the map is anisomorphism on two local patches y 6= 0 and x+z 6= 0. Let’s do the (harder) x+z 6= 0 case: on Ux+z

the equation y2z−x3 +xz2 = 0 becomes y2(1−x)−x(2x−1) = 0, whereas the fiber ϕ−1(E∩Ux+z)is affine variety in Ux+z ' A4 given by equations x(2x− 1) = yw and y(1− x) = w (so plugging inthe second equation to first) we have that P2 ⊃ E ∩ Ux+z ' Y ∩ Ux+z ⊂ P3 isomorphism.

I.5.12. (a): Follows immediately from Sylvester’s law of inertia (whose proof is essentially asmart Gram-Schmidt process). (b): Use that if f factors, it must be two linear (homogeneous)polynomials. (c): The singular locus is given by V (x0, . . . , xr) (cf. Ex.I.5.8). (d): Quite obvious,once in the right coordinates (think “reverse of projection map”).

I.5.13. The statement is local, so WLOG assume that Y is an affine variety with A := A(Y ). LetA′ be integral closure of A where A is a finite-type k-algebra domain. Theorem I.3.9A implies thatfor some f1, . . . , fm ∈ A′, the map A[f1, . . . , fm] → A′ is in fact surjective (hence isomorphism).Now, since fi = ai/bi for some ai, bi 6= 0 ∈ A, consider Ab1···bm . Clearly, A′ ⊂ Ab1···bm , and in fact,A′b1···bm ' Ab1···bm . Thus, Ab1···bm is integrally closed. This shows that the nonnormal points areproper.

To show that normal points are in fact open, suppose m ⊂ A corresponds to a normal point.Since (A′)m is the integral closure of Am, we have that the map A → A′ induces an isomorphismAm

∼→ (A′)m. By Ex.I.4.7, we have that there is an open set of Y containing m that is isomorphicto an open subset of Y , the normalization of Y .

I.6 Nonsingular curves

Skip for now.

I.7 Intersections in Projective Space

I.7.1. (a): Let Y ⊂ PN be the image of the Veronese embedding of Pn. Then S(Y ) ' k[x1, . . . , xn]d•,so that PY (z) = PPn(dz) =

(dz+nn

)= dn

n! zn + · · · . (b): If Y ⊂ PN is the image of the Segre embed-

ding of Pr × Ps, then S(Y ) '⊕

i≥0(k[x]i ⊗ k[y]i) as graded rings. Hence, PY (z) = PPr(z)PPs(z) =(z+rr

)(z+ss

)= 1

r!s!zr+s + · · · .

Lemma. Let S(Y ) and S(Z) be coordinate rings for projective varieties Y ⊂ Pn and Z ⊂ Pm. Thenthe coordinate ring for Y × Z as a subvariety of the Segre embedding of Pn × Pm is isomorphic to⊕

i≥0(S(Y )i⊗S(Z)i) as graded rings (k-algebras). In particular, PY (z) ·PZ(z) = PY×Z(z). Proof:The ring S(Y × Z) is the image of the map k[zij ’s]• → S(Y )• ⊗k S(Z)• where zij 7→ xi ⊗ yj .

I.7.2(c,d,e). (c): From 0→ S(−d)f→ S → S(H)→ 0, we have PH(z) =

(z+nn

)−(z−d+nn

)so that

pa(H) = (−1)n−1(1 −(−d+n

n

)− 1) = (−1)n (n−d)(n−d−1)···(n−d−n+1)

n! = (−1)2n (d−1)···(d−n)n! =

(d−1n

).

(d): Let f, g polynomials defining hypersurfaces H1, H2 of degree a, b. Then, from 0→ S/f(−b) g→S/f → S/(f, g) → 0, we have PH1∩H2(z) = PH1(z)− PH2(z − b) so that pa(H1 ∩H2) =

(a+b−3

3

)−(

a−33

)−(b−3

3

)= 1

6(3ab(a+ b)− 12ab+ 6) = 12ab(a+ b− 4) + 1. (e): Follows immediately from the

lemma above.

I.7.3. Let Y be given by a polynomial f , and suppose p = (p0 : p1 : p2) ∈ Y . Let L be the linea0x0 + a1x1 + a2x2 = 0 going through the point p (i.e. ~a · ~p = 0). WLOG, let p0 6= 0. Then, since

9

a0 = −a1p1+a2p2p0

, all the lines through the point p are given exactly by a1(x1x0 −p1p0

)+a2(x2x0 −p2p0

) = 0(where a1, a2 not both zero), and so we can work in the affine chart U0. Now, apply the followinglemma:

Lemma. If f(x, y) is a polynomial and P = (p, q) ∈ V (f) ⊂ A2, and L : a(x− p) + b(y − q) = 0is a line through (p, q), then the intersection multiplicity V (f) ∩ L at P is the multiplicity of theroot X = 0 of the polynomial f(X + p,−a

bX + q) (if b 6= 0). In particular, if V (f) is nonsingularat P , then I(V (f), L;P ) > 0 iff (a : b) = (fx(P ) : fy(P )).

Proof: The first statement follows from the fact that k[x](x) is a DVR. For the second statement,WLOG let (p, q) = (0, 0). Then f = fx(P )x + fy(P )y + (higher order). When using ax + by = 0to do the coordinate change, the linear term vanishes iff (a : b) = (fx(P ) : fy(P )). Otherwise, thelinear term survives and the multiplicity of zero remains 1.

I.7.4. By Bezout’s theorem and Ex.I.7.3 above, a line L intersects Y ⊂ P2 at exactly d points if itis not tangent to any point of Y and does not go through singular points of Y . Since dimY = 1,we have dim Sing Y = 0, so that Sing Y is finitely many points ~p1, . . . , ~pm. Thus, the set of linesthat go through singular points or is tangent to a point of Y is given by the union of the morphismof Ex.I.7.3 (which is a proper closed subset) and

⋃i V (~pi · ~x).

I.7.5(b). Note that if p ∈ Y ⊂ P2 is a point of multiplicity m, then a generic line through p meetsY at p with multiplicity m (i.e. there is an dense open subset U of P1 ' lines through p such thatif L ∈ U then i(Y,L; p) = m. Now, WLOG let P = (0 : 0 : 1) ∈ Y be the point with multiplicityd − 1. Then we can consider two maps: (1) ϕ : Y \ P → P1 which is the projection map and (2)ψ : P1 ⊃ U → Y \ P which maps a point L of U to the unique point that L meets Y other than P(uniqueness guaranteed by Bezout’s theorem and the above observation). In sum we have:

Y \ P ϕ→ P1 ψ99K Y \ P

(a : b : c) 7→ (a : b : 0) 7→ (a : b : c)

It is clear that these are rational (dominant) maps, and thus establishes a birational equivalence ofY \ P and P1.

I.7.6. Proposition I.7.6(a,b) implies that Y is in fact irreducible, so suppose Y is a variety. Also,r = 0, n case is trivial, so we assume 1 ≤ r ≤ n− 1.

First, suppose dimY = 1. Take P 6= Q ∈ Y and let L = PQ. If there exists R ∈ Y such thatR /∈ L, we can find a hyperplane H such that H ⊃ PQ and H 63 R. Then since deg Y = 1, Bezout’simplies that Y ∩H must be a degree 1 variety with dimension 0 (dimension by Krull). But Y ∩Hcontains two points P,Q and hence is a contradiction. Hence, PQ = Y .

Now, let dimY = r. By induction hypothesis, for any H 6⊃ Y , we have that Y ∩H is a linearvariety of dimension r− 1. By dimension argument, we can find two hyperplanes H1, H2 such thatZi = Y ∩Hi (i = 1, 2) are two different linear (r− 1)-planes. Moreover, by Krull’s height theorem,Z1∩Z2 = Y ∩H1∩H2 must have dimension r−2. Hence, Z1∩Z2 as a k-vector space has dimensionr−1 so that Z1 +Z2 as a k-vector space has dimension r+1. Thus, Z1Z2 := Z1 +Z2 as a projectivelinear variety has dimension r. Now, if there exists P ∈ Y such that P /∈ Z1Z2, then since Z1Z2

has dimension r < n we can find a hyperplane H such that H ⊃ Z1Z2 but H 63 P . Once againY ∩H must be a variety of dimension r − 1, but it contains Z1 and Z2, which is a contradiction.Hence, Y = Z1Z2, as desired.

Alternatively, one can show that for since Y ∩ H is linear (or Y ) for any hyperplane H, Y ∩linear variety is always linear. Hence, for any two points p, q ∈ Y , we have Y ∩ pq being a linear

10

variety (hence is pq). Thus, since any line through two points on Y is in Y , we have that Y mustbe linear.

I.7.7. Again, none of the answers online are satisfactory to me... Maybe I’m being stupid...

I.7.8. Follows immediately from Ex.I.7.7. and Ex.I.7.6. It’s just quite cool that a degree 2 varietyin Pn is thus in fact a complete intersection.

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Chapter II

Schemes

II.1 Sheaves

II.1.1 A bit of category nonsense...

that Hartshorne doesn’t really do.

An additive category is a category C such that (i) the Hom-sets are Abelian groups such thatcomposition is a Z-bilinear map, (ii) the zero object exists, and (iii) finite direct product / coproductexist.

A few easy facts regarding additive categories:

• finite direct product is isomorphic to coproduct.

• ker/coker of mono/epi-morphism is the 0 map (0→ Aϕ→ B if ϕ monomorphism)

• ker/coker of 0 map (A→ 0 / 0→ A) is itself, and conversely if ker / coker of a map is itselfthen the map is a 0 map.

• ker/coker is a mono/epi-morphism

What may not true is that: (∗) if ker(ϕ : A → B) = 0 or cokerϕ = 0, then ϕ is a monomorphismor epimorphism (respectively) Find an example of this?

An Abelian category is an additive category C such that (i) kernels and cokernels exist, and(ii) coimage (cokernel of kernel) is isomorphic to image (kernel of cokernel) for any morphism.Equivalently, one can say (ii) by saying (a) every monomorphism is kernel of its cokernel, and (b)every epimorphism is cokernel of its kernel.

The relevant diagram is:

ker θi−−−−→ A

θ−−−−→ Bπ−−−−→ coker θy x

coim θ∼−−−−→ im θ

Now the notion of exact sequences make sense, and the statement (∗) above is true. For example,

if Aϕ→ B

0→ 0 is exact, then B = ker 0 = imϕ = ker(B → cokerϕ), so that cokerϕ = 0 (so far

12

we didn’t use Abelian-ness). To conclude that ϕ is epimorphism, we note that the diagram abovebecomes:

ker θi−−−−→ A

ϕ−−−−→ Bπ−−−−→ 0y xId

coimϕ∼−−−−→ B

so that A→ coimϕ ' B is a cokernel map and hence an epimorphism.

Let’s show that sheaves of Abelian groups and more generally OX -modules form Abelian categories.Let’s not be careful about which one exactly and say the category ShX of sheaves on a topologicalspace X. That ShX is an additive category is easy to check. Moreover, that kernels and cokernelsexist is also easy.

N.B. Hartshorne defines =(ϕ : F → G ) by sheafifying the presheaf =ϕ(U) := =ϕ(U)(F (U)). Tosee that this matches the category theoretical definition, we note:Lemma.If π : F → G is a map of sheaf to a presheaf, and π+ : F → G + is the map π composedwith sheafification of G , then kerπ+ ' (kerπ)+.

The proof is just universal property nonsense. The lesson is that to say ker(π : F → G ) is asheaf, we need both F and G to be sheaves.

With these set, to show that coimage is isomorphic to image note that if F ' G as presheavesthen F+ ' G + naturally as sheaves. So, we can check on the presheaf coimage and image theisomorphism, reducing our case to statements about coimage and images in AbGrp, R-Mod, etc.

And... there we have it! Sheaves of Abelian groups or modules is an Abelian category, so thatexact sequences make sense, and kernels, cokernels, images, and coimages as defined in Hartshornematch categorical theoretical viewpoint. In fact, the terms “injective” and “surjective” is nowbetter replaced with “monomorphism” and “epimorphism.”

II.1.2 Back to exercises...

The above discussion essentially takes care of the following exercises: II.1.6, II.1.7, II.1.9.

II.1.2. (a): Let’s just do everything concretely. From kerϕi→ F

ϕ→ G , we have a natural

map (kerϕ)pip→ Fp

ϕp→ Gp. The map ip takes [(f, U)] ∈ (kerϕ)p and treats it as an elementof Fp. Injectivity is clear, and if [(f, U)] ∈ kerϕp then ϕ(f)|V = 0 for some V ⊂ U so that[(f, U)] = [(f |V , V )] ∈ (kerϕ)p since f |V ∈ kerϕ(V ). For imϕ, since im = ker of coker, use the factthat cokernel commutes with colimits and apply the results we just proved.

(b): The sheafification construction makes it clear that for a sheaf F , we have Fp = 0 ∀p ∈ Xiff F = 0. Our category theory definition makes it clear that ker = 0 or coker = 0 implies injectiveor surjective. Hence, via part (a), that ϕ is injective iff ϕp is injective ∀p follows. Note thatcokerϕ = 0 iff imϕ = G , and if (imϕ)p = imϕp = Gp for all p, then (cokerϕ)p = 0 for all p. (Herewe secretely use that cokerϕp = (cokerϕ)p.)

(c): Follows easily from (a).

II.1.3. (a): If the condition holds, then ϕp is surjective for all p. Conversely, if ϕ is surjective,then ϕp is surjective for all p, so for each sp we can find [(tp, Vp)] ∈ Fp that maps to sp such that(shrinking Vp if necessary) we have ϕ(tp) = s|Vp for each p.

(b): No idea yet.

13

II.1.6. (a): Since F/F ′ is the cokernel of F ′ → F , and F ′ → F is a monomorphism (which is akernel of a cokernel ). Piece together category stuff. (b): Same thing; piece together category stuff.

II.1.13.(Espace Etale). The topology on Spe(F ) can be described more concretely as: V ⊂ Spe(F )is open iff for every s : U → Spe(F ) the set s−1(V ) is open in U . This means that at each pointp ∈ X, the set of germs of continuous at p is exactly Fp This is mysteriously hard... Ask?

Now, an element t ∈ F+(U) is by definition a map t : U → Spe(F ) such that π t = IdUand locally at each point p ∈ V ⊂ U , the map t|V is equal to tp for tp ∈ F (V ), and hence t iscontinuous. Conversely, if t : U → Spe(F ) is a continuous section, then it is locally a section ofF by the observation above.

II.1.16.(Flasque sheaves) (a): First, note that if U is connected, then A(U) = A. If X is irreducible,then any open subset is connected.

(b): Let ϕ : F → F ′′. We need a lemma:

Lemma. Let t1 ∈ F (U1) and t2 ∈ F (U2) such that ϕ(t1)|U1∩U2 = ϕ(t2)U1∩U2 . Then there existst ∈ F (U1 ∪ U2) such that ϕ(t|Ui) = ϕ(ti), i = 1, 2. Proof: Since t1|U12 − t2|U12 ∈ kerϕ(U12), andkerϕ is flasque, the element lifts to t′ ∈ kerϕ(U1). Now, considering t1 − t′ and t2 (which agree onintersection now), by sheaf condition we obtain t satisfying the lemma.

Now, let s ∈ F ′′(U). Let Σ be a poset of pairs (Ui, ti ∈ F (Ui)) where ϕ(ti) = s|Ui ordered bycompatible inclusion. Given any chain, sheaf condition ensures that the union of the chain is theupper bound. Hence, Σ has a maximal element. Now, by Ex.II.1.3 there exists ti ∈ F (Ui) with⋃i Ui = U and ti 7→ s|Ui , which implies that the maximal element of Σ must be (U, t) for some

t ∈ F (U).(c): Easy using (b). (d): Immediate from definition of pushforward.

II.1.19(c). 0→ j!(F |U )→ F → i∗(F |Z)→ 0: check on stalks using parts (a) and (b).

II.1.20. (a): That H is a presheaf is clear. Now, if si ∈ H (Vi) for⋃i Vi = V are sections

agreeing on intersections, then it uniquely patches to s ∈H (V ), and the support is still containedin V ∩ Z since sp = (si)p for p ∈ Vi.

(b): It is clear that H = kerϕ where ϕ : F → j∗(F |U ).

II.2 Schemes

Lemma. Taking nilradical commutes with localization. Precisely, denote by N(A) the nilradicalof A, then N(S−1A) = S−1N(A). Proof: ⊃ is easy. For ⊂, if (a/s)n = 0, then tan = 0 ∃t ∈ S sothat (ta)n = 0 and hence ta ∈ N(A). Thus, a

s = tats ∈ S

−1N(A).

II.2.3.(Reduced schemes) (a): The above lemma implies that an affine scheme U = SpecA isreduced iff Ap is for all p ∈ U . Moreover, reduced-ness is clearly an affine local property. AlternateSoln: Using the definition that OX(U) := compatible s : U →

∐OX,p, if all OX,p are reduced

then OX(U) is for any U , and if U is reduced then so are OX,p.(b): Some fanciness: Let NX be the sheaf (an OX -module) associated to presheaf defined by

NX(U) := N(OX(U)). The lemma above implies that NX is a quasi-coherent sheaf and that

0→ NX → OX → OX/NX → 0

is an exact sequence of quasi-coherent sheaves, with OXred = OX/NX . This is all to say that ifU ⊂ X is an affine subscheme then OXred(U) = OX(U)/N(OX(U)), and thus letting A := OX(U)we have (U,OXred|U ) ' (SpecA/N(A),OSpecA/N(A)) (again by Lemma above). That there is a

14

morphism Xred → X is clear: the topological map is just homeomorphism (since quotienting bythe nilradical doesn’t change the topology in the affine patches) and the map of sheaves is the oneabove OX → OX/NX .

