Hartshorne Notes/Solutions

Christopher Eur
 This document was created to help the author study the book, and as a result is likely filled
with abundance of inelegance if not inaccuracies. Please use with caution. The solutions written
up here are: exercises that are really propositions, exercises that serve as good examples, exercises
that correct wrong intuitions, and exercises that just subjectively seem interesting. Some solutions
here and they come not from the author but from various sources which the author has failed
to 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 email
chrisweur@gmail.com.

1
Contents

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

II Schemes 12
II.1 Sheaves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
II.1.1 A bit of category nonsense... . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
II.1.2 Back to exercises... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
II.2 Schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
II.3 First properties of schemes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
II.4 Separated and proper morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
II.5 Sheaves of modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
II.6 Divisors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
II.7 Projective morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
II.8 Differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

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

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

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 IV.5 The Canonical Embedding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
IV.6 Classification of Curves in P3k . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

V Surfaces 50
V.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 + cy 2 + 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 use
k = k). Noting that (by)2 − 4acy2 = (b2 − 4ac)y2 , let D = b2 − 4ac. If D 6= 0, then C ' A1 \ {0}
since there is a linear change of coordinates so that F becomes XY = 1. If D = 0, then either
a = b = c = 0 or at least two among three are 0, or all nonzero, so that change of coordinates then
gives X = F or X 2 + DX + F = Y .
I.1.10(c,e). (c): Consider Spec k[x](x) and its generic points as the open subset. (e): Nagata has
an example of Noetherian ring that is infinite dimensional.
I.1.11. First, some easy general observations:
ϕ
Lemma 1: If A is a domain, then ker : B → A for any ring B is prime; i.e. im(ϕ# ) is irreducible.
Lemma 2: If ϕ : B → A is finite, then B/ ker ϕ ,→ A is finite, so dim A = dim im(ϕ# ) ≤ dim B.
The two lemmas immediately imply that the variety C ⊂ A3 defined by x = t3 , y = t4 , z = t5 is
a variety of dimension 1, and thus p := I(C) is prime of height 2 (∵ I.1.8A). Lastly, to see that it
requires three generators, consider solving for the kernel K of [t3 t4 t5 ] in k(t), and note that there
are three k[t]-linearly independent elements in K ∩ k[t].
I.1.12. x2 + y 2 (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, dim S(Y ) = dim A(Yi )[x±
i = tr degk S(Y )xi = tr degk A(Yi ) + 1 = dim Yi + 1.
]
Now, noting that dim Y = supi dim Yi , we have that dim S(Y ) = dim Y + 1, and in fact, we have a
stronger statement that dim S(Y ) = dim Yi + 1 for any Yi 6= ∅.
Remark: The statement is true even for Y a projective algebraic set (not just a projective
variety). However, the stronger statement is no longer true (as tr deg depends on having integral
domains).
I.2.7(b). Let Yi := Y ∩ Ui be one where dim is maximized so that dim Y = dim Yi . Then
dim Yi = dim(Yi ⊂ Ui ) = dim(Y ∩ Ui ) = dim Y (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) dim Y0 = n − 1 so that Y0 ' Spec k[x1 , . . . , xn ]/f0 for f0 irreducible. Now, Y = V (f0h ) can

4
be checked on all affine patches. Algebraic soln: Let’s use that S is a UFD. Let p be the prime such
that Proj S/p = Y , and so there is no (homogeneous) prime between (0) ( p. Take any nonzero
homogeneous element f ∈ p, and factor it (note that each factor is homogeneous too), so that at
least 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] )}. Let (y• ) ∈ Z(a), so that yI 6= 0 for some I ∈ I . WLOG assume 0 ∈ I. Then


d
xi yI\0∪i
we claim that = works (and this solution is unique up to scaling). To see this, note that
x0 yIQ Q
`∈J x` yI\0∪` yJ
for any J ∈ I , we have d
= `∈J d = (where last equality follows from (y• ) ∈ Z(a),
x0 yI y0d
and letting J = I we have y0d 6= 0). (c): With part (b), we have that D(yid )’s (i = 0, . . . , n) cover
k[y/y0d ’s]
Z. On D(y0d ), part (b) solution above shows that a ' k[x1 /x0 , . . . , xn /x0 ]. (d): The map
is [s, t] 7→ [s , s t, st , t ] so that on an affine patch Us it is [t, t2 , t3 ], which is the twisted cubic.
3 2 2 3

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 − z 2 (up to coordinate change). Hence, V (f ) = ν2 (P1 ) (the
degree 2 Veronese embedding of P1 ,→ P2 ). (e): Then A(Y ) ' O(Y ) ' k, and hence dim Y = 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 are
easy to check. (b): If ϕ is an isomorphism, its inverse morphism ψ gives natural inverse to ϕ∗p . For
the converse, use the fact that isomorphism is local on the target, by taking an affine cover of Y
and noting that Op ' Ap for affines. (c): If ϕ(X) = Y , then ϕ(ϕ−1 (U )) = U for any U ⊂ Y , so
that ϕ∗ is injective.
I.3.5. Let H = V (f ) ⊂ Pn with deg f = d. Then under νd : Pn ,→ PN (N = n+d


n − 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 dim Y ≥ 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) = [p0 q1 − p1 q0 , . . . p0 qn+1 − pn+1 q0 ],
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 a
morphism iff xxji ◦ 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 a
morphism. Proof: That a map is a morphism is a local property, so take distinguished affine covers
of 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 since
x × Y is irreducible (hence for each x ∈ X we have x × Y ⊂ Z1 or x × Y ⊂ Z2 ). Moreover, X1 and
X2 are closed since ideal membership test is an algebraic condition on the coefficients. But then
X = 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 for
K(X × Y ). (Use Vakil’s equivalence relation definitions for transcendental basis).