(c): We are looking at:

Y

Xfoo

∃!Yred

The topological map is clear. For the sheaf maps, we have

NY// OY

// f∗OX

OY red

::

where the top two maps compose to 0 map since X is reduced. Hence, by universal property ofcokernel, there exists a map OY red → f∗OX .

Lemma. Let X,Y be locally ringed spaces, and suppose there is a cover Ui of X with map

of locally ringed spaces (fi, f#i ) : Ui → Y such that fi|Uij = fj |Uji . Then the maps fi glue to a

morphism f : X → Y . Proof: The topological maps indeed glue to a topological map f : X → Y .For the map of sheaves, f# : OY → f∗OX is given via the following diagram: For any U ⊂ Y wehave

0 // OY (U)

∃!

//∏i OY (U ∩ Ui)

f#i

////∏i,j OY (U ∩ Ui ∩ Uj)

0 // OX(f−1(U)) //

∏i OX(f−1(U ∩ Ui)) ////

∏i,j OX(f−1(U ∩ Ui ∩ Uj))

Since f#|Ui = f#i , it is clear that f#

p is a local homomorphism for all p ∈ X.

Lemma. Let X,Y be locally ringed spaces, and ϕ,ψ : X → Y morphisms. Suppose there is acover Ui of X such that ϕ|Ui = ψ|Ui for all i. Then in fact ϕ = ψ. Proof: Use the ! part of theexistence of the downward map above.

Lemma. Let SpecA,SpecB ⊂ X be two affine opens in scheme X with nonempty intersection.Then for any p ∈ SpecA ∩ SpecB, there exists an open subset U of the intersection containingp such that SpecAf ' U ' SpecBg for some f ∈ A, g ∈ B. Proof: Take a distinguished affineopen of A in the intersection, and then a smaller one inside that is distinguished in B. Now useEx.II.2.16(a).

II.2.4. We construct the inverse map of α as follows. Cover X by affines Ui = SpecBi. Given aring map A→ OX(X), we have maps ϕi : A→ OX(Ui), which gives us morphisms fi : Ui → SpecA.Moreover, for any p ∈ Ui ∩ Uj take an affine Uij = SpecBij around p distinguished in both Ui andUj (possible by lemma above). Then the commutativity of the diagram

A −−−−→ OX(Ui)y yOX(Uj) −−−−→ OX(Uij)

15

imply (via lemma above) that fi’s agree on the intersections. Hence, they glue to a unique morphismf : X → SpecA.

II.2.7. We prove a more general version. Note that for any x ∈ X there is a natural morphismSpec OX,x → X. We claim that for any local ring (A,m), giving a morphism SpecA → X isequivalent to giving a point x ∈ X and a local ring map OX,x → A. That is, there is a canonicalmap Spec OX,x → X such that:

Spec OX,x

Spec(A,m)

∃!77

// X

Proof: If x ∈ X is the image of m under the map ϕ : SpecA → X, then for any open setx ∈ U ⊂ X and p ∈ SpecA, we have ϕ(p) ∈ U since continuity of ϕ implies that ϕ(p) ⊂ ϕ(p) sothat x ∈ ϕ(p). So, take U to be any affine SpecB ⊂ X. Then the map ϕ factors through SpecB, sothat we get a map B → A where m pulls back to the point corresponding to x. Hence, by universalproperty of localization, we get a natural map (of local rings) OX,x → A.

In the case where A is a field, the kernel of the map is mx so that the map extends to κ(x)→ A.

II.2.8. Let ϕ : Speck[ε]/ε2 → X. The claim in the previous exercise Ex.II.2.7. shows that thisis the same as giving a local homomorphism OX,x → k[ε]/ε2 for some x ∈ X. To see that x isrational over k, note that since X is a k-scheme OX,x, so that the extension of the local map aboveκ(x) → k must be an isomorphism. Lastly, note that local map OX,x → k[ε]/ε2 induces a map ofκ(x) = k-vector spaces mx/m

2x → (ε)/(ε2), whose dual map gives an image of ε∨ in Tx.

II.2.11. Later after NT or FT

II.2.14. (a): If every element of S+ is nilpotent, then for any p homogeneous prime and anyhomogeneous f ∈ S+ we have fn = 0 ∈ p so that p ⊃ S+. If there is an element f ∈ S+ that is notnilpotent, then S(f) is not a zero ring, such that D+(f) ⊂ ProjS• is nonempty.

(b): U = ProjT \ V (ϕ(S+)) by definition is open. Moreover, it has an affine open cover byD+(f) ⊂ U | f ∈ ϕ(S+) homogenous. Since S → T naturally extends to S(f) → T(ϕ(f)), wecan define morphisms ψf : D+(f) → ProjS for each f ∈ ϕ(S+), and these maps agree on theintersections since S(f) → (S(f)) gdeg f

fdeg g

' S(fg) commutes with ϕ.

(c): If p ⊃ S>d for some d, then p = S+ in fact. Hence, for any d > 0, we have an open coverD+(f) | f ∈ S>dhomogeneous of ProjS. This first shows that if S → T is a map of gradedrings that is isomorphism after high enough degrees, then U = ProjT , and moreover, the mapsϕf : D+(f) → ProjS is an isomorphism since S(f) → T(ϕ(f)) is (since deg f ≥ d0). Now useEx.II.2.17(a) below.

(d): Patch isomorphisms t(V ∩ Uxi) ' SpecS(xi) together.

II.2.15. (a): WLOG V is affine so that t(V ) = SpecA for A = k[x1, . . . , xn]/I, and note thatlocalization commutes with quotients. Now, if the residue field is k, then we have k → (A/p) →(A/p)p ' k so that k = A/p already. Now, if P is a closed point then Strong Nullstellensatz impliesthat κ(P ) = A/p is a finite (and hence algebraic) extension of k, and hence κ(P ) = k since k = k.

(b): We have a natural map of local rings OY,f(P ) → OX,P . Now, it is easy to check that ifB → A is a local k-algebra homomorphism and A/m ' k then B/n ' k.

(c): Injectivity is easy. Show the surjectivity for V,W affine using (a,b). Patch.

II.2.16(c,d). Just to get rid of trivialities, we assume that f is not nilpotent throughout.

16

(c): Write U ′i := Ui ∩ Xf , and let Ai = OX(Ui). Then since X since index i is finite, thereexists n and ai ∈ Ai’s such that fnb|U ′i = ai. Now, note that for any i, j we have (ai − aj)|Ui∩Uj

vanishes on Ui ∩Uj ∩Xf . Now, since Ui ∩Uj are all quasi-compact, and there are finitely pairwiseintersection, there exists m such that replacing ai by fmai we have ai = aj on intersections. Thus,fn+mb is an restriction of some a ∈ A by gluing together ai’s.

(d): Note that there is always a natural map Af → Γ(Xf ,OX) since f restricted to Xf isinvertible (glue inverse of f , which is just 1/f , from each affine pieces). Now, this map is injectiveif X is quasi-compact (∵ (b)), and is surjective if X further satisfies condition in (c) (∵ (c)).

II.2.17. (a): That f is a homeomorphism is clear, and that the sheaf map is isomorphism is clearsince stalk maps are isomorphisms.

(b): There is a natural morphism π : X → Spec OX(X). If f1, . . . , fn ∈ OX(X) such thatSpec OX(X)fi ’s cover Spec OX(X), then Xfi = π−1(D(fi))’s also cover X. And moreover, sinceXfi ’s are affine, we have Xfi∩Xfj = Spec OX(Xfi)fj so that the intersections are also quasicompact.

Hence, by Ex.II.2.16(d) we have Xfi ' Spec OX(X)fi , so that π|π−1(D(fi))’s are all isomorphisms.Applying (a) above, we have that X ' Spec OX(X).

In fact, the object Spec OX(X) has the following universal property: For any morphism X →SpecA, there exists a unique map Spec OX(X)→ SpecA such that

X

%%

// Spec OX(X)

∃!

SpecA

With this few, using Vakil’s ACL, it is easy to show the lemma below from (b) above.

Lemma. The property of a morphism being affine is LOCT.

II.3 First properties of schemes

II.3.7. Take an affine SpecB ⊂ Y and SpecA ⊂ f−1(SpecB) so that we have f |SpecA : SpecA→SpecB given by ϕ : B → A. Since X,Y are integral, A,B are domains. Moreover, since fis dominant, the generic point must go to the generic point, and hence ϕ is an injection. LetK(A) := Frac(A) = K(X) (likewise for B). Moreover, since f is of finite type, A = B[a1, . . . , am]for some ai ∈ A. Thus, the fiber over the generic point is B[a1, . . . , an]⊗BK(B) ' K(B)[a1, . . . , an].For this to be a finite set, dimK(B)[a1, . . . , an] = 0, and hence tr degK(A)/K(B) = 0. Thus, byNoether normalization, we have that K(A) is a finite extension of K(B). Localizing B by all thedenominators that show in the monic polynomials (in K(B)) that the ai’s satisfy, we have thatBf → Bf [a1, . . . , an] = Aϕ(f) is a finite ring map.

Now, since Y is irreducible, WLOG let Y = SpecB be affine, and moreover, shrinking SpecBif necessary, we can assume that f−1(SpecB) can be covered by finitely many affines Ui := SpecAi

such that Ui → Y is a finite morphism. We finish with the following lemma:

Lemma. Suppose f : X → SpecB is an morphism of integral schemes such that X is covered byfinitely many affines Ui := SpecAi such that Ui → SpecB is an integral morphism. Then thereexists b ∈ B such that f : Xb → SpecBb is an affine morphism.

Proof: Since X is irreducible, W :=⋂i Ui is nonempty. We can find an affine open V ⊂W that

is distinguished in each Ui. In other words, V = SpecAiai ∀i for some choices of ai ∈ Ai. Since aiis integral over B, let bi be the nonzero constant term of the monic polynomial in B[t] that ai is a

17

zero of. Note that ai ∈ p ⊂ Ai =⇒ bi ∈ p so that SpecAibi ⊂ SpecAiai . Thus, the inverse image

of SpecBb where b =∏i bi is SpecAib (for any i) so that Xb → SpecBb is a morphism of affine

schemes (and is still finite type / integral if f was).

II.3.10. (a): Categorical nonsense reduces to the case where we have SpecA→ SpecB and p ⊂ B.Then SpecA ×SpecB Specκ(p) = Spec(A ⊗B κ(p)) which is made by localization and a quotient,which are both topological inclusions.

(b): For the last part, we compute that k[s, t]/(s − t2) localized by multiplicative set S =k[s] \ 0 gives us k(s)[t]/(s− t2), which is a degree 2 extension of k(s).

II.3.11. We actually do (b) first then (a). We need a short lemma: If f : Y → X is a closedimmersion, and U ⊂ X, then the fiber product f−1(U) = U ∩ Y → U is also a closed immersion.Proof: That the map is homeomorphism onto a closed subset is clear, and map of stalks aresurjective since it doesn’t change from f#.

(b): Note that Y is quasicompact. We cover it by affines of the form D(fi) ∩ Y for fi ∈ A.For any point p ∈ Y , take an affine U 3 p. Topologically and scheme theoretically (by Ex.II.2.2),U = U∩Y for some U ⊂ X open. Now, take D(f) ⊂ U , and note that D(f)∩Y = U∩Yf = (U∩Y )f(by Ex.II.2.16). In other words, we have:

(U ∩ Y )f −−−−→ U ∩ Y −−−−→ Yy y yD(f) −−−−→ U −−−−→ X

where all squares are fiber diagrams and right arrows are open embeddings, so that all the downarrows are closed embeddings (by the short lemma above). Covering Y by these affines, takingfinite subcover, and then extending to cover of X, we have that f : Y → X is an affine morphism.Lastly, Ex.II.2.18(d) gives f# : OX(X) → OY (Y ) to be a surjection, so that OY (Y ) = A/I forsome I.

(c): We have f : Y → X and f ′ : Y ′ → X. Note that topologically, f and f ′ are the same. Forany U = SpecA ⊂ X, we have that f−1(U) = SpecA/

√I and (f ′)−1(U) = SpecA/I. Since there

is a natural map such that

A/√I Aoo

A/I

∃!

OO

which defines a map f−1(U) → (f ′)−1(U). These maps agree for any affines U, V ⊂ X, so theypatch up to give a morphism Y → Y ′.

(d): Topologically, Y = f(Z) clearly. If Z is reduced or quasicompact, we show that we canmake Y affine locally by constructing a quasicoherent sheaf of ideals of OX . For U ⊂ X affine,define I (U) := ker(OX(U) → OZ(f−1(U))). For g ∈ OX(U), that I (Ug) ' I (U)g follows fromthe diagram below:

I (U) −−−−→ OX(U) −−−−→ OZ(f−1(U)) −−−−→∏iAiy y y y

I (Ug) −−−−→ OX(Ug) −−−−→ OZ(f−1(Ug)) −−−−→∏i(Ai)g

The universal property follows easily from the construction.

18

In general, given a map f : Z → X, we have a map of sheaves OX → f∗OZ , and let I be thekernel. In the case when Z is qcqs, f∗OZ is in fact quasicoherent, so that I is already quasicoherent.In general, define IY :=

∑I ′⊂I QCoh I ′, which is QCoh, which defines closed subscheme Y with

the desired universal property.

Lemma. Let X be a finite type k-scheme. If l ⊃ k is an algebraic extension of k, then a l-valuedpoint Spec l → X (over k) is a closed point. In particular, the closed points of X are x ∈ Xsuch that κ(x) is an algebraic extension of k. Proof: For any map ϕ : k[x]/I → l, note thatA = k[x]/ kerϕ is a k-algebra domain of finite type that injects into l (an algebraic extension ofk) and hence is a field. Thus, the point that corresponds to a l-valued point is a maximal ideal inevery affine that contains it, so it is closed.

II.3.14. By the lemma, we may assume that k is algebraically closed. But in this case, for anyaffine piece U ⊂ X given by k[x]/I, there are k-valued points of U by Nullstellensatz.

II.3.16. We need show that every proper closed subset Y of X has property P. Suppose P doesnot hold for some Y0 ( X closed. Then there exists a closed proper subset Y1 ( Y such that P

does not hold, and so forth, so that we get a chain of closed subsets Y0 ) Y1 ) · · · . But Noetheriancondition implies that this chain must stabilize, which is a contradiction.

II.3.17–19. Some other time... When I need it...

II.3.20. (a): We first show that for any closed points p, q ∈ X, we have dim Op = dim Oq. Note thatthis is true for two closed points in an affine integral finite type k-scheme. Now, take affines U, Vcontaining p, q respectively, and take a closed point in an affine open in the intersection U ∩V , andthen note that number equality is transitive (lol). Now, if X = Z0 ) · · · ) Zd = P is a the maximalsequence of irreducibles giving the dimension of X, take and affine U = SpecA containing P . ThendimA = dimX (use X irreducible to argue that Zi ∩ U ) Zi+1 ∩ U ∀i), and dimA = dim Op.Combined with dim Op = dim Oq for any two closed points, we have that dim Op = dimX for anyclosed point p ∈ X.

(e): If SpecA ⊂ U , then from part (a) we know that dimX = dimA ≤ dimU ≤ dimX.(b): Take any SpecA ⊂ X. Then dimX = dimA = tr degk Frac(A) = tr degkK(X).(c): Suppose ξ is the generic point for an irreducible closed subset Z ⊂ X. Take Z = Z0 (

· · · ( Zc = X the maximal chain that gives c = codim(Z,X). Take an affine open SpecA aroundZ0, so it contains all points corresponding to Zi’s. If p ⊂ A corresponds to Z, then c = dim OX,p.We are done since codim(Y,X) := inf codimZ⊂Y (Z,X) = infdim Oξ(Z),X |ξ(Z) ∈ Y

(d): From the previous two parts, we can assume X affine. And for any irreducible closed Y ,this is true, hence true for general Y (dim is sup and codim is inf).

(f): No idea... ASK

II.3.21. For concreteness, let R = k[x](x) (in fact, R ' k[x](x) for k its residue field under ourcondition). Then R[t] ' k[x](x) ⊗k k[t]. Now, localizing by x gives k(x)[t] and quotienting by xgives k[t]. That is, SpecR[x] consists of a closed subscheme k[t] and an open subscheme k(x)[t].

II.3.22. Hold for now... Return to this later.

II.4 Separated and proper morphisms

II.4.1. Affine morphisms are separated. Finite maps are of finite type. And finite maps are closed,and are preserved under base change.

19

Lemma. Suppose X is reduced, and Z is a closed subscheme of X containing an open densesubset U of X. Then in fact X = Z. Proof: Since U is reduced, the scheme-theoretic image of Uis U red = X since U is dense and X is reduced. Thus, by universal property of scheme-theoreticimage, there exists a unique j such that

Z // X

X

j

OO

Id

>>

from which we can deduce that Z ' X.

Warning: This is NOT true if X is not reduced. For example, consider X = Speck[x, y]/(x2, xy),Z = Speck[x, y]/(x), and U = k[y±] '

(k[x, y]/(x2, xy)

)x+y' (k[x, y]/(x))x+y.

Lemma. Suppose f, g : X → Y are morphisms of S-schemes, and let h : X → Y ×S Y be the mapinduced from f, g. Denote by h−1(∆) (suggestively) the fiber product:

h−1(∆) −−−−→ Yy y∆

Xh−−−−→ Y ×S Y

Then f = g iff h−1(∆) = X. Proof: Universal property nonsense.

II.4.2. (a): Since Y is separated over S, the fiber (as in lemma above) h−1(∆) → X is a closedembedding. Moreover, we have chain of fiber diagrams:

U −−−−→ h−1(∆) −−−−→ Y∥∥∥ y y∆

U −−−−→ Xh−−−−→ Y ×S Y

so that by the first lemma above h−1(∆) = X, and hence by the second lemma f = g, as desired.(b): X not reduced: consider k[x, y]→ k[x, y]/(x2, xy) by x 7→ x, y 7→ y and x 7→ x2 = 0, y 7→ y.

Y not separated: map A1 to upper and lower affine line with double origin.

Lemma. (Magic square) Let Y → Z,X1 → Y,X2 → Y be morphisms. Then

X1 ×Y X2 −−−−→ X1 ×Z X2y yY

∆−−−−→ Y ×Z Y

is a fiber diagram. This is super useful in this chapter! Proof: (Intense) category theory nonsense

II.4.3. Let U, V ⊂ X be affine opens. Noting that U ×X V ' U ∩ V , we have by the magic squarelemma a fiber diagram:

U ∩ V −−−−→ U ×S Vy yX

∆−−−−→ X ×S X

20

where the horizontals are closed embeddings and verticals are open embeddings. Since U ×S V isaffine, so is U ∩ V which is an open subscheme of X.

II.4.4. Note that we have f : Z → X → Y and g : Y → S where g f is proper and g is separated.Hence, f is proper. Thus, f(Z) is closed in Y . So, denote by f(Z) the scheme-theoretic image ofZ in Y . I have no idea why f(Z) need be proper over S...

By the way, why do we need quasicompact for the proof of Lemma II.4.5.???

II.4.5.(a,b): Note that SpecK(X) → X is the generic point, so that the closure of image is infact X. Thus, if R is a valuation ring of K, then Lemma II.4.4. specializes to say: to give a mapSpecR→ X such that

SpecK //

X

SpecR

;;

commutes is equivalently to giving a point x ∈ X such that R dominates OX,x. Moreover, if werequire R to be a valuation ring of K/k, then giving a map SpecR→ X so that

SpecK //

X

SpecR

99

// Spec k

commutes is equivalent to giving a point x ∈ X such that OX,x dominates R. That is, if X isseparated/proper, then the center x for any valuation ring R of K/k is unique/uniquely exists.

(c) This is HOID.(d) Let a ∈ OX(X) such that a /∈ k. Since k = k, note that a is transcendental over k so

that treating a ∈ OX(X) ⊂ K(X), we have k[a−1](a−1) ' k[x](x), a local ring. Hence, there existsvaluation ring R of K/k that dominates k[a−1](a−1). By properness, there exists x ∈ X such thatR dominates OX,x. However, a ∈ OX,x, so that OX,x → R gives a, a−1 ∈ R so that a−1 /∈ mR,which is a contradiction.

II.4.6. We in fact show this for affine integral schemes. If f : SpecA → SpecB is induced byϕ : B → A, then since f is proper ϕ is of finite type. Thus, to show finite, we need show that it isintegral. Replacing B by ϕ(B), we assume that ϕ is injective. Suppose R ⊃ B is a valuation ringof K = K(A). Then R ⊃ A by properness. Since the integral closure of B in K, denoted Bcl, is theintersection of all valuation rings of K containing B, we thus have Bcl ⊃ A. Thus, every elementof A is integral over B, as desired.

II.4.7. Let p ∈ X and q = σ(p). By the additional condition there is an affine U ⊂ X containingp, q. Note that V := σ(U) is also affine containing p, q. Thus, since X is separated, U ∩ V is anaffine containing p, q such that σ(U ∩ V ) = U ∩ V . Hence, we first consider the case when X isaffine SpecC[x1, . . . , xn]/I.

No fucking clue. How is this done?? For example, just take C[x, y]→ C[x, y] via x 7→ ix, y 7→ iy.

II.4.8. (d): Universal property nonsense crazy diagram.(e): The magic diagram (lemma above) gives us a fiber diagram:

XΓf−−−−→ X ×Z Y

f

y yY

∆−−−−→ Y ×Z Y

21

Combined with the diagramX ×Z Y −−−−→ Yy y

X −−−−→ Z

we have that if X → Z and ∆ have property P, then so does X → Y .(f): Note that the natural map out of Xred in Ex.II.3.11(c) is in fact a closed embedding. Now,

consider the diagram:Xred

&&

// X ×Y Yred

// Yred

X // Y

II.4.9. We have closed embeddings i : X → PnY and j : Y → PmZ . Now, note that PnY ' PnZ ×Z Y .

So, we have an closed embedding Xi→ PnZ ×Z Y

Id×j−→ PnZ ×Z PmZ → Pnm−m−nZ .

II.4.10(c). This is like,, impossible...

II.4.11. Later when studying CA

II.5 Sheaves of modules

Lemma. For G an OX -modules, we have a natural HomOX(O⊕IX ,G ) ' Hom(O⊕IX (X),G (X)).

Proof: → map is clear. For →, for any U , the image of the standard basis ei of OX(U)⊕I isdetermined by ei of OX(X)⊕I → G (X)→ G (U) since ei restricts to ei.

II.5.1(a,d). (a): There is a natural map E → E ∨∨ as follows: For any U ⊂ X, given s ∈ E (U),define evs : E ∨|U → OU by evs(W )(ϕ) = ϕ(W )(s|W ) for W ⊂ U and (ϕ : E |W → OW ) ∈ E ∨|U (W ).To show that this map is an isomorphism when E is locally free, we can assume E is in fact free sinceisomorphism is local condition. But then, by the lemma above, it suffices to show that M →M∨∨

is an isomorphism for M a free OX(X)-module of finite rank, but this is true.(d): We will work at the presheaf-tensor level. First, there is a natural map f∗F ⊗OY

E →f∗(F ⊗OX

f∗E ) as follows: For each V ⊂ Y , the RHS is F (f−1V ) ⊗f∗f−1OY (V )

f∗f−1E (V ), but since

(f−1, f∗) are adjoints there are natural maps OY (V ) → f∗f−1OY (V ) and E (V ) → f∗f

−1E (V ),which gives us a map

F (f−1V ) ⊗OY (V )

E (V )→ F (f−1V ) ⊗f∗f−1OY (V )

f∗f−1E (V )

Now, for the isomorphism, since we can work locally, assume that E ' O⊕nY . The above map isjust F (f−1V )⊕n → F (f−1V )⊕n where the left is a OY (V )-module and the right as f∗f

−1OY (V )-module restricted to OY (V )-module. This is clearly an isomorphism.

II.5.6. (a): For p ∈ X, we have mp 6= 0 ⇐⇒ fm 6= 0 ∀f ∈ A \ p ⇐⇒ p ⊃ Ann(m).(b): If p ∈ SuppM , then Mp 6= 0 =⇒ fM 6= 0 ∀f ∈ A/p ⇐⇒ p ⊃ Ann(M) (note that ⇐=

may not hold if M is not finitely generated!). If M if finitely generated, then M = (m1, . . . ,mr)and Ann(M) =

⋂i Ann(mi) so that if p ⊃ Ann(M), then p ⊃ Ann(mi) for some i so that for any

f ∈ A \ p we have fmi 6= 0 hence Mp 6= 0. (We actually did not need Noetherian).(c): This follows from part (b).

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(d): First, H 0Z (F ) is quasicoherent, as it is the kernel of F → j∗(F |U ) (as X is Noetherian,

so is U , and so F |U quasicoherent implies that j∗(F |U ) is also). To finish, we show that over anaffine X = SpecA now, ΓI(M) = ΓZ(X,F ) as modules (where M = Γ(X,F )). Well, for m ∈M ,Supp(m) = V (annm) ⊂ V (I) implies that I ⊂

√annm so that some power of I lies in annm (as

I is finitely generated)—i.e. m ∈ (0M : I∞). Conversely, if m ∈ (0M : I∞), then In ⊂ annm forsome n so that V (In) = V (I) ⊃ V (annm) = suppm.

(e): Follows from the same argument in (d).

II.5.7. (a): Can assume X affine so that F = M . Suppose Mp is free, i.e. A⊕np∼→ Mp. This map

can actually made so that it lifts to A⊕n →M . Since M is coherent, we have a exact sequence:

A⊕mα→ A⊕n

β→M → 0

where imαp = 0. But since imα is finitely generated, there exists f such that (imα)f = 0. Hence,

localizing the sequence above by f , we obtain A⊕nf∼→Mf .

(b): Follows from part (a).(c): Generally, there is a map F ⊗ F∨ → OX . When F is invertible, so is F∨ and this

map is an isomorphism. For the converse, suppose G exists. Then for an affine patch we haveMp ⊗Np ' Ap at each point p. Now, by applying κ(p) to both sides and using Nakayama, we notethat both Mp, Np are of the form Ap/I, so that the only way ' holds is if Mp ' Np ' Ap. Thus,we conclude that F ,G are both locally free of rank 1 by part (a,b). Does the converse not hold ifwe don’t have coherence?

II.5.8. Notation: Let’s denote M ⊗A κ(p) by M |p.(a): We show that x : ϕ(x) ≥ ` is closed on every trivializing affines. (so reduce to X affine

and M finitely generated A-module). That is, we have an exact sequence

A⊕mα→ A⊕n

β→M → 0

With rank-nullity, one concludes that dimM |p equals n minus the rank of the matrix [α|p]. Now,rk[α|p] ≤ n− ` iff all (n− `+ 1)-minors vanish. That is, if p ∈ V ((n− `+ 1)-minors of [α]).

(b): The rank is constant if X is connected.

(c): Suppose ϕ = n constant. We again work affine locally, which gives us A⊕mα→ A⊕n

β→M → 0 again, where now [α|p] = 0 for all p. That is, all the entries of [α] are in the nilradical ofA, which is zero since A is reduced.

II.5.9. (a): For each f ∈ S1 and d ∈ Z≥0, we have a natural map Md → Γ(Xf , M(d)) = M(d)(f)

which glues (as M(d) is a sheaf) to a map αd : Md → Γ(X, M(d)).(b): Do it w M or J?

II.5.10. (a):

II.5.11. We work affine locally: there is a natural isomorphism (S ×A T )(f⊗g)∼→ S(f) ⊗A T(g).

Same kind of argument goes for O(1) part (except we need check transition maps too now).

II.5.14.

II.5.16. Later when I need it.

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II.6 Divisors

On products and Weil class group. Let X satisfy (∗). We first discuss X × An and X ×k Ank(for X a k-variety):

Observation 1: Consider ϕ# : A → A[t] (where A is integral domain) and the associated mapϕ : SpecA[t] → SpecA. If p ∈ SpecA, then p[t] ∈ SpecA[t], and in fact, A[t]/p[t] ' (A/q)[t].In other words, ϕ−1(V (p)) = V (p[t]). Moreover, p[t] is of codimension 1, since A[t]p[t] ' Ap[t]pp[t]

and if∑ait

i/∑biti ∈ A[t]p[t] then the valuation is maximi : uπmi = ai. Thus, the map

ϕ∗ : WDiv(SpecA) → WDiv(SpecA[t]) is a well-defined homomorphism that also sends principaldivisors to principal ones. Moreover, note that this is an injection at the level of class groups.

Observation 2: For q ∈ SpecA[t], codimA[t] q ≥ codimA ϕ(q) (codimension can at most decreaseunder ϕ), since any chain (0) ( p ( · · · ( ϕ(q) gives a chain (0) ( p[t] ( · · · ( ϕ(q)[t] ⊂ q. Hence,if q is of codimension 1, then ϕ(q) is either codimension 1 prime p or (0) in A. When ϕ(q) = [(0)],then q ∩ A = 0 so that A[t]q ' K[t]q⊗K where K = Frac(A), and since K[t] is a UFD, we haveq⊗A K = (f) for an irreducible f ∈ K[t].

Observation 3: First, we note that the two observations above works the same for any (finite)number of variables. Now, note that the fiber of the generic point of the map X × An → X (orX × Pn → X) is AnK(X) (or PnK). In other words, we have

AnK(or PnK) −−−−→ X × An(or Pn)y ySpecK(X) −−−−→ X

Thus, we have a well-defined map WDivAnK → WDivX × An via the top map (and locally it isf ∈ K[x] treated as a polynomial in Af [x] for some f ∈ A). This map again sends principal divisorsto principal divisors. In the case of X × Pn, we thus get a map ClPnK → Cl(X × Pn).

From the three observations we get the following summary

Lemma. A prime divisor X × A1 either is Y × A1 for Y ⊂ X of codimension 1, or correspondsto P where P ∈ A1

K is a closed point. By the same reasoning (with induction), a prime divisor ofX × An either is Y × An, or corresponds to V (f) ⊂ AnK .

Lemma. X × Pn as two types of prime divisors: Y × Pn for Y prime divisor in X, and onescorresponding to hypersurfaces of PnK .

Theorem. It thus follows that ClX × An = ClX and ClX × Pn = ClX ⊕ Z.

II.6.1. X×Pn is indeed Noetherian, separated, and integral (see TIL:7/20/2016), and since X×Anis regular in codimension 1, so is X × Pn (regularity is a local condition).

Now, let π1 : X × Pn → X and π2 : X × Pn → Pn be two projections. Let H = V (x0) ⊂ Pn bea prime divisor, and note that X × Pn \X ×H ' X × An. This gives us a exact sequence:

ClPn ' Z→ ClX × Pn → ClX × An → 0

where the first map sends [H] to [X × H], which is the map π∗2. this map is injective since if[X × H] = 0, then for f ∈ K(X × Pn) such that div f = X × H, f restricted to SpecA × Ui forany i 6= 0 and for any SpecA ⊂ X is just x0/xi ∈ A[

xjxis], but this is impossible. Moreover, note

that composing ClX × An ∼→ ClX and π∗1, we get a map σ that is a section of the second map inthe sequence above. Hence the above sequence is short exact and splits.

24

II.6.2. (a) There’s basically nothing to prove...(b): Let f ∈ K(Pn

k), which can be written as p/q where p, q ∈ k[x]d for some d. Since

components of (f) does not contain X, both p and q does not vanish when restricted to k[x]/I(X),so that f is still a rational function f of X. Thus by definition in (a) we have (f).X = (f). Now,since X is not contained in every hyperplane of Pn, and every divisor in Pn is equivalent to ahyperplane, we get a map ϕ : ClPn → ClX that is well-defined.

(c): We need show that νYi(f i) = µpi(S/I(X) + (f)) where pi is the homogeneous prime inS := k[x]• corresponding to Yi. Combining the two facts in TIl 7/22/2016, we have that if p′i isthe prime corresponding to Yi ∩ Ui in the coordinate ring A(X ∩ Ui), then νYi(f i) = νp′i(f i) =

µp′i(A(X ∩ Ui)/I(X ∩ Ui) + (f)) = µpi(S/IX) + (f)), as desired.Now, for deg(D.X) = degD · degX, it suffices to show for D = V a hypersurface. But

deg(V.X) =∑

i i(X,V ;Yi) deg Yi = (deg V )(degX) from the general Bezout’s theorem.(d): If D is a principal divisor on X, then it is of the form p/q where p, q ∈ k[x]d (∃d) such that

p, q /∈ I(X). Thus, we have degD = 0 from the equation in part (c). Hence, we get a well-definedmap ClX → Z. Composing with the map ϕ in part (b), and using part (c), we get a commutativediagram:

ClPnk

ϕ−−−−→ ClX

'ydeg

ydeg

Z × degX−−−−−→ ZAnd in particular, ϕ must be injective!

II.6.3 (a): Let S(V ) := k[x]•/I(V )• the graded coordinate ring of V . Then A(X) = k[x]/I(V ) andS(X) = k[x, y]•/I(V )•. Thus, on Vi = V ∩Uxi , we have that π−1(Vi) = Speck[x/xi, y/xi]/I(Vi) 'Vi × A1. This in fact gives X \ P a structure of A1-bundle over V . Now, use the fact in TIL8/26/2016 about An-bundles over X in regards to class groups.

(b): [V ] ∈ ClX is given by y = 0 (continuing with S(X) = k[x, y]•/I(V )•). If ` is a linear formin k[x]• that doesn’t vanish on V (i.e. /∈ I(V )•), then y/` ∈ K(X) such that [y = 0] ∼ [` = 0],and the latter is equal to the class of V.H pullbacked by π∗. Now, consider the exact sequenceClV (y) → Cl(X \ P ) → Cl(X \ P ) → 0. Under the isomorphism in (a), the first map becomesZ → ClV, 1 7→ H.V and the second map becomes π∗ followed by restriction to X \P ⊂ X. Hence,we obtain a short exact sequence

0→ Z→ ClV → ClX → 0

(c) Follows immediately from part (b) and II.6.2 as S(Y ) without grading is the same asA(C(Y )) = A(X).