5
  
x0 y0 · · · x0 yn
I.3.16. (a/b): Writing the coordinates of PN as  ... .. .. , it is clear what the equations

. . 
xn y0 · · · xn yn
in I(X) and I(Y ) become in these coordinates. (c): work via distinguished affine patches by noting
that 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 over
Am for all m, and hence f ∈ Am for all m, and thus f ∈ A. (e): If A0 ⊃ A is integral closure of A,
and A ,→ B where B is integrally closed in K(B), then A0 ⊂ B.
I.3.18(b). That the twisted quartic is normal is easy (on affine patches, it is actually isomorphic
2 2
to A1 ). However, it is not projective normal since t2 u2 = zw ∈/ S(Y ) and ( zw )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 dim Y ≥ 2 guarantees
that p 6= m and moreover there exists m 6= m0 ⊃ p. Thus,
\
Am = Ap
p⊂m, ht p=1

f ∈ O(Y \ P ), then f ∈
T
by algebraic Hartog’s lemma. Now, if T m0 6=m Am0 . Moreover, f ∈ Am by
our observation above, and hence f ∈ m0 Am0 = A = O(Y ).

I.4 Rational maps

I.4.4(c) Yx ' V (y 2 = 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 any
two curves in P2 intersect, whereas P1 × P1 has a family of lines whose members don’t intersect.
I.4.6. (a,b): For a1 a2 a3 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 Cl P2 = Cl U = Cl P2 \ S as
points are of codimension 2 in P2 . So, it ϕ extends, then the linear system on U is the same as
that of the original, i.e. k{xy, 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 isomorphism
∼ ∼
θ : OY,q → OX,p gives θe : K(Y ) → K(X), which gives a rational map X → Y by functions
e i ) = θ(yi ) ∈ OX,p . So the rational map is defined on an open set containing p. To see that
θ(y
p 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 map
of local rings OY,q → OX,p that extends ϕ∗ : O(Y ) → O(X). Then ϕ(p) = q. Proof: Suppose
ϕ(p) = q 0 6= q. If f ∈ O(Y ) vanishes on q then ϕ∗ (f ) must vanish on p. Now, restricting to U ⊂ Y
affine if necessary (so that q, q 0 correspond to two different maximal ideals) we can find a function
f ∈ OY,q such that f (q) 6= 0 but f (q 0 ) = 0 (f is a unit). Then ϕ∗ (f )(p) = f (q 0 ) = 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. The
curve C 0 = C \ (0, 0) is equal to C \ V (x) or C \ V (y). In the ring k[x, y]/(x3 − y 2 ), note that

6
inverting x makes y invertible and vice versa, so the in either case the coordinate ring k[C 0 ] is
k[x± , y± ]/(x3 − y2 ). We now compute the fiber
ϕ−1 (C 0 ) −−−−→ X ⊂ A2 × P1
 
 
y y
C0 −−−−→ 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 − y 2 , x ut − y) = k[x± , y ± , ut ]/(x − ( ut )2 , y − ( ut )3 ) so that the
closure of ϕ−1 (C 0 ) 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( ut )2 − 1, y( ut ) − 1), so that
the closure does not contain a fiber over (0, 0). In fact, all the points of C e belong to Ut , and the
map C e → C is given by k[x, y]/(x3 − y 2 ) ,→ k[x, y, ]/(x − ( )2 , y − ( )3 ) ' k[ u ]. In Ex.I.3.2. we
u u u
t t t t
showed 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 a
null 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, it
is dimk (grm M ) = dimk M/mM + dimk mM/m2 M + · · · (if M is Artinian, then mN M = 0 for
N >> 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 y n is nonzezo in g. Then
O0 /(f, g) = mn. Proof: The monomials are exactly {xα y β : 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 I
dont’ 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 (the
nonzero homogeneous part of f of least degree) has nonzero xr coefficient. If a 6= 0 (so WLOG
a = 1), then do change of coordinates x0 = x + by, y 0 = y. Then the equality holds iff fr0 has nonzero
y 0r coefficient (the cases when the coefficient is zero is a polynomial equation in b, so there are only
finite 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 of
degree d. Now, note that for p = (px , py , 0) ∈ Y ∩ L ∩ Ux , we have Op /(f, z) ' (k[ xy ]/f0x )p where
f0x is f dehomogenized and then z set to 0. Note that f0x is g x , the dehomogenization of g. Since p
p
is a point on Y ∩ L, it corresponds to a root of f0x so that `(Op /(f, z)) = power of ( xy − pxy ) in the
x
linear factorization of f0 , i.e. the power of xpy − ypx in the linear factorization of g. From this it
is clear that, (L · Y ) = d.
I.5.5. xd + y d + z d works as long as d 6= p. For d = p, consider f = xy p−1 + yz p−1 + zxp−1 . Then
[df ] = [y p−1 + (p − 1)zxp−2 , z p−1 + (p − 1)xy p−2 , xp−1 + (p − 1)yz p−2 ]. Setting [df ] = 0, and noting
that char = p, we have that the singular points are given by f = 0 and xy p−1 = yz p−1 = zxp−1 . The
only solution is x = y = z = 0 if p 6= 3. Actually, the better thing to do is this: f = xp +y p +z p +xyz.