(d) The map Spec OP → X is given by(k[x, y]/I(V )

)m

→k[x, y]/I(V ) (where m = (x)).Indeed, this gives us a map ClX Cl(Spec OP ) and a section map Cl(Spec OP ) → ClX. Now,if p ⊂ A(X) of height 1 is not contained in m, then we claim that it is actually principal (see TIL8/19/2016). I use Fulton’s intersection theory here... Is there a way to avoid this???

II.6.4. As f is square-free, (z2 − f) is irreducible in k(x)[z], so that K := k(x)[z]/(z2 − f) isan Galois extension of k(x) of degree 2. It is also the quotient field of A = k[x, z]/(z2 − f). Weshow that A is the integral closure of k[x] in K (and hence is integrally closed in K). Considerα = g + hz ∈ K where g, h ∈ k(x) that is integral over k[x]. Note that α satisfies the monicpolynomial q(t) := t2− 2gt+ (g2−h2f) ∈ k(x)[t], so the minimal monic polynomial p(t) of α (overk[x]) must divide q(t). If p(t) is of degree 1, then α ∈ k[x] already so we are done. If p(t) is ofdegree 2, then p(t)|q(t) implies that p(t) = q(t). This implies that 2g ∈ A hence g ∈ k[x], and

25

moreover, g2 − h2f ∈ A so that g2 − h2f ∈ k[x] so that h2f ∈ k[x], implying h ∈ k[x] also since fis square-free.

Lemma. A short lemma if you wish to avoid the degree 1 and 2 argument: Let A be integrallyclosed in its quotient ring K, and L ⊂ K be an algebraic extension. If α ∈ L is integral over A,then the minimal monic polynomial p(t) ∈ A[t] of α over A is the same as the minimal polynomialq(t) ∈ K[t] of α over K (after scaling it to monic). Proof: Clearly, q(t)|p(t) in K[t] by minimalityof q(t). But if p(t) factors into two monic polynomials f, g, then both f, g has coefficients that areintegral over A, and hence in A as A is integrally closed.

II.6.5. (b) Let R := Spec k[x0, . . . , xr]/(−x0x1 + x22 + · · ·+ x2

r) with r 6= 3 and X := SpecR. LetH := V (x1). Since x1 is a nonzerdivisor in R, we have dimR/x1 = dimR − 1, and the minimalprimes of R/x1 have height 1 over R, hence the irreducible components of X ∩ H are to primedivisors of X, and for r 6= 3 we actually have X ∩ H has one irreducible component. This onlypreserves set-theoretic information, and lets denote the divisor obtained in this way as [X ∩s H].Moreover, note that X \ H ' Spec k[x±1 , x2, . . . , xn] so that ClX \ H = 0. Hence, we have asurjective Z ClX sending 1 7→ [X ∩s H]. We analyze the relationship of [X ∩s H] and [X ∩H],the divisor associated to the scheme-theoretic intersection, for various values of r ≥ 2. (Note that[X ∩H] = 0 in the class group since it is principal—it is given by function x0).

When r = 2, we get Spec k[x, z]/(z2), whose unique minimal prime (z) has z2 valued to 2 so thatwe get [X ∩ V (x1)] = 2[V (y, z)]. However, when r ≥ 4, we get instead Spec k[x0, x2, . . . , xr]/(x

22 +

· · ·+ x2r), and (x2

2 + · · ·+ x2r) is prime so that we just get [X ∩ V (x1)] = [V (x2

2 + · · ·+ x2r)]. That

is, if r = 2, we have 2[X ∩sH] = [X ∩H] = 0. However, if r ≥ 4, then [X ∩sH] = [X ∩H] so thatthe map Z ClX is just a zero map!

Warning: The usual excision exact sequence only works when the subtracting set is irreducible!Witness the case r = 3 here. A fix is as follows: if Z is a proper closed subset of X, with irreduciblecomponents Z1, . . . , Zr, then Z⊕r → ClX → ClX \ Z → 0 where the first map is ei 7→ [Zi] (where[Zi] = 0 if codimX Zi > 1).

II.6.8. (a): Pullback of locally free is locally free, and pullback respects tensors. Lastly, pullbackof structure sheaf is the structure sheaf.

(b): Let f : X → Y be a finite map of nonsingular curves. Denote by f∗D : ClY → ClX themap just to distinguish it from f∗ : PicY → PicX. We need show that

ClY −−−−→ CLXy yPicY −−−−→ picX

is commutative. As both horizontal maps are group homomorphisms, it suffices to check for [Q] ∈ClY where Q ∈ Y is a point. As Q is locally principal, take a affine cover of Y where [Q] isprincipal in each. On affines V not containing Q, the diagram is clearly commutative. On U 3 Q,let’s say U = SpecB and f−1(U) = SpecA, and h ∈ B exactly carves out Q (i.e. B/h ' κ(Q)).We have f# : B → A, and denoting h′ := f#(h). Note that SpecA/h′ gives the divisor f∗D(Q) onX. Now, we also see that the pullback is A⊗B B · h−1 = A · (h′)−1, as desired.

(c): Pretty much the same proof as part (b).

II.6.10. Note the following lemma:

Lemma. If 0 → F0 → · · · → Fr → 0 is a long exact sequence of coherent sheaves on X, thentheir alternating sum (as an element in G0(X)) is zero.

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(b): If 0 → F ′ → F → F ′′ → 0 is an SES, then 0 → F ′ξ → Fξ → F ′′ξ → 0 is an SESof K-modules so that rk F ′ + rk F ′′ = rk F . Hence, the map G0(X) → Z is well-defined. Forsurjectivity, note that O⊕nX = n ∀n ≥ 0.

(a): If R is a PID, and X = SpecR, then the rank map G0(X)→ Z is in fact an isomorphism.

Suppose M is torsion R-module (i.e. rk M = 0), then there exists a resolution 0 → Rm → Rn →M → 0 of M (as R is a PID) and rkM = 0 forces m = n, so that as an element of G0(X), we have

M = OnX − On

X = 0.(c): Later. Just memorize for now.

II.6.12. As X is projective, we embed X in some Pnk and consider F to come from a finitelygenerated graded S-module. Setting deg F = χ(F ) − (rk F ) · χ(OX) works. (1) follows by R-R,and (2) by use of Hilbert polynomial, and (3) by additiveness of Euler characteristic and rank.

II.7 Projective morphisms

II.7.1. Let (A,m, k) be a local ring, and consider a surjective map of A-modules ϕ : A A. Sincetensoring is right exact, we have ϕk : A ⊗ k ∼→ A ⊗ k so that ϕ(1) /∈ m. Hence, the map ϕ ismultiplication by a unit in A, hence an isomorphism.

II.7.2. Since si and tj span the same space, there exists j0, . . . , jn ∈ 0, . . . ,m such thatspan(tj1 , . . . , tjn) = V (otherwise, dimV > n+ 1 and thus a contradiction). WLOG let jk = k, andlet L = V (y0, . . . , yn) ⊂ Pm. Since t0, . . . , tn generate V ⊂ Γ(X,L ), we have that ψ−1(Pm \L) =

X and we can the composition with the projection π : Pm \ L→ Pn to get ψ : Xψ→ Pm \ L π→ Pn.

Now, each tj can be expressed as a linear combination of si’s, and moreover, by constructionthere exists (exists! this is a bit subtle) an invertible (n + 1) × (n + 1) matrix M such that(s0 · · · sn)M = (t0 · · · tn).

Lemma. Let X be a k-scheme. Then a map ϕ : X → Pnk , given by s0, . . . , sn ⊂ Γ(X,L ) for a linebundle L on X, is essentially determined by the associated linear series V := span(s0, . . . , sn) ⊂Γ(X,L ). More precisely,

(a) There exists a projection map π : Pnk \ L → PV from a linear subspace L of (projective)dimension n− dimV − 1 such that π ϕ = ψ where ψ : X → PV .

(b) There is a map α : PV → Pnk such that α ψ = ϕ.

As a corollary, if s0, . . . , sm and t0, . . . , tn ⊂ Γ(X,L ) generate two base-point-free linearsystems V,W , with V ⊂ W and m ≤ n, then there exists a linear projection π : Pnk − L → Pmkwhich when composed with an automorphism ι of Pmk we get ϕs = ι π ϕt.

II.7.3. (a): Note that the intersection of n hyperplanes in Pn is non-empty. Now, let d besuch that ϕ∗(OPm(1)) ' OPn(d), and give Pm coordinates x0, . . . , xm. Consider the linear systemV := span(ϕ∗x0, . . . , ϕ

∗xm) ⊂ Γ(Pn, ϕ∗(O(1))). If dimV ≤ n, then V cannot be base-point-freeunless d = 0, and hence ϕ(Pn) = pt. Thus, to have ϕ(Pn) 6= pt, we need dimV ≥ n + 1, andso n ≤ m necessarily. FSOC, if dimϕ(Pn) ≤ n − 1 in this case, then we can find a sequence(H1, . . . ,Hn) of hyperplanes in Pm such that ϕ(Pn) ∩ H1 ∩ · · · ∩ Hn = ∅. This would mean thatϕ∗Hi’s form a base-point-free linear system of dimension at most n, which is a contradiction.

(b): Use the generalization (Lemma) above.

II.7.4. (a): If suchX admits a ample invertible sheaf L , then there exists an immersion i : X → PnAby L n.

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(b): PicX = Z⊕ Z/(1, 1). And the only case when O(a, b) is base-point-free is (0, 0), i.e. OX .

III.7.5. Note that that if F ,G are globally generated coherent sheaves on X, then F ⊗ G is alsoglobally generated. For the proofs below, let F be a coherent sheaf on X.

(a): (F ⊗L n)⊗M n is globally generated for some n.(b): M ⊗L n−1 is globally generated for some n, so by part (a) above M ⊗L n is ample.(c): Take n large enough so that both F ⊗L n and M n is globally generated.(d): Note that while M is a priori not finitely globally generated, we can assume so since X

is Noetherian and M is coherent. Now, let On+1X L and Om+1

X M give maps ι : X → PnAand ϕ : X → PnA where ι is an immersion factoring through X

ι→ X → PnA (the first map isopen embedding and the second map is closed embedding). As PnA is separated, the graph mapΓϕ : X → X × PmA is a closed embedding. Hence, X → PnA × PmA ' Pmn+m+n

A is a locally closedembedding, and X Noetherian means it is the same as an immersion. Now, pulling back O(1) onPmn+m+n exactly gives us L ⊗M , and thus L ⊗M is very ample.

(e): Follows from (d) as Lm is globally generated for large enough m.

II.7.6. (a): If i : X → Pnk is the embedding, then we have L (D) = i∗OPn(1), so that dim Γ(X,L n) =

dim Γ(X, i∗OPn(n)) = Γn(OX), the n-th degree of Γ•(OX). Now, OX ' S/I where S = k[x]• andI = Γ•(IX/Pn

k). Now, we have (S/I)n ' Γn(OX) for all n >> 0 by Ex.II.5.9.

(b): We have a map deg : ClX → Z, and if D is torsion, then degD = 0 indeed. Thus,if L (nD) has a global section s, then div s = 0, so that L (div s) ' OX , i.e. nD = 0 or r|n.Otherwise, L (nD) has no global section.

II.7.7(c). Denote by ~p(2) := (p20, p

21, p

22, p0p1, p1p2, p2p0) for a point [p0 : p1 : p2] ∈ P2

k. Then,the linear system V is ker ~p(2), so the linear system has rank 4 (dimension 5). Now, WLOG letP = [0 : 0 : 1]. Then, the map ϕ : P2

k \P → P4 is given by sections (x2, y2, xy, xz, yz) of Γ(X,O(2)).On Uy and Ux, this map is clearly a closed embedding.

We blow-up the point P to get X; locally on Uz, we get X ∩Uz = V (xv − yu). More precisely,we have the graded ideal I = (xv − yu) in (k[x, y])[u, v]• with u, v have weight 1, whose Projis A2

k ×k P1k. Uy, Ux parts doesn’t change at all, so we have X = V (xv − yu) ⊂ P2 × P1 where

(xv − yu) is an ideal of k[x, y, z]• ⊗k k[u, v]•. Now, the map ϕ extends to ϕ : X → P4 where[0 : 0 : 1]× [u : v] 7→ [0 : 0 : 0 : u : v] (on Uz, the map is [x : y : 1] 7→ [x2 : y2 : xy : x : y], and whichif WLOG x 6= 0 if [x : y2/x : y : 1 : y/x], so that since v/u = y/x, we obtained the map as aboveas (x, y) limits to (0, 0)). To check that ϕ is in fact a closed embedding on Uz (it is already on Uxand Uy), we note that on Uz, we have ϕ(x, y, z, u, v) = [ux, vy, vx = uy, u, v], and one checks easilythat on affine patches Uu and Uv, the map is indeed a closed embedding.

Lastly, to show deg = 3, note that a hyperplane V (x2) ⊂ P4 pulls back to three lines: x/y =0, y/x = 0, and the exceptional fiber.

II.8 Differentials

II.8.1. (a): Using a Proposition from Eisenbud’s book, we only need a map of k-algebras B/m→B/m2 that splits B/m2 B/m. Well, as B/m2 is complete local ring, we have κ(B) ' K →B/m2 → κ(B), a splitting as desired by Cohen structure theorem.

Alternate Soln: Let’s do this more by hand. Note that given the choice axiom, for map ofvector spaces V → W to be injective it is sufficient to show that the induced dual map W ∗ → V ∗

is surjective. We will thus show that Homκ(B)(ΩB/k ⊗ κ(B), κ(B))δ∗→ Homκ(B)(m/m

2, κ(B)) is

28

surjective. Note that Homκ(B)(ΩB/k ⊗ κ(B), κ(B)) ' Derk(B, κ(B)). The map δ∗ sends D ∈Derk(B, κ(B)) to D : m/m2 3 b ⊗ b

′ 7→ db ⊗B b′ 7→ D(b)b

′or equivalently b 7→ D(b) where b

is any lift of b ∈ m/m2. Given h ∈ Homκ(B)(m/m2, κ(B)), we now wish to construct h such

that δ∗(h) = h. Well, Leibniz rule implies that giving a derivation B → κ(B) is the same asgiving a derivation B/m2 → κ(B) (as m2 is in the kernel as map of Abelian groups). As B/m2

is complete local ring, there is a field K → B/m2 such that K∼→ κ(B). We can thus build

h ∈ DerK(B/m2, κ(B)) ⊂ Derk(B/m2, κ(B)) = Derk(b, κ(B)) as in TIL 9/27/2016.

(b): Note that as B is a localization of finite type k-algebra, let’s say B = Ap, we havedimB + tr degk κ(B) = dimAp + tr degk κ(p) = dimAp + dimA/p = dimA = tr degkK where K isthe quotient field.

Now, If k is perfect, then κ(B) is separably generated, so that by part (a) we have

0→ m/m2 → ΩB/k ⊗ κ(B)→ Ωκ(B)/k → 0

If ΩB/k is free of rank dimB + tr degk κ(B), then m/m2 has κ(B)-dimension dimB so that B isregular. If B is regular, then the proof is exactly the same as Proposition II.8.8 (noting that as kis perfect, tr degkK = rk ΩK/k).

(c): Follows immediately from the previous parts.(d): We don’t really need algebraic closure, just that k is perfect. For (ΩX/k)x to have OX,x-

rank dimX, since rkK(X)(ΩX/k ⊗ K(X)) = dimX always, from part (a) we see that we onlyneed ΩX/k|x := ΩX/k ⊗ κ(x) to have κ(x)-rank = dimX. Let’s now work affine locally, and sayX = SpecA(X) where A(X) := k[x1, . . . , xn]/I so that we have

I/I2 J→ k[x]dx1, . . . , dxn ⊗ k[x]/I → ΩX/k → 0

where J is the (co)Jacobian matrix mod I. Tensoring by ⊗A(X)κ(x), we get:

I ⊗ k[x]/I ⊗ κ(x)J(x)−→ k[x]⊕n ⊗ κ(x)→ ΩX/k|x → 0

So that we exactly need the rank of J(x) to be n − dimX. But its rank is always ≤ n − dimX,since otherwise, we would have rk ΩX/k|x < dimX. Hence, that rank ≤ n− dimX − 1 is a closedcondition implies that rank = n− dimX is an open condition.

Generalization: If X =⋃i Zi is not integral, for x ∈ Zi not in any other irreducible components,

replace dimX with dimZi and the same argument pulls through. If X = Spec k[x1, . . . , xn]/I, thenn− dimZi = codimAn Zi = ht Ip where V (p) = Zi.

II.8.2. We can reduce to V being finite dimensional. Let’s say dimk V = s and rk E = r. For x ∈ Xclosed point, consider the map ϕx : V → E |x := Ex⊗κ(x). ϕx is surjective, and as a k-vector spacekerϕx thus has dimension s−r[κ(x) : k]. Thus W ⊂ X×k |V | defined by W := (x, s) : sx ∈ mxExis an algebraic subset whose dimension is ≤ supdimX+ s− r[κ(x) : k] ≤ n+ s− r ≤ s−1. Thus,the second projection X × |V | → |V | show that the image of W is proper in |V |. Thus, there existss /∈ p2(W ), i.e. s /∈ mxEx ∀x ∈ X closed. As closed points are dense in finite type k-schemes, thisalso means that s doesn’t vanish on any x ∈ X (not necessarily closed). Defining a map OX → Eby 1 7→ s, over any x ∈ X, we have sOX,x = OX,x so that

0→ OX,x → Ex → E ′x → 0

is a SES of free modules. Thus, we get the desired SES of locally free sheaves.

Alternate Soln See TIL 10/6/2016. The main task is to interpret V → E |x as OsX E

29

II.8.3. (a): A bit of CA review: if A→ B,A→ A′ and B′ = B⊗AA′, then ΩB′/A′ ' ΩB/A⊗AA′ viad(b⊗a′)↔ db⊗a′. Now, in our case, this implies that ΩX×SY/X ' p

∗2ΩY/S and ΩX×SY/Y ' p

∗1ΩX/S .

We thus get a commutative diagram

p∗2ΩY/S

'

&&p∗1ΩX/S

//

'&&

ΩX×SY/S//

ΩX×SY/X// 0

ΩX×SY/Y

so that the horizontal line is in fact a split short exact sequence. Thus, ΩX×SY/S ' p∗1ΩX/S⊕p∗2ΩY/S .

(b): Let Xr, Y s be smooth k-varieties, so that ΩX/k,ΩY/k are locally free of rank r, s. Then

ωX×Y =∧r+s ΩX×Y/k ' p∗1ωX ⊗ p∗2ωY (check affine locally for pedantic details).

(c): Recall that PX×Y = PXPY (with X × Y embedded via Segre embedding), and hencepa(X × Y ) = pa(X)pb(Y ) + (−1)spa(X) + (−1)rpa(Y ). If X = Y = nonsingular plane cubic curve,then pa(X) = 1 − PX(0) = 1, and pa(X ×X) = −1. However, pg(X) = 1 and pg(X ×X) = 1, asωX ' OX(3− 2− 1) = OX .

II.8.4. (a): If Y is a complete intersection with I = (f1, . . . , fr) for fi ∈ S, then indeed Y =V (f1) ∩ · · · ∩ V (fr) and each V (fi) is a hypersurface. For the converse, we need a lemma:

Lemma: Let H be a hypersurface (locally principal closed subscheme of codimension 1) in Pn.Then H = V (f) for some f ∈ S homogeneous (not necessarily irreducible). Proof: First, note thatif A is a UFD, and I ⊂ A a (f)-primary ideal for f ∈ A irreducible, then I = (fm) for some m;if m is the least such that fm ∈ I, then any g ∈ I factors to fkg1 · · · g`, and as no power of gi’sare in I (∵ f does not divide them) fk ∈ I so that k ≥ m. Second, if X ⊂ An a closed subschemesuch that I(X)fj is principal for some open cover D(fj)j of An, then Ass(I(X)fj ) = ∅ or allassociated primes of I(X)fi ’s are codimension 1 by the unmixedness theorem, so that all associatedprimes of I(X) is codimension 1. Thus, as k[x1, . . . , xn] is a UFD, this implies that the primarydecomposition of I(X) is

⋂i(g

nii ) = (

∏i gnii ) for irreducible gi’s so that I(X) is in fact principal.

Thus, if H is a hypersurface in Pnk , then H∩Uk = V (g(k)) so that I(H) = g for some g homogeneouspolynomial in S.

Thus, if Y = H1 ∩ · · · ∩Hr as schemes, then Y = V (f1) ∩ · · · ∩ V (fr) = V (f1, . . . , fr) for somehomogeneous polynomials f1, . . . , fr, so that I(Y ) = (f1, . . . , fr).

(b): Let X = SpecS/I be the affine cone of Y . Its codimension in An+1 is still codimX Y . Itis indeed a (locally) complete intersection, so that A(X) is Cohen-Macaulay. To check that X isregular in codimension 1, note that the cone point P is of codimension ≥2 since dimY ≥ 1, and asX \ P is a A1-bundle over Y , we have that A(X)p ' OY,y[t] for some y ∈ Y for any codimension1 points p in X.

(c): Follows immediately from II.5.14(d). The surjection implies that OY (Y ) is of k-rank atmost 1, so that Y is connected.

(d): Suppose X ⊂ Pnk is codimension r ≤ n − 2 nonsingular subvariety. Consider the d-uple

embedding of νd : Pnk → P(nd)−1

k . By Bertini’s theorem, there exists a hyperplane section H suchthat H 6⊃ νd(X) and H ∩ νd(X) is nonsingular variety, so that X ∩ ν−1

d H is a nonsingular varietyof codimension r − 1. Use this to induct and make the desired complete intersection.

(e): Use II.8.20 repeatedly.

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(f): We have ωX = OX(d − n − 1). If d < n + 1, then Γ(X,ωX) = 0 (as a global sectionwould give an effective divisor of negative degree). If d ≥ n + 1, apply − ⊗ OP(d − n − 1) tothe SES 0 → Op(−d) → OP → OX → 0 (which stays SES by part (c)) to get that h0(ωX) =h0(OP(d− n− 1))− h0(OP(−n− 1)) =

(d−1n

)(plug in z = d− n− 1 into PP(z) =

(z+nn

)). Hence, in

this case pg(X) = pa(X).(g): Moreover, if we have nonsingular complete intersection X of type (d, e) in Pn, (once again,

let’s just consider d+ e− n− 1 ≥ 0 case) then the following free resolution:

0→ S(−d− e)→ S(−d)⊕ S(−e)→ S → S/I → 0

and its twist by d+ e− n− 1, combined with the fact that X is also projectively normal (part (b),and hence (c) applies), we have that pg(X) =

(d−1n

)+(e−1n

)−(d+e−1n

). When n = 3, we get the

desired formula.

II.8.5. Later...

31

Chapter III

Cohomology

III.1 Derived Functors

Know your homological algebra.

III.2 Cohomology of Sheaves

III.2.1. (a): Let’s denote the set of two k-points P,Q as Y (Y is a closed subscheme of A1k). Note

that U = A1 \ Y is actually irreducible (and hence connected) in Zariski topology. Now, from

0→ ZU → Z→ ZY → 0, we get 0→ Z→ Z→ Z⊕ Z ϕ→ H1(ZU )→ H1(Z)→ · · · , and the imageof the map ϕ is not zero since Z→ Z⊕ Z cannot be surjective.

III.2.3. (a): Note that if ϕ : F → G is a map of sheaves, then ΓU∩Y (U,F ) → ΓU∩Y (U,G ) iswell-defined, as if s ∈ ΓU∩Y (U,F ), then ϕ(U)(s)P = ϕP (sP ) = ϕP (0) = 0. We have shown thatH 0Y (−) : Sh(X) → Sh(X) is a functor (where Sh(X) is sheaves whose objects are at least abelian

groups). In particular, ΓY (X,−) : Sh(X)→ AbGrp is a functor—let’s call it F.

To see that this is left exact, let 0 → F ′α→ F

β→ F ′′ → 0 be a SES of sheaves on X. ThatFα is injective and Fβ Fα = 0 is clear. Now, if s ∈ ΓY (X,F ) maps to 0 in ΓY (X,F ′′), then byexactness sp ∈ F ′p ∀p ∈ X, so that s ∈ F ′(U) in particular, and its support is still in Y .

(b): We need show that ΓY (X,F ) → ΓY (X,F ′′) is surjective. Given s′′ ∈ ΓY (X,F ′′), con-sidering as an element of F ′′(X), there exists s ∈ F (X) such that s 7→ s′′ (by II.1.16). Now, ass|U ∈ F (U) maps to s′′|U = 0 ∈ F ′′(U), we have s|U ∈ F ′(U). Extend s|U to t ∈ F ′(X) (possiblesince F ′ is flasque), then s− t has support in Y since s|U − t|U = 0, and s− t still maps to s′′ sincet mapst to 0.

(c): Exactly the same as the proof of Proposition III.2.5.(d): From Ex.II.1.20(b), we have that 0 → H 0

Y (F ) → F → j∗(F |U ) is exact, and moreover,

if F is flasque, the last map on any open V is F (V )res F (V ∩ U). Now, combine this with that

Γ(X,−) is left exact to obtain the SES desired.(e): Take an injective solution I • of F , to obtain a flasque resolution I |•U of of F |U . Then

from the previous part, we have chain complex of short exact sequences:

0→ ΓY (X,I •)→ Γ(X,I •)→ Γ(U,I |•U )→ 0

so that applying the homology functor and using snake lemma, we get the desired LES.(f): Let U = X \ V . Note the exact sequence 0 → FU → F → FV → 0. We claim that

FU := j!(F |U ) satisfies H iY (X,FU ) = 0 for all i. Well, following the proof of III.2.2, one can

32

construct injective resolution I • of FU such that (I i)p = 0 for all p ∈ V ⊃ Y Is this actuallytrue?. Hence, no I i has a global section supported in Y , and thus H i

Y (X,FU ) = 0 for all i asdesired.

III.3 Cohomology of Noetherian Affine Scheme

III.3.1. Let j : Xred → X be the closed embedding. If X is affine, then for any quasi-coherentsheaf G on Xred, we have H i(Xred,G ) = H i(X, j∗G ) which is zero for all i > 0 as X is affineand Noetherian-ness implies that j∗G is quasi-coherent. Now, suppose Xred is affine, and letF ∈ QCohX and N the sheaf of nilpotents. Note that we have a filtration F ⊃ N ·F ⊃ · · · (whichis actually a finite filtration as X is Noetherian). The successive quotients are N i ·F/N i+1 ·F 'N i ·F ⊗OX/N = j∗j

∗(N i ·F ) so that their nonzero cohomologies all vanish since Xred is affine.Once again, use LES to climb up the chain and conclude that nonzero cohomologies of F hence allvanish.

III.3.2. If X is affine, then so are its closed subsets. For the converse, we show a stronger statement:

Lemma: Let X be reduced Noetherian, Y, Z closed integral subschemes of X that are both affine.Then Y ∪ Z is also affine.

Proof: First, replace X by Y ∪ Z (by reduced structure on Y ∪ Z, which makes no harm asX is reduced). If one of Y,Z is contained in another, then we are done, so suppose not, so thatIY IZ ⊂ IY ∩IZ = 0. Note that for any F ∈ QCohX , the higher cohomologies of F ⊗ OX/IY

vanish. Now, let F ∈ QCohX . From 0 → IM/IJM → M/IJM → M/IM → 0 we have a SES0→ IY F ⊗OX/IZ → F → F ⊗OX/IY → 0 (as F ·OX/IY IZ = FOX/0 ' F ). By LES, wehave that higher cohomologies of F all vanish, and hence X is affine.

III.3.3. Let X = SpecA. We have shown that ∼: A-Mod∼→ QCohX is an equivalence of categories.

And moreover, for Y = V (a), we have m ∈ ΓY (M) ⇐⇒ V (Annm) ⊂ Y ⇐⇒ a ⊂√

Annm ⇐⇒m ∈ (0 :M a∞) = Γa(M), so that as functors we have ΓY ∼= Γa. This makes all (a,b,c) partsclear.

III.3.4. (a): If depthaM ≥ 1, then there is f ∈ a a nonzerodivisor of M , which implies that(0 :M a∞) = 0 since no nonzero element of m can kill a power of f . If M is further finitelygenerated, then depthaM = 0 implies that a is contained in some associated prime (of which thereare finitely many as M is f.g.). This associated prime is Annm for some 0 6= m ∈ M , so thatm ∈ Γa(M), and so Γa(M) 6= 0.

(b): We induct on n. n = 1 is part (a) (well, n = 0 is empty-statement, so assume n ≥ 1 and

begin at n = 1). Consider the SES 0→Mf→M →M/f → 0 where f ∈ a is a nonzerodivisor of M .

Then deptha(M/f) = depthaM − 1, so that depthaM ≥ n ⇐⇒ deptha(M/f) ≥ n− 1. Moreover,

we have the LES from the SES: · · · → H ia(M)

f→ H ia(M) → H i

a(M/f) → H i+1a (M) → · · · . Hence,

we have H ia(M) = 0 ∀i < n ⇐⇒ H i−1

a (M/f) = 0 ∀i < n: The =⇒ direction is easy; for ⇐=direction, note that by Exer.III.3.3.(c), for i < n− 1, H i

a(M) is zero when localized at p 6⊃ a, andfor p ⊃ a, Nakayama implies that H i

a(M) is zero, and for i = n − 1, note that ×f cannot be aninjective map as the support of every element of H i

a(M) is contained in V (a) ⊂ V (f) (so somepower of f annihilates that element). By induction hypothesis, we are done.

III.3.5. Let’s generalize to P just a point in X, and U−P means U−P. For any U 3 p, we havea LES by Exer.III.2.3.(e): 0 → H0

P(U,OU ) → H0(U,OU ) → H0(U − P,OU ) → H1

P(U,OU ) → · · · .

Now, depth OP ≥ 2 ⇐⇒ H0P

= H1P

= 0 from Exer.III.3.4.(b).

33

III.4 Cech Cohomology

III.3.4.1. If f : X → Y is an affine morphism, then choose a affine cover V = Vi of Y . ThenU := Ui := f−1(Vi) is an affine cover X such that C•(V, f∗F ) = C•(U,F ), and hence by III.4.5we have H i(X,F ) ' H i(Y, f∗F ) ∀i.

III.3.4.2. (a): Let V := SpecB ⊂ Y and U := SpecA = f−1(SpecB) ⊂ X. Then K(A) is afinite extension of K(B) (∵algebraic+f.g.), and thus if [K(A) : K(B)] = r, then shrinking V ifnecessary (thus shrinking U), we have that Br ' A as B-modules. Letting M := OU , we see thatM is coherent since X is separated (and U is affine). Moreover, define a map α : Or

Y → f∗(OU ) byselecting r sections in Γ(Y, f∗(OU )) = OU (U) = A that span A as a B-module as given by Br ' A.

(b): Just follow the hint, and for isomorphism restrict to an affine (since quasi-coherent) anduse the fact that localization commutes with Hom.

(c): Suffices to show that Yred is affine, so replace Y by Yred, and hence X by Xred. Now, weneed only show that each irreducible components of Y is affine, so assume Y is integral. But thenf is a closed surjective map, we can also assume that X is integral.

We will now show that Y is affine by Noetherian induction. Well, if Z ⊂ Y is a closed subsetof Y , then f : f−1(Z) → Z is again a finite surjective morphism where f−1(Z) is affine since it isclosed subscheme of X. Hence, we reduce to showing that Y is affine given that every closed subsetof Y is affine.

Now, let F be any coherent sheaf on Y , and let β : f∗G → F r be as in part (b). ViaCech cohomology and III.4.5, we note that H i(Y,F r) ' (H i(Y,F ))r. Thus, as H i(Y, f∗G ) =H i(X,G ) = 0 ∀i > 0 (X is affine), it remains to show that H i(Y,F r) ' H i(Y, f∗G ) ∀i > 0.

Consider the exact sequence 0→ kerβ → f∗G → F r → cokerβ → 0. As f∗G and F r are bothcoherent (since f is finite), kerβ and cokerβ are both coherent, and hence their supports, let’s sayY1, Y2, are both closed. Thus, we have kerβ = (j1)∗(j1)∗(kerβ) and similarly for cokerβ. Hence,as Y1 and Y2 are affine (by Noetherian hypothesis), the higher cohomologies of kerβ and cokerβvanish. Now, splitting the sequence above into 0 → kerβ → f∗G → imβ → 0 and 0 → imβ →F r → cokerβ → 0, and applying LES, we obtain H i(Y, f∗G ) ' H i(Y, imβ) ' H i(Y,F r) for i > 0,as desired.

III.4.3. We have 0 → k[x±, y] × k[x, y±] → k[x±, y±] → 0 where the map is (f, g) 7→ f − g. Thekernel is thus k[x, y], and the image (as k-vector space) is all monomials xαyβ where at least oneof α, β is non-negative. Hence, H0(X,OX) = k[x, y] and H1(X,OX) = kxiyi : i, j < 0.

III.4.4. (a): For each p, we define λp : Cp(A,F ) → Cp(B,F ) by sending α ∈ F (Ui0,...,ip) 7→∏J α|UJ

where the J ’s are (j0, . . . , jp) such that λ(jk) = ik. As λ is an index map (preserving the

well-ordering), one checks easily that λp’s form a map of chain complexes C•(A,F )→ C•(B,F ).In fact, in the same way, we have a map λ• : C •(A,F )→ C •(B,F ).

(b): If F → I • is an injective resolution of F , then we have maps ϕA : C •(A,F )→ I • andϕB : C •(B,F )→ I • induced from the identity map on F . Moreover, ϕB is homotopic to ϕA λ•,and hence the maps H i(A,F )→ H i(B,F ) are compatibles with maps H i(A,F )→ H i(X,F ).

(c): Following the hint and using LES, we need show that lim−→AD•(A)

∼→ lim−→AC•(A,F ). Well,

the injectivity always holds, and for surjectivity, use Exer.II.1.3(a).

III.4.5. Define a map PicX → H1(X,O∗X) as follows: Given L , take any trivializing coverA = Ui. As L is a line bundle, on Ui ∩ Uj , we have transition maps ϕji : OUi∩Uj

∼→ OUi∩Uj ,which is a multiplication by an element of OX(Ui ∩ Uj)∗, so that we have an element L (A) inC1(A,O∗X). That the transition maps satisfy the cocycle condition implies that L (A) ∈ ker d1 :C1 → C2. Now, if L (A) ∈ im d0, then L ' OX (can construct an explicit isomorphism by

34

OX(Ui)×fi→ OX(Ui) where fi ∈ O∗X(Ui)). We thus have an element of H1(A,O∗X). Lastly, the

isomorphism lim−→mfAH1(A,O∗X)

∼→ H1(X,O∗X) implies that we can consider L (A) as an element of

H1(X,O∗X) and that the map PicX → H1(X,O∗X) is indepndent of the choice of the cover A. Tosee that this is a group homomorphism, given L ,M , taking common refinement of each trivializingcover.