7
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, we
have f = 0 and xu = y, so that plugging in we have x2 u + g(x, xu) = 0. But x2 |g(x, xu) since
deg g > 2 so that we have x2 (u + g(x, xu)/x2 ) = 0. The x2 = 0 gives the exceptional divisor
E, and u + g(x, xu)/x2 = 0 gives the pullback of our curve. Here, if x = 0 then u = 0 (since
deg g > 2), so on this affine patch the blow-up curve meets E at u = 0 (i.e. once). Moreover, the
point (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 ∂x
∂f ∂
(x0 , . . . , xj = 1, . . . , xn ) = ∂x f (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 i

I.5.7. (a): Let X = V (f ). That X \ {0} is nonsigular follows easily from the lemma above, and
moreover, that X is in fact singular at (0, 0, 0) follows from deg f > 1.
(b): X e is possibly singular at points on X e ∩ E (where E is the exceptional divisor). Now, let
3 2
(x, y, z, t : u : v) be coordinates on A × P and consider X e ∩ Ut . The equations for Bl0 (A3 ) is
xu = 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) = xd f (1, u, v). Now xd = 0
gives the exceptional divisor E, and f (1, u, v) gives the strict transform. X e ∩ Ut is nonsingular on
points (x, u, v) = (0, u, v) satisfying f (1, u, v) = 0, since X is singular only at (0, 0, 0). Likewise for
patches 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
∂fi
are homogeneous (hence so are ∂x j
’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 the
rank 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, so
consider p in the intersection. Then Op /f is not an integral domain. However, V (f ) is a nonsingular
curve, 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 on
the dimension of A. If dim A = 0, then A ' k so A is a domain. Now, if dim A = n, take a
regular sequence (x1 , . . . , xn ) (exists by regular-ness). Now, consider A0 = A/x1 . By Krull’s height
theorem, dim A0 = n − 1, and moreover m/m2 ' (m/m2 )/(k · x1 ), so that A0 is in fact regular local
ring as well. Thus, A0 is a domain. Now, suppose a, b ∈ A nonzero such that ab = 0. By Krull’s
intersection theorem, we can write a = a0 xm 0 n 0 0 / (x ). Since x is a nonzero
1 and b = b x1 where a , b ∈ 1 1
0 0
divisor, we thus have a0 b0 = 0. However, this implies that a0 , b 6= 0 but a0 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 -modules
n → m, from which we get a map n/n2 → m/m2 of OY,q -modules and hence OY,q /n ' k-vector
spaces. 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 by
yi 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)/(y 2 ) which comes from restriction of
k[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 y 2 z−x3 +xz 2 ,
the equation for the elliptic curve E ⊂ P2 . For any points (x : y : z) ∈ E satisfying such that y 6= 0

8
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 reads
2 = 0). Thus, ϕ : Y \ P → E \ [1, 0, −1] is a bijection. Moreover, we know check that the map is an
isomorphism 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 y 2 z − x3 + xz 2 = 0 becomes y 2 (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 in
the 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 a
smart 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 ). Let
A0 be integral closure of A where A is a finite-type k-algebra domain. Theorem I.3.9A implies that
for some f1 , . . . , fm ∈ A0 , the map A[f1 , . . . , fm ] ,→ A0 is in fact surjective (hence isomorphism).
Now, since fi = ai /bi for some ai , bi 6= 0 ∈ A, consider Ab1 ···bm . Clearly, A0 ⊂ Ab1 ···bm , and in fact,
A0b1 ···bm ' Ab1 ···bm . Thus, Ab1 ···bm is integrally closed. This shows that the nonnormal points are
proper.
To show that normal points are in fact open, suppose m ⊂ A corresponds to a normal point.
Since (A0 )m is the integral closure of Am , we have that the map A ,→ A0 induces an isomorphism
∼
Am → (A0 )m . By Ex.I.4.7, we have that there is an open set of Y containing m that is isomorphic
to an open subset of Ye , 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• ,
n
so that PY (z) = PPn (dz) = dz+n = dn! z n + · · · . (b): If Y ⊂ PN is the image of the Segre embed-


n
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) =
L
z+r z+s 1 r+s



r s = r!s! z + ···.
Lemma. Let S(Y ) and S(Z) be coordinate rings for projective varieties Y ⊂ Pn and Z ⊂ Pm . Then
n m
L coordinate ring for Y × Z as a subvariety of the Segre embedding of P × P is isomorphic to
the
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 .
f
I.7.2(c,d,e). (c): From 0 → S(−d) → S → S(H) → 0, we have PH (z) = z+n − z−d+n



n n so that
−d+n (n−d)(n−d−1)···(n−d−n+1) (d−1)···(d−n)
n−1 n 2n = d−1



pa (H) = (−1) (1 − n − 1) = (−1) n! = (−1) n! n .
g
(d): Let f, g polynomials defining hypersurfaces H1 , H2 of degree a, b. Then, from 0 → S/f (−b)
→
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−3 b−3