III.4.7. Cech is actually cumbersome. Just use LES of 0→ OP2(−d)→ OP2 → OX → 0.

III.4.8. (a): By Exer.II.5.15, a quasi-coherent sheaf F on X is a direct limit of coherent ones, butdirect limit commutes with cohomology.

(b): Let X be open subscheme of a projective scheme Y ⊂ Pnk . Suppose H i(X,E ) = 0 ∀i > nfor any locally free coherent sheaf E . Now, let F be a coherent sheaf on X; by Exer.II.5.15there exists a coherent sheaf F ′ on Y such that F ′|X = F . By Serre’s theorem II.5.18 wehave Or

Y F ′ ⊗ OY (m) for some r,m. Tensoring by OY (−m) (which is exact as OY (−m) islocally free), we obtain a SES 0 → K ′ → E ′ → F ′ → 0, where E ′ is locally free of rank r, andK ′ is coherent, as E ′ is coherent (and Y Noetherian). Restricting to X (an open subscheme),we obtain a SES 0 → K → E → F → 0, from which we obtain a LES that implies thatH i(X,F ) ' H i+1(X,K ) ∀i > n. Grothendieck vanishing implies that Hd+1(X,K ) = 0, andhence Hd(X,F ) = 0 where d = dimX. But this implies that any coherent sheaf of X hasvanishing dth cohomology, so applying this to K we have that Hd−1(X,F ) = 0. Continuing thison, we conclude that H i(X,F ) = 0 ∀i > n, as desired.

(c): If X is covered by r + 1, then is no cohomology beyond r in the Cech complex.(d): For k = k case, project from a generic point. For general case, see homogeneous Noether

normalization in Eisenbud Comm. Alg.(e): Y ⊂ Pnk is a set-theoretic intersection of codim r if Y = V (f1, . . . , fr) (that is, I(Y ) =√

(f1, . . . , fr) in the coordinate ring). Thus, X \ Y =⋃ri=1(X \ V (fi)), but X \ V (fi) = D+(fi) is

affine (since Veronese embedding is an closed embedding for any ring).

III.4.9. If Y is a complete intersection (of codimension 2) in X = A4k, then cd(X \ Y ) ≤ 1.

However, LES of Exer.III.2.3 implies that H2(X \ Y,OX\Y ) ' H3Y (X,OX). Setting P = Y1 ∩ Y2,

and suppressing some notations, we have by Mayer-Vietoris the following sequence:

→ H3P → H3

Y1 ⊕H3Y2 → H3

Y → H4P → H4

Y1 ⊕H4Y2 → · · ·

Now, as depth of P is 4, H3P = 0. Moreover, Using Exer.III.2.3 on X \ Y1 and X \ Y2 (note that

both admit cover by 2 affines so that cd(X \ Yi) ≤ 1), we have H`(X \ Yi) ' H i+1Yi∀` > 0. This

implies that H`Yi

= 0 ∀` ≥ 3. Thus, we now have that H3Y ' H4

P . But H4P 6= 0 as depth of P is

exactly 4. This implies that Y cannot be a complete intersection.

III.4.11. Include F in an injective sheaf I , so that we have 0 → F → I → G → 0. By takingthe derived LES of Γ(V, ·) from this SES for every V , and noting that I is injective and F has allvanishing higher cohomologies, we have that H i(V,G |V ) = 0 ∀i > 0 as well. Now, the rest of theproof is exactly the same as III.4.5.

III.5 The Cohomology of Projective Space

III.5.1. Alternating sums of dimensions of LES is zero.

Lemma. Let A be a Noetherian ring of finite dimension, and f ∈ A such that f /∈⋃

p∈Min(A) p.Then, dimA/f = dimA − 1. Proof: Topologically, SpecA is the union of irreducible components

35

V1 ∪ · · · ∪ Vr, and SpecA/f is hence union of Vi ∩ V (f). But for each i, the dimension of Vi ∩ V (f)is dim(A/pi)/f = dim(A/pi)− 1 by Krull’s PIT as f is a nonzerodivisor in A/pi.

III.5.2. (a): Embed ι : X → Pnk by OX(1), and thus reduce to X = Pnk by Exer.III.4.1 as ι∗Fis coherent (∵ ι is a closed embedding). Moreover, by II.5.15, we have M∼ := Γ∗(F )∼ ' F ,and by Serre’s lemma II.5.17, we have that M is a finitely generated graded S-module. Now, weinduct on the dimension of Supp F (where dim= −1 means the support is empty). Note thatit is a closed subscheme carved out by the sheaf of ideals I :=

√AnnM

∼; for each Ui, we have

I (Ui) =√

AnnS(xi)M(x0) =

√AnnM (x0). Now, base case is the the support being empty, in which

case we have a zero sheaf, so that χ(M(n)) = 0 ∀n ∈ Z. Now, suppose Supp M 6= ∅. Then as an S-module, the irrelevant ideal (x0, . . . , xn) is not a minimal homogeneous prime of M , so that there isa linear form, WLOG say x0, such that x0 /∈

⋃homog.Min(M) p by prime avoidance. This gives a map

of sheaves 0→ K → M(−1)→ M → G → 0. Moreover, on each distinguished affines Ui, we have

0→ K → Mi

x0xi→ Mi → Mi/(

x0xi

)Mi → 0 (for i 6= 0; the i = 0 case is trivial). As x0xi

is a not in anyminimal prime of Mi := M(xi), we have that SuppMi/(

x0xi

)Mi ⊂ V (AnnM + (x0xi )) has dimension 1less than SuppMi by the Lemma above. The same holds for SuppK. Thus, by induction hypothesis,χ(G (n)) and χ(K (n)) are polynomials in n, and χ(F (n))+χ(G (n+1)) = χ(F (n+1))+χ(K (n+1))from Exer.III.5.1, so that we have that first-difference χ(F (n+ 1))−χ(F (n)) is a polynomial, andhence χ(F (n)) is a polynomial.

(b): By III.5.2, we have χ(F (n)) = dimkH0(X,F (n)) for n >> 0. But by definition ofM := Γ∗(F ), we have H0(X,F (n)) = Mn.

III.5.3.(a): pa(X) = (−1)r(χ(OX) − 1) = −(−1)r +∑r

j=0(−1)r+j dimkHj(X,OX) which is∑r

j=1(−1)r+j dimkHj(X,OX) =

∑r−1i=0 H

r−i(X,OX)(b): If X = ProjS/I, then by Exer.III.5.2(b) we have PX(n) = χ(OX(n)).(c): If f : X → Y is a birational map, then it actually extends to all of X (I.6.8) and is a

surjective finite map (II.6.8). Thus, by III.4.1, we conclude that the pa(X) = pa(Y ).

(Super useful) Lemma. Let X be a quasicompact separated scheme over A Noetherian, F acoherent sheaf on X, and A → B a flat map. Then H i(X,F ) ⊗A B ' H i(XB,F ⊗ B) as B-modules. Proof: Flatness implies that ⊗B is an exact functor. Thus, applying this to the Cechcomplex, and using FHHF theorem, we have our desired isomorphisms.

III.5.5. (a): We use descending induction on q. If q = r, there is nothing to prove. Suppose q ≥ 1,and say Y = ProjS/(f1, . . . , fr−q). Then S is Cohen-Macaulay, so dimC(Y ) = q+1 = r+1−(r−q)implies that (f1, . . . , fr−q) is a regular sequence. Hence, Y ′ = ProjS/(f1, . . . , fr−q−1) is also acomplete intersection of dimension q + 1. Now, we have 0 → OY ′(−deg fr−q) → OY ′ → OY → 0.Now, the statement follows from part (c) below.

(b): k = H0(X,OX) H0(Y,OY ), and hence Y must be connected. Note that as X = Prk,we still have OX(X) = k even if k is not necessarily algebraically closed. In fact, as OY (Y ) is anonzero (nonzero follows from base-changing to k and using the lemma above) k-algebra, we havethat OY (Y ) ' k as well.

(c): We again use descending induction on q. If q = r, then our statement is exactly III.5.1.Suppose true for Y ′ of dimension q′ > 1, and let Y be as in (a) (of dimension q = q′ − 1). Thenagain, we have the SES 0→ OY ′(−d)→ OY ′ → OY → 0 where −d = deg fr−q, tensoring by OY ′(n)and taking the LES, we have H i(Y,OY (n)) ' H i+1(Y ′,OY ′(n− d)) = 0 ∀0 < i < q′ − 1.

(d): Combine (b) and (c), and use same proof for Exer.III.5.3(a).

36

Lemma (Kunnuth formula) If X,Y are quasicompact separted k-schemes, and F ,G are quasi-coherent sheaves on X,Y respectively, then H`(X ×k Y,F G ) '

⊕i+j=`H

i(X,F )⊗k Hj(Y,G ).

III.5.6. (a): The Kunnuth formula above just kills this (and intuition too).(b1): Recall that if Y is a effective Cartier divisor (i.e. locally principal subscheme), then

L (−Y ) ' IY . Now, suppose Y is such that L (Y ) ' OQ(a, b) with a, b > 0. Then we have0 → OQ(−a,−b) → OQ → OY → 0 so that by applying LES and part (a) we have OQ(Q) OY (Y ). We have OQ(Q) = k since Γ∗(OQ|P1

k×kU0) ' k[v/u][x, y]• (only x, y has weight 1) and

Γ∗(OQ|P1k×kU1

) ' k[u/v][x, y]•. Hence, as OY (Y ) is again nonzero, we have Y is connected.

(b2): OQ(a, b) with a, b > 0 is very ample and gives an embedding i : Q → Pa × Pb → Pab+a+b.Let N = ab+ a+ b. Take a hyperplane H in PN such that i(Q)∩H is regular at each point (existsby Bertini’s theorem). Taking i−1, we get a curve that is connected (from (b1)) and regular at eachpoint (and hence irreducible as well).

(b3): If |a− b| ≤ 1 as well, then by part (a), we have that H0(Q,OQ(n)) H0(Y,OY (n)) ∀n sothat Y is projectively normal (Y is already normal as it is nonsingular). Now, if |a− b| > 1, thenby twisting appropriately we are in the third case of (a) so that the surjection no longer holds.

(c): Again, we have 0 → OQ(−a,−b) → OQ → OY → 0. And from 0 → OP3(−2) → OP2 →OQ → 0, we have that χ(OQ) =

(33

)−(

13

)= 1. Hence, we have 1 − χ(OY ) = χ(OQ(−a,−b)). By

Kunnuth formula, we have χ(F G ) = χ(F )χ(I ), and moreover, χ(OPr(n)) =(n+rr

)so that we

end up with (−a+ 1)(−b+ 1) = ab− a− b+ 1, which is the answer.

Note on Q. The Segre embedding P1 × P1 → P3 gives Q = k[x, y, z, w]•/(xw − yz). Keep in

mind the map

[a0

a1

]×[b0b1

]7→[x = a0b0 z = a0b1y = a1b0 w = a1b1

]. Thus, the family of lines P1 ×

[b0b1

]are given

by VQ(b1x − b0z, b1y − b0w) as [b0 : b1] varies, and likewise we have

[a0

a1

]× P1 corresponding to

VQ(a1x− a0y, a1z − a0w). Thus, (x, y), (z, w) are in the same family of lines, and (x, z), (y, w) arein another family of lines.

Now, let [sn : sn−1t : stn−1 : tn] define the rational nth curve C on Q. We claim that itstype is (1, n − 1). Well, let Y = VP3(wn−2y − zn−1), and consider Y ∩ Q. As a divisor, [Y ∩Q] = (n − 1, n − 1) since Y ∼ (n − 1)H for a hyperplane H in P3. But also, we claim that[Y ∩Q] = [C]+(n−2)[V (z, w)]. Well, Y ∩Q∩Uw = VUw(y−zn−1, x−yz) = VUw(y−zn−1, x−zn),on this open set, we have a codimension one prime corresponding to the curve C (of order 1). Now,on Uy, we have (wn−2−zn−1, xw−z) = (wn−2(1−wxn−1), xw−z), and hence we have [z = w = 0]of order 2, and another component that is an open part of C (already counted). Hence, we have[Y ∩ Q] = [C] + (n − 2)[z = w = 0] as desired, and thus (n − 1, n − 1) = [C] + (n − 2, 0) so that[C] = (1, n− 1).

Alternate computation: Another way to solve this is to compute the intersection of C along ageneral line in each family. The intersection of C with [z = w = 0] (i.e. P1 × [1 : 0]) has degreen− 2; consider k[sn, sn−1t, stn−1, tn]/(stn−1, tn) and count monomials. Likewise, intersection of Cwith [y = w = 0] has degree 1 (by same sort of counting).

III.5.7. Later for review

37

III.6 Ext Groups and Sheaves

III.7 The Serre Duality Theorem

III.8 Higher Direct Images of Sheaves

III.9 Flat Morphisms

Note. A few things to note about flat maps:

• An A-module M is faithfully flat over A if it is flat and M ⊗A N = 0 =⇒ N = 0 for anyA-module N . Note that M is faithfully flat iff M ⊗A N = 0 =⇒ N = 0 for any N cyclicmodule iff M/mM 6= 0 for any m ⊂ A maximal. If ϕ : A→ B is faithfully flat map of rings,then SpecB → SpecA is surjective. In fact, ϕ is strongly injective in the sense that forI ⊂ A we have Iec = I.

• A flat local homomorphism of rings is faithfully flat. This implies the Going-Down theorem(a different proof is given below also).

• If A ⊂ B is a faithfully flat injection of domains whose quotient fields are the same, thenA = B; well, if b ∈ B, then b = a1

a2, and so a2b = a1 ∈ A, and by faithfully flatness we have

a2B∩A = (a2)A so that a2b = a2a′ for some a′ ∈ A, and thus a2b = a2a

′ = a1 ⇒ a′ = a1a2

= b.As a result, surjective birational map of varieties is flat iff it is an isomorphism. This showsthat the normalization map is flat iff it is actually an isomorphism.

• Flat map is a nice relative notion, as follows: If f : A→ B is an algebra, and M is a B-module,then M is flat over A iff Mq is flat over Af−1(q) for all q ∈ SpecB. Proof: Let I ⊂ A be anideal. Consider I ⊗AM →M . Note that this map is injective as a map of A-modules iff it isinjective as a map of B-modules (set-wise, they are the same map!). Now, for any maximalideal m ⊂ B and p := f−1(m) ⊂ A, note that (I ⊗AM)⊗B Bm ' (I ⊗AM)⊗B Bm ⊗A Ap 'Ip ⊗AMm ' Ip ⊗Ap Mm → Mm. The key observation here is as follows: if M is a A-module,for which S ⊂ A (a multiplicative subset) acts on M by isomorphisms, then M is canonicallyan S−1A-module; put in another way, if M is a S−1A-module, then AM ⊗A S−1A ' M(restriction of scalars, then extending).

• Fun exercise: When is R→ R/I flat (for R Noetherian)? (Answer: almost never nontrivially).If R → R/I is flat, then TorR1 (R/J,R/I) = I ∩ J/IJ = 0 for any J , but setting J = I wehave I/I2 = 0. Hence, I ⊗ A/I = 0 =⇒ Supp(I) ∩ Supp(A/I) = ∅, but SpecA =Supp(I) ∪ Supp(A/I) from 0→ I → A→ A/I → 0.

III.9.1. We first prove the following Going-down theorem for flat maps:

Lemma (Going-down). Let ϕ : A→ B be a flat map of (Noetherian) rings. Then Going-Downproperty holds for the map SpecB → SpecA; given p = ϕ−1(q) and p ⊃ p′, there exists q′ ⊂ Bsuch that ϕ−1(q′) = p′. Proof: Base change to A/p′ (i.e. A/p′ → B/p′B) so that we can assumeA domain and p′ = 0. Let q′ be a minimal prime of B contained in q. Now, ϕ−1(q′) = 0 since0 6= a ∈ A is a nonzero divisor and remains nonzerodivisor in B as A → B is flat. But a minimalprime consists of zero divisors.

The lemma above implies that f(U) is stable under generization. Combined with the fact thef(U) is also constructible, we have that f(U) is open.

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III.9.2. We have the twisted cubic X parametrized as [s3, s2t, st2, t3], and P = [0 : 0 : 1 : 0]. Wedo the local computation where t 6= 0, so we have Xa given by

x = u3, y = u2, z = au

(where u = s/t). Eliminating u, we have the ideal 〈x2 − y3, a3x− z3, a2y − z2, ay2 − xz, ax− yz〉,and setting a = 0, we get 〈x2 − y3, z2, xz, yz〉. The underlying set is indeed the cuspidal cubicV (x2 − y3), and when x 6= 0 we have z = 0 from xz = 0 so that it is reduced everywhere except atthe cusp (0, 0, 0) on which we have z2 = 0.

III.9.3. (a): Immediate from (III.9.7).(b): The fiber over each point on the plane Y is two points except at the point over which the

two planes of X meet. That’s where things must go wrong. Let the map X → Y be given byS := k[s, t]→ R := k[x, y, z, w]/(x, y)∩ (z, w) where s 7→ x+ z, t 7→ y+w. Then s⊗ y− t⊗ x is anonzero element of (s, t)⊗S R that maps to 0 in R.