1 1
3 − 3 = 6 (3ab(a + b) − 12ab + 6) = 2 ab(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 line
a0 x0 + a1 x1 + a2 x2 = 0 going through the point p (i.e. ~a · p~ = 0). WLOG, let p0 6= 0. Then, since

9
a0 = − a1 p1p+a
0
2 p2
, all the lines through the point p are given exactly by a1 ( xx10 − pp01 ) + a2 ( xx20 − pp20 ) = 0
(where a1 , a2 not both zero), and so we can work in the affine chart U0 . Now, apply the following
lemma:
Lemma. If f (x, y) is a polynomial and P = (p, q) ∈ V (f ) ⊂ A2 , and L : a(x − p) + b(y − q) = 0
is a line through (p, q), then the intersection multiplicity V (f ) ∩ L at P is the multiplicity of the
root X = 0 of the polynomial f (X + p, − ab X + q) (if b 6= 0). In particular, if V (f ) is nonsingular
at 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 = 0
to do the coordinate change, the linear term vanishes iff (a : b) = (fx (P ) : fy (P )). Otherwise, the
linear 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 it
is not tangent to any point of Y and does not go through singular points of Y . Since dim Y = 1,
we have dim Sing Y = 0, so that Sing Y is finitely many points {~ p1 , . . . , p~m }. Thus, the set of lines
that go through singular points or is tangent to a S
point of Y is given by the union of the morphism
of 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 meets
Y at p with multiplicity m (i.e. there is an dense open subset U of P1 ' {lines through p} such that
if L ∈ U then i(Y, L; p) = m. Now, WLOG let P = (0 : 0 : 1) ∈ Y be the point with multiplicity
d − 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:

ϕ ψ
Y \ P → P1 99K 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 of
Y \ 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 dim Y = 1. Take P 6= Q ∈ Y and let L = P Q. If there exists R ∈ Y such that
R∈ / L, we can find a hyperplane H such that H ⊃ P Q and H 63 R. Then since deg Y = 1, Bezout’s
implies that Y ∩ H must be a degree 1 variety with dimension 0 (dimension by Krull). But Y ∩ H
contains two points P, Q and hence is a contradiction. Hence, P Q = Y .
Now, let dim Y = r. By induction hypothesis, for any H 6⊃ Y , we have that Y ∩ H is a linear
variety of dimension r − 1. By dimension argument, we can find two hyperplanes H1 , H2 such that
Zi = 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 dimension
r − 1 so that Z1 + Z2 as a k-vector space has dimension r + 1. Thus, Z1 Z2 := Z1 + Z2 as a projective
linear variety has dimension r. Now, if there exists P ∈ Y such that P ∈ / Z1 Z2 , then since Z1 Z2
has dimension r < n we can find a hyperplane H such that H ⊃ Z1 Z2 but H 63 P . Once again
Y ∩ H must be a variety of dimension r − 1, but it contains Z1 and Z2 , which is a contradiction.
Hence, Y = Z1 Z2 , 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 must
be 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 variety
in Pn is thus in fact a complete intersection.

11
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 that
composition is a Z-bilinear map, (ii) the zero object exists, and (iii) finite direct product / coproduct
exist.
A few easy facts regarding additive categories:

• finite direct product is isomorphic to coproduct.

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

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

then 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 coker ϕ = 0, then ϕ is a monomorphism
or 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:
i θ π
ker θ −−−−→ A −−−−→ B −−−−→ coker θ
 x
 
y 
∼
coim θ −−−−→ im θ
Now the notion of exact sequences make sense, and the statement (∗) above is true. For example,
ϕ 0
if A → B → 0 is exact, then B = ker 0 = im ϕ = ker(B → coker ϕ), so that coker ϕ = 0 (so far

12
we didn’t use Abelian-ness). To conclude that ϕ is epimorphism, we note that the diagram above
becomes:
i ϕ π
ker θ −−−−→ A −−−−→ B −−−−→ 0
 x
 
y Id
∼
coim ϕ −−−−→ B
so that A → coim ϕ ' 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 topological
space X. That ShX is an additive category is easy to check. Moreover, that kernels and cokernels
exist is also easy.
N.B. Hartshorne defines =(ϕ : F → G ) by sheafifying the presheaf =ϕ(U ) := =ϕ(U )(F (U )). To
see 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 composed
with sheafification of G , then ker π + ' (ker π)+ .
The proof is just universal property nonsense. The lesson is that to say ker(π : F → G ) is a
sheaf, 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 presheaves
then F + ' G + naturally as sheaves. So, we can check on the presheaf coimage and image the
isomorphism, 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 that
exact sequences make sense, and kernels, cokernels, images, and coimages as defined in Hartshorne
match categorical theoretical viewpoint. In fact, the terms “injective” and “surjective” is now
better 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.
i ϕ
II.1.2. (a): Let’s just do everything concretely. From ker ϕ → F → G , we have a natural
ip ϕp
map (ker ϕ)p → Fp → Gp . The map ip takes [(f, U )] ∈ (ker ϕ)p and treats it as an element
of Fp . Injectivity is clear, and if [(f, U )] ∈ ker ϕp then ϕ(f )|V = 0 for some V ⊂ U so that
[(f, U )] = [(f |V , V )] ∈ (ker ϕ)p since f |V ∈ ker ϕ(V ). For im ϕ, since im = ker of coker, use the fact
that 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 ∈ X
iff F = 0. Our category theory definition makes it clear that ker = 0 or coker = 0 implies injective
or surjective. Hence, via part (a), that ϕ is injective iff ϕp is injective ∀p follows. Note that
coker ϕ = 0 iff im ϕ = G , and if (im ϕ)p = im ϕp = Gp for all p, then (coker ϕ)p = 0 for all p. (Here
we secretely use that coker ϕp = (coker ϕ)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 0 is the cokernel of F 0 → F , and F 0 → F is a monomorphism (which is a
kernel of a cokernel ). Piece together category stuff. (b): Same thing; piece together category stuff.
II.1.13.(Espace Étalé). The topology on Spé(F ) can be described more concretely as: V ⊂ Spé(F )
is open iff for every s : U → Spé(F ) the set s−1 (V ) is open in U . This means that at each point
p ∈ 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 → Spé(F ) such that π ◦ t = IdU
and locally at each point p ∈ V ⊂ U , the map t|V is equal to tp for tp ∈ F (V ), and hence t is
continuous. Conversely, if t : U → Spé(F ) is a continuous section, then it is locally a section of
F 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 00 . 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 exists
t ∈ F (U1 ∪ U2 ) such that ϕ(t|Ui ) = ϕ(ti ), i = 1, 2. Proof: Since t1 |U12 − t2 |U12 ∈ ker ϕ(U12 ), and
ker ϕ is flasque, the element lifts to t0 ∈ ker ϕ(U1 ). Now, considering t1 − t0 and t2 (which agree on
intersection now), by sheaf condition we obtain t satisfying the lemma.
Now, let s ∈ F 00 (U ). Let Σ be a poset of pairs (Ui , ti ∈ F (Ui )) where ϕ(ti ) = s|Ui ordered by
compatible inclusion. Given any chain, sheaf condition ensures that the union of the chain is the
upper
S bound. Hence, Σ has a maximal element. Now, by Ex.II.1.3 there exists ti ∈ F (Ui ) with
i U i = 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
S
agreeing on intersections, then it uniquely patches to s ∈ H (V ), and the support is still contained
in V ∩ Z since sp = (si )p for p ∈ Vi .
(b): It is clear that H = ker ϕ where ϕ : F → j∗ (F |U ).

II.2 Schemes
Lemma. Taking nilradical commutes with localization. Precisely, denote by N(A) the nilradical
of A, then N(S −1 A) = S −1 N(A). Proof: ⊃ is easy. For ⊂, if (a/s)n = 0, then tan = 0 ∃t ∈ S so
that (ta)n = 0 and hence ta ∈ N(A). Thus, as = ta −1
ts ∈ S N(A).
II.2.3.(Reduced schemes) (a): The above lemma implies that an affine scheme U = Spec A is
reduced iff Ap is for all p ∈ U . Moreover, reduced-ness is clearly an
`affine local property. Alternate
Soln: 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 if
U ⊂ X is an affine subscheme then OXred (U ) = OX (U )/N(OX (U )), and thus letting A := OX (U )
we have (U, OXred |U ) ' (Spec A/N(A), OSpec A/N(A) ) (again by Lemma above). That there is a

14
morphism Xred → X is clear: the topological map is just homeomorphism (since quotienting by
the nilradical doesn’t change the topology in the affine patches) and the map of sheaves is the one
above OX → OX /NX .
(c): We are looking at:
f
Y o X

 } ∃!
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 of
cokernel, 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 , fi# ) : 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 we
have
/ OY (U ) / // Q
i OY (U ∩ Ui ) i,j OY (U ∩ Ui ∩ Uj )
Q
0
∃! fi#
  
/ OX (f −1 (U )) / // Q
i OX (f i,j OX (f
−1 (U ∩ U )) −1 (U ∩ U ∩ U ))
Q
0 i i j

Since f # |Ui = fi# , it is clear that fp# 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 a
cover {Ui } of X such that ϕ|Ui = ψ|Ui for all i. Then in fact ϕ = ψ. Proof: Use the ! part of the
existence of the downward map above.
Lemma. Let Spec A, Spec B ⊂ X be two affine opens in scheme X with nonempty intersection.
Then for any p ∈ Spec A ∩ Spec B, there exists an open subset U of the intersection containing
p such that Spec Af ' U ' Spec Bg for some f ∈ A, g ∈ B. Proof: Take a distinguished affine
open of A in the intersection, and then a smaller one inside that is distinguished in B. Now use
Ex.II.2.16(a).
II.2.4. We construct the inverse map of α as follows. Cover X by affines Ui = Spec Bi . Given a
ring map A → OX (X), we have maps ϕi : A → OX (Ui ), which gives us morphisms fi : Ui → Spec A.
Moreover, for any p ∈ Ui ∩ Uj take an affine Uij = Spec Bij around p distinguished in both Ui and
Uj (possible by lemma above). Then the commutativity of the diagram

A −−−−→ OX (Ui )
 
 
y y
OX (Uj ) −−−−→ OX (Uij )

15
imply (via lemma above) that fi ’s agree on the intersections. Hence, they glue to a unique morphism
f : X → Spec A.
II.2.7. We prove a more general version. Note that for any x ∈ X there is a natural morphism
Spec OX,x → X. We claim that for any local ring (A, m), giving a morphism Spec A → X is
equivalent to giving a point x ∈ X and a local ring map OX,x → A. That is, there is a canonical
map Spec OX,x → X such that:

Spec OX,x
7
∃!