(c): The picture is that the fibers over each point on the plane Y is a fuzz point of fuzz-size 2everywhere except over the origin (fuzz-size 3). Thus, again, x⊗z−y⊗w becomes zero but is nonzeroin (x, y)⊗k[x,y]A(X). So, f is not flat. However, Xred = Spec k[x, y, z, w]/(z, w) ' Spec k[x, y] = Yindeed, and moreover, as (z, w) is annihilator of z or w, and (z, w) is a minimal prime of X, ifp = Ann(r) for some r ∈ (x, y), then p ⊂ (z, w) in fact. So X has no embedded points!

III.9.4.

39

Chapter IV

Curves

N.B. Few things to note about divisors and line bundles on a locally factorial integral k-schemes.

• For let U ⊂ X be affine open. Then for any f ∈ OX(U) ⊂ K(X), we have that deg(div f |U ) =0 ⇐⇒ f is a unit in OX(U). (This follows from algebraic Hartog’s lemma). Furthermore,for any f ∈ f(X), we have div f = 0 ⇐⇒ f is constant (i.e. f ∈ k). Proof: Since f is aunit in every affine piece, in particular f ∈ OX(Ui) for all i for any cover Ui of X, we havef ∈ OX(X) = k.

• Recall that we had L (D)|Ui = OX · f−1i where Ui is any cover of X with D|Ui principal

on Ui, i.e. D|Ui = div fi for fi ∈ OX(Ui). As X is integral in our case, we can also defineL (D) as: Γ(U,L (D)) = f ∈ K(X)× : div f |U + D|U ≥ 0 ∪ 0. On a cover Ui whereD is locally factorial, it is easily seen that these two are the same, hence isomorphic. Thus,we’ll use these two notions interchangeably when no confusion should arise; a big caution onthis however: For f ∈ Γ(X,L (D)) ⊂ K(X), the notation div f can be ambiguous—f can beconsidered either as a rational function or as a section of a line bundle. To distinguish thesetwo situations, we write div f if considering f as a rational function, and (s)0 if considerings as a global section of a line bundle.

N.B. Also, a few things about nonsingular (integral) curves over k = k:

• Let X be a nonsingular integral curve and suppose we have a morphism ϕ : X \P → Y whereY is a proper k-scheme. Then ϕ extends uniquely to ϕ : X → Y .

• Furthermore, if X,Y are nonsingular complete curves, then K(Y ) → K(X) induces a dom-inant rational map ϕ : X → Y , which in fact uniquely extends to a surjective finite map ofdegree [K(X) : K(Y )]. (This is just combining (I.4.4) and (II.6.8)).

• As a special case of above, a nonconstant rational function f ∈ K(X) induces k(f) ⊂ K(X),which induces f : X → P1

k where f(0) = div f+ and f(∞) = div f−. As a result a nonsingularcurve X is rational iff there exists P 6= Q ∈ X such that P ∼ Q.

IV.1 Riemann-Roch Theorem

IV.1.1. Let n ≥ g+1. Then l(nP ) = n−g+1+l(K−nP ) ≥ 2, and thus L (nP ) has a nonconstantglobal section, say f , which we can consider as an element ofK(X). As div f+nP ≥ 0, and div f 6= 0(since f is nonconstant), we have that f has a pole of order at least 1 and most n on P . Moreover,f restricts to a regular function on X \ P as Γ(X \ P,L (nP )) = OX(X \ P ).

40

IV.1.2. We induct on r. r = 1 is the previous exercise. Now, suppose we have a rational functionf ∈ K(X) that is regular on X \P1, . . . , Pr−1 and poles at P1, . . . , Pr−1. Let nr be the order of fat Pr. Let n ≥ g + 1, so that Γ(X,L (nP )) ≥ 2, and let 1, f ∈ Γ(X,L (nP )) as rational functions.Then by choosing c ∈ k carefully (as k is infinite), we can make g := f + c · 1 that does not vanishon any of P1, . . . , Pr−1. As g is nonconstant, we also have that g has a pole of order at least oneat Pr. Now, take fgnr+1 as the function that is regular on X \ P1, . . . , Pr with poles exactly onP1, . . . , Pr.

IV.1.3. By (I.6.10), X is quasiprojective. Let’s say X → X. As D := X \X is a finite collectionof points, there exists a rational function f ∈ K(X) such that f regular except with poles on D.This f gives a map k[t] → OX(X) so that we have f : X → A1

k → P1k. And this map extends

uniquely to f : X → P1k where points of D are exactly ones that map to the point at infinity.

IV.1.4. Just follow hint.

IV.1.5. By R-R, dim |D| ≤ degD is equivalent to l(K −D) ≤ g, which always holds as l(K) = gand D is effective. Now, note the following lemma:

Lemma. X is rational iff there exists a point P such that l(P ) ≥ 2. Proof: If X is rational, thenl(P ) = 2 for any P ∈ X in fact. Conversely, if l(P ) ≥ 2, then there is a nonconstant rationalfunction f ∈ Γ(X,L (P )) ⊂ K(X) such that f has a pole of order 1 at P . This induces a degree 1map X → P1, and hence X is rational.

Thus, if l(K −D) = g, then either D = 0, or l(K − P ) = g − 2 + l(P ) ≥ g for some P so thatl(P ) ≥ 2, which implies X is rational.

IV.I.6. Consider the divisor D := (g+1)P for any P ∈ X. Then the nonconstant rational functionin Γ(X,L (D)) gives a map X → P1 of degree equal to the order of the pole of f at P , which is≤ g + 1.

IV.I.7. (a): We have the following lemma:

Lemma. If l(D) ≥ 2 and degD = 2 for a divisor D on a curve X of genus g ≥ 1, then L (D)is base-point-free. Proof: Let s, t be two linearly independent global sections of L (D). Then(s)0 = P + Q and (t)0 = P ′ + Q′, with P,Q distinct from P ′, Q′ (∵ if P = P ′, say, thenQ − Q′ ∼ 0, so that g ≥ 1 forces Q = Q′, implying s = ct for some c ∈ k×). Thus, we haveV (s) ∩ V (t) = 0 so that L (D) is base-point-free, and s, t defines a degree 2 map X → P1.

With the lemma, the rest of (a) is just R-R.(b): Let i : X → Q = P1 × P1 be of type (g + 1, 2). Consider the map ϕ : π2 i. Any point of

P1 pulls back to a line in Q that intersects X with degree 2. Hence, ϕ : X → P1 is a degree 2 map.

IV.1.8. (a): Note that for a point P on a 1-dimensional integral k-scheme X, we have thatP ∈ Xreg iff OX,P is integrally closed. Now, as Xreg an open subset of X, if f : X → X is thenormalization of X (which is a finite map), then OX → f∗OX

is an isomorphism at stalks P ∈ Xfor all but finitely many points. Hence, we have an SES

0→ OX → f∗OX→∑P∈X

OP /OP → 0

Now, since Euler characteristic is additive and f is affine, we have χ(OX) +∑

P∈X δP = χ(OX

) sothat 1− χ(OX) = pa(X) gives the desired equation.

(b): pa(X) implies by above equation that that OP is integrally closed for all P , and hence Xis already nonsingular of genus 0, so that X ' P1.

41

(c): Proof later, just examples now: For the cusp, note that dimk k[t]/k[t2, t3] = 1, and for thenode, note that dimk k[t]/k[t2 − 1, t3 − t] = 1.

I.1.9. (a): As suppD ∈ Xreg, the divisor D is locally principal, and hence defines a Cartier divisorD on X so that we can consider the line bundle L (D). Note that χ(L (D)) = degD + 1 − paholds when D = 0. Now, we show that the equation holds for D iff it holds for D + P for anyP ∈ Xreg. Consider the SES 0 → L (−P ) → OX → κ(P ) → 0 and tensor by L (D + P ) to get0→ L (D)→ L (D + P )→ κ(P )→ 0, and now use the additivity of Euler characteristic.

(b): As X is integral, Cartier class group is the same as Picard group. Embed ι : X → Pn viasome line bundle O(1) (which corresponds to a very ample Cartier divisor). Given a line bundleL , note that L (n) globally generated for some n by (II.5.17). Moreover, by Exer.II.7.5(d) we havethat L (n+ 1) is then also very ample, so that we have L ⊗O(n+ 1) is very ample (and O(n+ 1)is very ample).

(c): We claim that if L is a very ample line bundle, then it is ' L (D) where suppD ∈ Xreg.Well, take a hyperplane L ∈ O(1) (that does not meet any of singular points of X) and pull it backto a section s of L . Then (s)0 contains none of the singular points.

(d): Yes indeed.

IV.2 Hurwitz’s Theorem

IV.2.2. (a): A complete linear system of |K| gives a degree 2 map f : X → P1. Moreover, byHurwitz formula, we have that degR = 2 · 2 − 2 − 2(0 − 2) = 6. Note that if f is ramified at P ,then 1 < eP ≤ 2 so that eP = 2. Hence, R = P1 + · · · + P6 for 6 distinct points Pi, each withramification index 2 (a tame ramification as char k 6= 2).

(b): Let h = (x−α1) · · · (x−α6) ∈ k[x], and by change of coordinates on x if necessary, assumeαi 6= 0. As αi’s are distinct, h is square-free, so that K = k(x)[z]/(z2−h) is a degree 2 finite Galoisextension of k(x). As K is separable degree 1 over k, we can consider the projective nonsingularcurve Y := CK whose function field is K. K ⊃ k(x) then gives a map f : Y → P1 of degree 2. Onone chart of this map, we have f−1(U0)→ Spec k[x] via k[x] → k[x, z]/(z2 − h), as k[x, z]/(z2 − h)is the integral closure of k[x] in K. On the level of points, this map is (λ,±

√h(λ)) 7→ λ for λ ∈ k.

So on this chart, the map f is ramified exactly over α1, . . . , α6. Now, note the diagram (wherez′ := x−3z/

√α1 · · ·α6)

k[x−1, z′]/(z′2 −∏i(x−1 − α−1

i )) −−−−→ k[x±, z]/(z2 − h) ←−−−− k[x, z]/(z2 − h)x x xk[x−1] −−−−→ k[x±] ←−−−− k[x]

The leftmost is justified as

k(x)[z]

(z2 − h)=

k(x−1)[z′]

((x6α1 · · ·α6)(z′2 − h))=k(x−1)[z′]

(z′2 − h)

where h =∏i(x−1 − α−1

i ). So on the other affine patch, the map f is exactly ramified over thepoints αi as well. Thus, f : Y → P1 is a degree 2 map with degR = 6, so that g(Y ) = 2 by theHurwitz formula. Now, the map f is given by a divisor D on Y . Then degD = 2 and l(D) ≥ 2implies (via R-R) that l(K −D) ≥ 1. But deg(K −D) = 0 and hence l(K −D) 6= 0 iff K ∼ D.

Lemma. For k a field (not necessarily k = k), we have AutPnk ' PGLn+1(k). Proof: Let

f : Pnk∼→ Pnk be an isomorphism. Then f∗ : PicPn = Z → PicPn = Z is an isomorphism so that

42

f∗O(1) ' O(1). Let si := f∗xi ∈ O(1) for i = 0, . . . , n + 1. Writing si =∑

j aijxj , we note thatV (s0, . . . , sn+1) = ∅ iff A = (aij) is an invertible matrix, thus we get an element of PGLn+1(k). Itis clear that this gives an isomorphism of groups AutPn ' PGLn+1(k). Note that the map givenby A where A(e1, . . . en+1) = (s0, . . . , sn+1) is actually just A[~p].

(c): Let P1 =

[ab

], P2 =

[cd

], P3 =

[ef

]. As P1, P2 are distinct, by scaling a, b and c, d

appropriately, we can assume that a + c = e and b + d = f . The fractional linear transformation(which really is just matrix multiplication) z 7→ az+b

cz+d then is an automorphism of P1k that sends

0,∞, 1 to P1, P2, P3.(d): Yes, indeed.(e): Just combine all the parts.

IV.2.3. Before delving into the problems, let’s state (or review) some lemmas regarding nonsingularplane curves of degree d. Let X = Proj k[x, y, z]•/(f)•.

Lemma. Note that the map X → (P2)∗ is given by p 7→ [Df(p)] where [Df ] = [fx : fy : fz] (fxdenotes ∂f/∂x). That is, it is given by graded ring map T : SX →k[x∗, y∗, z∗] where x∗ 7→ fx(likewise for y, z), as T (x∗, y∗, z∗) carves out empty scheme in X as X is nonsingular.

Lemma. Let’s generalize Euler’s formula. Let S = k[x1, . . . , xn], f ∈ S homogeneous of degree d,and T = T •(Sn) = S[t1, . . . , tn] the tensor algebra where ti’s don’t commute, xi’s commute, and xi’scommute with tj ’s. Consider two k-linear maps D : T → T and ~x∗ : T → T , where D is of degree 1and ~x∗ is of degree −1, defined as: D := (t1 ·∂x1 + · · ·+tn ·∂xn) ·− and ~x∗ := (x1 ·t∗1 + · · ·+xn ·t∗n) ·−where t∗i is the “dual operator” to ti in the sense that t∗i (tj1 · · · tjk) = tj2 · · · tjk if i = j1 and 0otherwise. Then, we have (as long as ` < d)

~x∗ D = (x1∂x1 + · · ·+ xn∂xn) · − and hence ~x∗D`+1f = (d− `)(D`f)

In particular, ` = 0 is Euler’s formula, and ` = 1 gives [x y z]

H

= (d− 1)[Df ].

Proof. The subtlety is to realize that (t∗i (tj(−))) = (∂t∗i tj)(−); this is why ∗ti is defined somewhatweirdly. After that, the proof is pretty much the proof of Euler’s formula.

Lemma (Inflection point). Let X = V (f) ⊂ P2k (not necessarily smooth), and L = V (ax +

by + cz) a line (denote by ~n = [a : b : c]). Let mP the intersection multiplicity of X and L at apoint P ∈ P2

k (0 if not in the intersection). Our goal is to give a criterion for mP ≥ m in terms ofDmf(P ) and ~nm(P ).

m = 1 is easy: check that f(P ) = ~n · P = 0.m = 2 is still easy: check that Df(P ) = λ~n for some λ ∈ k (can be zero!)m = 3 is where things get tricky... So, let’s do this more systematically...

WLOG suppose c 6= 0 (we then homogenize by c later so this isn’t a problem). Let g(s, t) :=f(s, t,−a

c s −bc t) = (p2s − p1t)

mP (as − bt)′things. Then we see that mP ≥ m iff ∂g∂si∂tj

(p1, p2) =0 ∀i+ j ≤ m.

We immediately recover m = 1,m = 2 case. For m = 3 case, (after some rather long computa-tion), we obtain the following:

Define

[a bc d

]

[e fg h

]:= de + ah − cf − bg. Let I, J ⊂ 1, 2, 3 of size 2, and given a 3 × 3

matrix M , denote by MI,J the square submatrix given by I, J . Let [~n2] be the outer product

43

of ~n with itself, and [D2f ] be the Hessian matrix of f . Then mP ≥ 3 if m ≥ 2 already and([~n2]I,J [D2f ]I,J)(P ) = 0 ∀I, J .

Okay! That was quite some detour...

(a): WLOG, let L : z = 0, and identify L ' Proj k[x, y]• = P1. Then the map ϕ : X → P1 isgiven p 7→ [Df(p)] × [0 : 0 : 1] = [fy(p) : −fx(p)] (the cross product, which is always nonzero forp ∈ X since L is not tangent to X). In other words, the map is given by ϕ# : k[x, y]→ k[x, y, z]/fvia x 7→ fy, y 7→ −fx (as ϕ#(x, y) carves out V (fx, fy, f), and L not being tangent to X impliesthat this is empty).

Now, let p ∈ X, so that ϕ(p) = [fy(p) : −fx(p)] = V (fx(p)x+ fy(p)y) ⊂ P1k. The function pulls

back to ϕ#(fx(p)x + fy(p)y) = fx(p)fy − fy(p)fx, so let h := fy(p)fx − fx(p)fy. The map ϕ isramified at p iff V (h) meets V (f) at p with multiplicity > 1 (note that p ∈ V (h) indeed). This isequivalent to stating that [Dh(p)] = λ[Df(p)] ∃λ ∈ k. Well, we have

[Dh(p)] =

Hf(p)

fy(p)−fx(p)

0

So, if p ∈ L to begin with, [fy(p) : −fx(p) : 0] = p so that the generalized Euler’s formula (seeLemma above) does the job.

Now for p = [a : b : 1] /∈ L case: The condition is equivalent to saying that [Dh(p)]×[Df(p)] = 0,which when computed out says

f2y fxx−2fxfyfxy+f2

xfyy, −fxfyfxz+f2xfyz−fyfzfxx−fxfzfxy, fzfyfxy−fxfzfyy−f2

y fxz+fxfyfyz

(evaluation at P is ommitted for convenience). This corresponds to (I, J) = ([12], [12]), ([12], [13]), ([12], [23])in the inflection point lemma above where ~n = Df(p). We could have at the very beginning arrangedso that L is z = 0 while p = [0 : 0 : 1] (i.e. do the coordinate change X = x−az, Y = y−bz, Z = z),so we can assume a = b = 0. Then fz(p) = fzz(p) = 0 (both repeated Euler’s formula), so that theequation for (I, J) = ([23], [23]) is also automatically satisfied, and so p is an inflection point.