Spec(A, m) /X

Proof: If x ∈ X is the image of m under the map ϕ : Spec A → X, then for any open set
x ∈ U ⊂ X and p ∈ Spec A, we have ϕ(p) ∈ U since continuity of ϕ implies that ϕ({p}) ⊂ ϕ(p) so
that x ∈ ϕ(p). So, take U to be any affine Spec B ⊂ X. Then the map ϕ factors through Spec B, so
that we get a map B → A where m pulls back to the point corresponding to x. Hence, by universal
property 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 : Spec k[]/2 → X. The claim in the previous exercise Ex.II.2.7. shows that this
is the same as giving a local homomorphism OX,x → k[]/2 for some x ∈ X. To see that x is
rational over k, note that since X is a k-scheme OX,x , so that the extension of the local map above
κ(x) ,→ k must be an isomorphism. Lastly, note that local map OX,x → k[]/2 induces a map of
κ(x) = k-vector spaces mx /m2x → ()/(2 ), whose dual map gives an image of ∨ 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 any
homogeneous f ∈ S+ we have f n = 0 ∈ p so that p ⊃ S+ . If there is an element f ∈ S+ that is not
nilpotent, then S(f ) is not a zero ring, such that D+ (f ) ⊂ Proj S• is nonempty.
(b): U = Proj T \ V (ϕ(S+ )) by definition is open. Moreover, it has an affine open cover by
{D+ (f ) ⊂ U | f ∈ ϕ(S+ ) homogenous}. Since S → T naturally extends to S(f ) → T(ϕ(f )) , we
can define morphisms ψf : D+ (f ) → Proj S for each f ∈ ϕ(S+ ), and these maps agree on the
intersections since S(f ) → (S(f ) ) gdeg f ' S(f g) commutes with ϕ.
f deg g
(c): If p ⊃ S>d for some d, then p = S+ in fact. Hence, for any d > 0, we have an open cover
{D+ (f ) | f ∈ S>d homogeneous} of Proj S. This first shows that if S → T is a map of graded
rings that is isomorphism after high enough degrees, then U = Proj T , and moreover, the maps
ϕf : D+ (f ) → Proj S is an isomorphism since S(f ) → T(ϕ(f )) is (since deg f ≥ d0 ). Now use
Ex.II.2.17(a) below.
(d): Patch isomorphisms t(V ∩ Uxi ) ' Spec S(xi ) together.
II.2.15. (a): WLOG V is affine so that t(V ) = Spec A for A = k[x1 , . . . , xn ]/I, and note that
localization 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 implies
that 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 if
B → 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 Ui0 := Ui ∩ Xf , and let Ai = OX (Ui ). Then since X since index i is finite, there
exists n and ai ∈ Ai ’s such that f n b|Ui0 = 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 pairwise
intersection, there exists m such that replacing ai by f m ai we have ai = aj on intersections. Thus,
f n+m b 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 is
invertible (glue inverse of f , which is just 1/f , from each affine pieces). Now, this map is injective
if 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 clear
since stalk maps are isomorphisms.
(b): There is a natural morphism π : X → Spec OX (X). If f1 , . . . , fn ∈ OX (X) such that
Spec OX (X)fi ’s cover Spec OX (X), then Xfi = π −1 (D(fi ))’s also cover X. And moreover, since
Xfi ’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 →
Spec A, there exists a unique map Spec OX (X) → Spec A such that

X / Spec OX (X)

∃!
% 
Spec A

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 Spec B ⊂ Y and Spec A ⊂ f −1 (Spec B) so that we have f |Spec A : Spec A →
Spec B given by ϕ : B → A. Since X, Y are integral, A, B are domains. Moreover, since f
is dominant, the generic point must go to the generic point, and hence ϕ is an injection. Let
K(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 ]⊗B K(B) ' K(B)[a1 , . . . , an ].
For this to be a finite set, dim K(B)[a1 , . . . , an ] = 0, and hence tr deg K(A)/K(B) = 0. Thus, by
Noether normalization, we have that K(A) is a finite extension of K(B). Localizing B by all the
denominators that show in the monic polynomials (in K(B)) that the ai ’s satisfy, we have that
Bf → Bf [a1 , . . . , an ] = Aϕ(f ) is a finite ring map.
Now, since Y is irreducible, WLOG let Y = Spec B be affine, and moreover, shrinking Spec B
if necessary, we can assume that f −1 (Spec B) can be covered by finitely many affines Ui := Spec Ai
such that Ui → Y is a finite morphism. We finish with the following lemma:
Lemma. Suppose f : X → Spec B is an morphism of integral schemes such that X is covered by
finitely many affines Ui := Spec Ai such that Ui → Spec B is an integral morphism. Then there
exists b ∈ B such that f : Xb → Spec Bb is an affine morphism.
T
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 = Spec Aiai ∀i for some choices of ai ∈ Ai . Since ai
is 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 Spec Aibi ⊂ Spec Aiai . Thus, the inverse image
of Spec Bb where b = i bi is Spec Aib (for any i) so that Xb → Spec Bb is a morphism of affine
Q
schemes (and is still finite type / integral if f was).
II.3.10. (a): Categorical nonsense reduces to the case where we have Spec A → Spec B and p ⊂ B.
Then Spec A ×Spec B Spec κ(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 closed
immersion, 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 are
surjective 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 Ue 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
e
(by Ex.II.2.16). In other words, we have:

(U ∩ Y )f −−−−→ U ∩ Y −−−−→ Y
  
  
y y y
D(f ) −−−−→ U −−−−→ X

where all squares are fiber diagrams and right arrows are open embeddings, so that all the down
arrows are closed embeddings (by the short lemma above). Covering Y by these affines, taking
finite 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 for
some I.
(c): We have f : Y → X and f 0 : Y 0 → X. Note that√ topologically, f and f 0 are the same. For
any U = Spec A ⊂ X, we have that f −1 (U ) = Spec A/ I and (f 0 )−1 (U ) = Spec A/I. Since there
is a natural map such that
√
A/ I o A
O
∃!
}
A/I

which defines a map f −1 (U ) → (f 0 )−1 (U ). These maps agree for any affines U, V ⊂ X, so they
patch up to give a morphism Y → Y 0 .
(d): Topologically, Y = f (Z) clearly. If Z is reduced or quasicompact, we show that we can
make 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 from
the diagram below:

I (U ) −−−−→ OX (U ) −−−−→ OZ (f −1 (U )) −−−−→

Q
i Ai
   
   
y y y y
I (Ug ) −−−−→ OX (Ug ) −−−−→ OZ (f −1 (Ug )) −−−−→ i (Ai )g
Q

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 the
kernel. In the case when ZPis qcqs, f∗ OZ is in fact quasicoherent, so that I is already quasicoherent.
In general, define IY := I 0 ⊂I QCoh I 0 , 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-valued
point Spec l → X (over k) is a closed point. In particular, the closed points of X are x ∈ X
such that κ(x) is an algebraic extension of k. Proof: For any map ϕ : k[x]/I → l, note that
A = k[x]/ ker ϕ is a k-algebra domain of finite type that injects into l (an algebraic extension of
k) and hence is a field. Thus, the point that corresponds to a l-valued point is a maximal ideal in
every 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 any
affine 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 does
not 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 Noetherian
condition 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 that
this is true for two closed points in an affine integral finite type k-scheme. Now, take affines U, V
containing p, q respectively, and take a closed point in an affine open in the intersection U ∩ V , and
then note that number equality is transitive (lol). Now, if X = Z0 ) · · · ) Zd = P is a the maximal
sequence of irreducibles giving the dimension of X, take and affine U = Spec A containing P . Then
dim A = dim X (use X irreducible to argue that Zi ∩ U ) Zi+1 ∩ U ∀i), and dim A = dim Op .
Combined with dim Op = dim Oq for any two closed points, we have that dim Op = dim X for any
closed point p ∈ X.
(e): If Spec A ⊂ U , then from part (a) we know that dim X = dim A ≤ dim U ≤ dim X.
(b): Take any Spec A ⊂ X. Then dim X = dim A = tr degk Frac(A) = tr degk K(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 Spec A around
Z0 , 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) = inf{dim 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 our
condition). Then R[t] ' k[x](x) ⊗k k[t]. Now, localizing by x gives k(x)[t] and quotienting by x
gives k[t]. That is, Spec R[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 dense
subset U of X. Then in fact X = Z. Proof: Since U is reduced, the scheme-theoretic image of U
is U red = X since U is dense and X is reduced. Thus, by universal property of scheme-theoretic
image, there exists a unique j such that

ZO /X
>
j
Id
X

from which we can deduce that Z ' X.

Warning: This is NOT true if X is not reduced. For example, consider X = Spec k[x, y]/(x2 , xy),
Z = Spec k[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 map

induced from f, g. Denote by h−1 (∆) (suggestively) the fiber product:

h−1 (∆) −−−−→ Y

 
 
y y∆
h
X −−−−→ 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 closed
embedding. Moreover, we have chain of fiber diagrams:

U −−−−→ h−1 (∆) −−−−→ Y

 
 
y y∆
h
U −−−−→ X −−−−→ 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 X2
 
 
y y
∆
Y −−−−→ 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 square
lemma a fiber diagram:
U ∩ V −−−−→ U ×S V
 
 
y y
∆
X −−−−→ X ×S X

20
where the horizontals are closed embeddings and verticals are open embeddings. Since U ×S V is
affine, 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 of
Z 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 Spec K(X) → X is the generic point, so that the closure of image is in
fact X. Thus, if R is a valuation ring of K, then Lemma II.4.4. specializes to say: to give a map
Spec R → X such that
Spec K /X
;


Spec R
commutes is equivalently to giving a point x ∈ X such that R dominates OX,x . Moreover, if we
require R to be a valuation ring of K/k, then giving a map Spec R → X so that
Spec K /
9X