(b): Finish other parts later...

IV.3 Embeddings in Projective Space

Note. Let D be a divisor of degree d and rank r := dim |D| = h0(X,L (D))−1 on a (nonsingular)curve X. Some facts to know in a chart:

44

d\r -1 0 1 2 3

< 0 Not effec-tive

can’t; is⇐=

⇐= ⇐= ⇐=

0 e.g.P − Qon g > 0

D ∼ 0 ⇐= ⇐= ⇐=

1 ... D ∼P ∃P

Is very ample, so ι :X∼→ P1 via L (D) '

ι∗O(1)

⇐= ⇐=

2 ... ... Either: (1) b-p-fhence hyperellipticor (2) rational

L (D) makes X the 2-uple embedding of P1

⇐=

3 ... ... Either: (1) b-p-fhence trigonal, or (2)hyperelliptic or ra-tional

Either: (1) very amplehence hyperelliptic (infact elliptic) or (2) onlyb-p-f so rational (mapsto a singular plane cu-bic) or (3) not b-p-fhence rational

L (D) makes Xthe 3-uple em-bedding of P1

IV.3.1. If degD ≥ 5 = 2 ·2+1, then it is very ample. Now, if D is very ample on a g = 2 curve X,then dim |D| ≥ 3 as there are no smooth planes curves of genus 2. Moreover, we have degD ≥ 3; ifdegD = 2, note either one of the following: (1) if E ∼ D effective, then dim |D − E| = 0 implyingthat dim |D| ≤ 2, which is a contradiction, (2) degD = 2 implies by Bezout’s Theorem that X ⊂ Hfor some linear space H of dimension 2 (just take any three points on X) and hence X is a conicin P2 and hence rational. Now using R-R, we have that degD = dim |D| − dim |K −D|+ 2− 1 ≥3− dim |K −D|+ 1, but dim |K −D| = −1 since deg(K −D) ≤ 2− 3 = −1. Hence, degD ≥ 5, asdesired.

IV.3.2. (a): Adjunction formula KX = (X +KP2)|X = (4H − 3H)|X = H|X = H ·X where H isa hyperplane section in P2 (which is a line).

(b): By R-R, we have dim |D| = dim |K −D|. Now, we showed in part (a) that KX is the veryample line bundle giving the embedding X → P2 (i.e. ωX ' OX(1)), and hence dim |K − D| =2− 2 = 0 by (IV.3.1).

(c): For X to be hyperelliptic, there must exist an effective divisor D of degree 2 and rank 1(this looks only necessary, but is actually sufficient if g ≥ 1). But in part (b), we showed that everydegree 2 effective divisor has rank 0.

IV.3.3. Suppose X =⋂iHi where degHi = di. Since ωX = OX(

∑i di − n− 1) and 2g − 2 > 0 if

g > 2, we have that∑

i di − n− 1 := m for some m > 0. If Pn → PN is m-tuple embedding, thenωX is OPN (1) pulled back to X, so that it is very ample. Now, if g = 2, we know that K is notvery ample, and hence geus 2 curves is never a complete intersection.

IV.3.4. (a): Toric geometry FTW.(b): If d < n, then take any d + 1 distinct points on X, which determines a linear subspace

of dimension at least d. By Bezout’s Theorem, X must be contained in this linear subspace,and hence X is degenerate. Thus, if X is non-degenerate, then d ≥ n. Now, if d = n, so thatdim |D| = degD = n, then by (Ex.IV.1.5), we conclude that g(X) = 0, and OX(1) ' OP1(d) underthe isomorphism X ' P1. Thus, X is a rational normal curve.

45

(c): Yes.(d): A curve of degree 3 in Pn is already contained in some P3 ⊂ Pn. So it is either plane elliptic

curve, or twisted cubic in P3.

IV.3.5 (a): Suppose X ′ := ϕ(X) is regular. Then the birational map X ′ 99K X extends to afull morphism X ′ → X, and thus the projection is an isomorphism X ' X ′. Thus, we have thefollowing:

X // q

""

P3 \O

P2

If L is a line of P2, which pulls back to a hyperplane section H of P3, we see that |X.H| = |X.L|.Well, this is convoluted way of just saying that OX(1) ' OX′(1). But h0(OX(1)) ≥ 4, whileh0(OX′(1)) ≤ h0(OP2(1)) = 3 since X ′ ⊂ P2 is a complete intersection (hence by (Ex.III.5.5.)we have H0(P2,OP2(1)) H0(X ′,OX′(1))). This is a contradiction. Hence, we conclude that anon-planar curve in P3 cannot be embedded in P2.

(b): As X is the normalization of X ′, we have that pa(X′) = pa(X) +

∑P δP . Note that

degX = degX ′ = d as projection is a degree 1 map, and hence g =(d−1

2

)−∑

P δP <(d−1

2

)as X ′

is singular.(c): Since Xred

0 ' ϕ(X), from part (a) we have pa(Xred0 ) = pa(ϕ(X)) > pa(X). However, by

(III.9.10) we have pa(X0) = pa(X1) = pa(X) so that Xred0 6' X0. Hence X0 is not reduced.

IV.3.6. (a): Again, by selecting 5 distinct points of X, by Bezout’s Theorem we can assume thatX ⊂ P4. If X ⊂ P2 then it is a plane curve of degree 4 and genus 3. If X ⊂ P4, by (Ex.IV.3.4)above we have X is the rational normal quartic. For the X ⊂ P3 case, we generalize:

Lemma: A degree d + 1 curve X in Pd is either rational or elliptic (d > 1 indeed). Proof: LetD be the effective divisor giving the embedding X ⊂ Pd, so that dim |D| ≥ d and degD = d + 1.Then R-R implies that g ≤ h0(K −D) + 1. But if D is a special divisor, then dim |D| ≤ 1

2 degDby Clifford’s theorem, which is only possible if d ≤ 1. Hence, h0(K −D) = 0, and thus g ≤ 1.

Hence, either X ⊂ P3 as a elliptic curve (in which case, it is a complete intersection of twoquadrics as shown below), or X ⊂ P3 is rational.

(b): Consider the LES associated to 0 → IX(2) → OP3(2) → OX(2) → 0. As h0(OP3(2)) =(52

)= 10 and h1(OP3(2)) = 0, and moreover h0(OX(2)) = 2 ·4 (since the deg is 8 > 2 ·1−2 = 0), we

conclude that h0(IX(2)) ≥ 2. In other words, if IX is the homogeneous ideal of X, then there existf1, f2 ∈ IX (not the same). If any of the two factors into linear factors (since they are quadrics),X would be degenerate, so that f1 and f2 are irreducible and don’t share any factors. Now, byBezout’s theorem, C := V (f1, f2) is a degree 4 curve containing X, which is also of degree 4, andhence by (I.7.6) we conclude that C is irreducible and hence C = X.

IV.3.7. Any plane curve X of degree 4 with a single node is not a projection of a smooth curve Xin P3. This is because 3 = pa(X) = pa(X) + 1 so that pa(X) = 2, but curves of degree 4 in P3 areeither g = 0, 1, or 3. Now, V (xyz2 + x4 + y4) ⊂ P2 is a plane curve with a single node at [0 : 0 : 1].

IV.3.8.

IV.3.9. For general H, H ∩X is d distinct points by Bertini’s theorem (or, observe that the familyof tangent lines to X in G(1, 3) is of dimension 1, so that planes containing tangent lines is at mostof dimension 2). Now, we show that the family of multisecant lines of X in G(1, 3) is of dimension1 as well. stuck here...

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IV.3.10. We say that P1, . . . , Pm ⊂ Pn are in general position if there is no linear space Lm−2 ofdimension m− 2 such that P1, . . . , Pm ∈ Lm−2. Now, let Vm ⊂

∏mi=1 Pn be subset of points not in

general position. Vm is a closed subset since it is the vanishing locus of m×m minors of a n×mmatrix. In particular, when m = n, we have that Vn is irreducible closed subset. If X ⊂ Pn, then(∏ni=1X) ∩ Vn is proper closed subset of

∏ni=1X, as X is irreducible and thus the intersection is

proper, or every set of n points in X is non-general position in which case X ⊂ Pn−1.

IV.3.11.

IV.4 Elliptic Curves

IV.4.1. Essentially a repeat of the proof of (IV.4.6.)...

IV.4.2. Let degD = d ≥ 3 and fix an embedding i : X → Pd−1 given by |D|. Since Xis nonsingular, it is normal, and hence to show projective normality is equivalent to showingH0(OPd−1(n)) H0(OX(n)) ∀n ≥ 0. Now, from the SES 0 → IX → OPd−1 → OX → 0, weneed show that H1(IX(n)) = 0 ∀n ≥ 0. This holds for n = 0, 1 since I0, I1 = 0 since i(X) isnonempty and nondegenerate (where I = Γ∗(IX)). Now,

IV.5 The Canonical Embedding

IV.5.1. Curves of genus 0 and 1 admits a degree 2 map to P1 and can be embedded as a com-plete intersection. So really, by hyperelliptic we mean g ≥ 2 automatically. For g ≥ 2 completeintersections, the canonical divisor is very ample (Ex.IV.3.3), and hence by (IV.5.2) it cannot behyperelliptic.

IV.5.3. The same reasoning as (Ex.IV.2.2) shows that the hyperelliptic genus g curves form anirreducible family of dimension 2g− 1. So hyperelliptic genus 4 curves are a family of dimension 7.

Now, for nonhyperelliptic genus 4, we have a canonical embedding X → P3 that is a completeintersection of an (irreducible) quadric and a (irreducible) cubic by (IV.5.2.2). For the quadric,we have

(3+2

2

)− 1 = 9 dimensional space, and the subsequence cubic that does that vanish on the

quadric chosen is of dimension(

3+33

)− 4 − 1 = 15. Lastly, PGL(3) has dimension 15, so we have

9 + 15− 15 = 9.Now, for the nonhyperelliptic genus 4 to have a unique g1

3, the quadric is a rank 2 quadric,which forms a family of dimension 8—it’s really a (irreducible) hypersurface in (P9)∗ given by thedeterminant of the symmetric matrix.

IV.5.4. We know that genus 4 curve X canonically embedded is a complete intersection of aquadric Q and a cubic C. Now, this irreducible quadric is either the nonsingular quadric P1 × P1

or the singular cone. In the first case, there are two g13’s coming from the two family of lines (note

that g13 is equivalent to three collinear points with the canonical embedding of X by R-R); these

g13’s are actually different since one collapses the three points along one family of lines while the

other keeps them distinct (the maps X → P1 are from the projections P1×P1 → P1). Furthermore,the sum of these two g1

3’s is the canonical divisor. In the singular conic case, there is a unique g13

since the lines in the family are all linear equivalent, and by the following lemma, and two timesthis g1

3 also gives the canonical divisor in this case (as the pullback of hyperplane is the double lineon the singular cone).

Lemma. LetX ⊂ Y closed embedding of schemes satisfying (∗)+locally-factorial of (II.6). SupposeD,D′ ∈ DivX such that D = X ∩ E, D′ = X ∩ E′ for some codimension-1 integral subschemes

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E,E′ of Y . Then E ∼ E′ on Y implies that D ∼ D′ on X. Proof: By assumption E,E′ does notcontain X. Let f ∈ K(Y ) be the rational function such that div f = E − E′. Then affine locallyf = a/b where b does not vanish on X. Hence, it makes sense to map f to a rational functionf ∈ K(X) by a/b (if A → A/p induces X → Y , then we have Ap → κ(p)). Then f now inducesD ∼ D′ since intersecting E (for e.g.) is the same is looking at where a vanishes.

Now for the actual problem:

(a): If X has two g13’s, then it is necessarily on the nonsingular quadric when canonically

embedded, so project away from a point P on the curve to get X \ P → P2 that is injectiveexcept at over two points in the image where P is a part of three collinear points. Thus, theclosure of the image X ′ is birational to X, and has two nodes, and thus by (Ex.IV.1.8) we havepa(X

′) = p(X) + 2 = 6 and hence degX ′ = 5.(b): If X is on the quadric singular cone however, and we take a point P on X, then there

is a plane through P that meet two other points Q,R with multiplicity 2 each (meets P withmultiplicity 2 as well) where P,Q,R are collinear. Hence the projection has a tacnode and is ofdegree 5 (note that δP = 2 for tacnode as twice blow-up resolves the singularity). To only havenodes, we need project from a point not on the singular cone. But then the number of secantsthrough the point is 6 By some intersection theory stuff....

IV.6 Classification of Curves in P3k

Note. Suppose X ⊂ Pn a curve and and P ∈ Pn such that projection from P induces X → X ′ abirational map. Then the degree of X ′ is one less than X if P ∈ X and same as X if P /∈ X.

Lemma. Let Y be a complete intersection in Pnk , and X ⊂ Y ⊂ Pnk projective k-variety that is aclosed subscheme of Y . Suppose dimX = dimY = 1 and degX = deg Y . Then X = Y . Proof:Y is ACM, and hence equidimensional, and moreover C(Y ) the affine cone is reduced since it is2-dimensional CM ring (hence normal). Now, dimX = dimY implies that X is an irreduciblecomponent of Yred = Y , and thus deg Y = degX implies that Y has no other components, thusY = X.IV.6.1. Let X ⊂ P3 be a rational curve of degree 4. Then taking LES of 0→ IX(2)→ OP3(2)→OX(2)→ 0 and counting dimensions gives us h0(IX(2))−h1(IX(2)) = 1, and hence h0(IX(2)) ≥1. Now, X cannot be contained in a plane since it is rational, and hence, if h0(Ix(2)) ≥ 2, sothat X lies on two different irreducible quadrics F1, F2 (irreducible since X is not in a plane), thenY := V (F1, F2) is a 1-dimensional k-scheme of degree 4 containing X and hence X = Y . But thiswould mean that X is a genus 1 curve. Hence, X lies on a unique quadric. If it lies on the singularquadric, then a projection from a point on the quadric but not on the curve gives a birational mapto X ′ with a tacnode. But then, pa(X

′) + 2 = 0 + 2 = pa(X′), which is not possible.

IV.6.2. The same argument as (Ex.IV.6.1) gives h0(IX(3)) ≥ 4, and hence X lies on a cubicsurface. [s5, s4t, st4, t5] lies on a quadric surface, but [s5, s4t+ s3t2, s2t3 + st4, t5] does not.

IV.6.3. Once again, one can compute that h0(IX(2)) ≥ 1, so X lies on a quadric surface (a uniqueone since if X is in two quadric surfaces then its degree is at most 4). Now, given an abstract genus2 curve X, note that any divisor D of degree 5 is very ample since 2g + 1 = 5 here, and has rank5− 2 = 3, so |D| gives an embedding X → P3.

Now, a (not supposed to be correct) proof that X cannot be embedded on a singular quadric:If X is on a singular quadric Q, then X ∩ H for a plane H tangent to Q is X ∩ 2L where 2L isthe double line. Thus, degX ∩ 2L must be even but degX ∩H = 5 by Bezout. (This faulty proofshows that doing intersection theory on singular surfaces gets really tricky!)

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However, here is an example of genus 2 curve of degree 5 in P3 lying on a singular quadric. LetS(X) := S/I have a free resolution:

0 −−−−→ S(−4)2

x yy zw2 z2

−−−−−−−→ S(−3)2 ⊕ S(−2)

I2(M)−−−−→ S −−−−→ S/I −−−−→ 0

For nonsingular quadric, any curve of type (2, 3) on P1 × P1 will do.

IV.6.4. Suppose X a genus 11 degree 9 curve in P3. X cannot lie on a quadric surface since noneof the cases in (IV.6.4.1) work out. Now, let’s consider the sequence 0 → IX(3) → OP3(3) →OX(3) → 0. As OX(3) is nonspecial, then we have h0(IX(3)) ≥ 20− (27− 11 + 1) = 3. Thus, Xlies in two independent cubics that are necessarily irreducible (since X does not lie in a plane ora quadric). But then X is the intersection of these cubic surfaces, in which case the genus will be10. Thus, X of degree 9 and genus 11 does not exist in P3.

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Chapter V

Surfaces

V.1 Geometry on a Surface

V.1.1. Just plug into Riemann-Roch.

V.1.2. Note that PX(t) = χ(OX(tH)) = 12 tH(tH −K) + pa + 1 = 1

2H2t2 − (H.K)t+ pa + 1, and

that 2π − 2 = H.(H +K) = H2 +H.K.

V.1.3. (a): (Note that we can’t just use adjunction formula since D may be singular). Take χ of theSES 0→ OX(−D)→ OX → OD → 0 to get 1−χ(OD) = 1 +χ(OX(−D))−χ(OX) = 1

2D.(D+K).(b): Follows from (a).(c): Follows from (a).

V.1.4. (a): As X ⊂ P3 is a hypersurface of degree d, we have KX = dH − 4H where H is thehyperplane class (restricted to X). By adjunction formula, −2 = C.(C + K) = C.(C + K) =C.(C + (d− 4)H) = C2 + d− 4 (as C.H = 1).

(b): Set X = V (f = xd + xzd−1 + yd + ywd−1). Note that Sing(f) = V (dxd−1 + zd−1, dyd−1 +wd−1, xzd−2, ywd−2) = ∅.

V.1.5. (a): (d− 4)2H2 = d(d− 4)2.

(b):

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