 
Spec R / Spec k

commutes is equivalent to giving a point x ∈ X such that OX,x dominates R. That is, if X is
separated/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 exists
valuation ring R of K/k that dominates k[a−1 ](a−1 ) . By properness, there exists x ∈ X such that
R 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 : Spec A → Spec B is induced by
ϕ : B → A, then since f is proper ϕ is of finite type. Thus, to show finite, we need show that it is
integral. Replacing B by ϕ(B), we assume that ϕ is injective. Suppose R ⊃ B is a valuation ring
of K = K(A). Then R ⊃ A by properness. Since the integral closure of B in K, denoted B cl , is the
intersection of all valuation rings of K containing B, we thus have B cl ⊃ A. Thus, every element
of 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 containing
p, q. Note that V := σ(U ) is also affine containing p, q. Thus, since X is separated, U ∩ V is an
affine containing p, q such that σ(U ∩ V ) = U ∩ V . Hence, we first consider the case when X is
affine Spec C[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:
Γf
X −−−−→ X ×Z Y
 

fy

y
∆
Y −−−−→ Y ×Z Y

21
Combined with the diagram
X ×Z Y −−−−→ Y
 
 
y 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

&  /Y

X

II.4.9. We have closed embeddings i : X → PnY and j : Y → Pm n n

Z . Now, note that PY ' PZ ×Z Y .
i Id ×j
nm−m−n
So, we have an closed embedding X → PnZ ×Z Y −→ PnZ ×Z Pm
Z ,→ PZ .
II.4.10(c). This is like,, impossible...
II.4.11. Later when studying CA

II.5 Sheaves of modules

⊕I ⊕I
Lemma. For G an OX -modules, we have a natural HomOX (OX , G ) ' Hom(OX (X), G (X)).
Proof: → map is clear. For →, for any U , the image of the standard basis ei of OX (U )⊕I is
determined by eei of OX (X)⊕I → G (X) → G (U ) since eei 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 since
isomorphism 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 −1 V ) ⊗ f∗ f −1 E (V ), but since
f∗ f −1 OY (V )
are adjoints there are natural maps OY (V ) → f∗
(f −1 , f∗ ) f −1 O Y (V ) and E (V ) → f∗ f −1 E (V ),
which gives us a map

F (f −1 V ) ⊗ E (V ) → F (f −1 V ) ⊗ f∗ f −1 E (V )
OY (V ) f∗ f −1 OY (V )

Now, for the isomorphism, since we can work locally, assume that E ' OY⊕n . The above map is
just F (f −1 V )⊕n → F (f −1 V )⊕n where the left is a OY (V )-module and the right as f∗ f −1 OY (V )-
module restricted to OY (V )-module. This is clearly an isomorphism.
II.5.6. (a): For p ∈ X, we have mp 6= 0 ⇐⇒ f m 6= 0 ∀f ∈ A \ p ⇐⇒ p ⊃ Ann(m).
(b): If p ∈ Supp M , then Mp 6= 0 =⇒ f M 6= 0 ∀f ∈ A/p ⇐⇒ p ⊃ Ann(M ) (note that ⇐=
may not hold if T 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 f mi 6= 0 hence Mp 6= 0. (We actually did not need Noetherian).
(c): This follows from part (b).

22
 (d): First, HZ0 (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 an
affine X = Spec A now, ΓI (M ) = ΓZ (X, F ) as modules (where M = Γ(X, F )). Well, for m ∈ M ,
√
Supp(m) = V (ann m) ⊂ V (I) implies that I ⊂ ann m so that some power of I lies in ann m (as
I is finitely generated)—i.e. m ∈ (0M : I ∞ ). Conversely, if m ∈ (0M : I ∞ ), then I n ⊂ ann m for
some n so that V (I n ) = V (I) ⊃ V (ann m) = supp m.
(e): Follows from the same argument in (d).
f. Suppose Mp is free, i.e. A⊕n ∼
II.5.7. (a): Can assume X affine so that F = M p → Mp . This map
⊕n
can actually made so that it lifts to A → M . Since M is coherent, we have a exact sequence:
α β
A⊕m → A⊕n → M → 0

where im αp = 0. But since im α is finitely generated, there exists f such that (im α)f = 0. Hence,
∼
localizing the sequence above by f , we obtain A⊕n
f → 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 have
Mp ⊗ Np ' Ap at each point p. Now, by applying κ(p) to both sides and using Nakayama, we note
that 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 if
we 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

α β
A⊕m → A⊕n → M → 0

With rank-nullity, one concludes that dim M |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.
α β
(c): Suppose ϕ = n constant. We again work affine locally, which gives us A⊕m → A⊕n →
M → 0 again, where now [α|p ] = 0 for all p. That is, all the entries of [α] are in the nilradical of
A, 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
f(d)) = M (d)(f )
which glues (as Mf(d) is a sheaf) to a map αd : Md → Γ(X, M
f(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.

23
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 map
ϕ : Spec A[t] → Spec A. If p ∈ Spec A, then p[t] ∈ Spec A[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]
P i P
and if ai t / bi ti ∈ A[t]p[t] then the valuation is maxi {mi : uπ mi = ai }. Thus, the map
ϕ∗ : WDiv(Spec A) → WDiv(Spec A[t]) is a well-defined homomorphism that also sends principal
divisors to principal ones. Moreover, note that this is an injection at the level of class groups.
Observation 2: For q ∈ Spec A[t], codimA[t] q ≥ codimA ϕ(q) (codimension can at most decrease
under 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 have
q ⊗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 (or
X × Pn → X) is AnK(X) (or PnK ). In other words, we have

AnK (or PnK ) −−−−→ X × An (or Pn )

 
 
y y
Spec K(X) −−−−→ X

Thus, we have a well-defined map WDiv AnK → WDiv X × An via the top map (and locally it is
f ∈ K[x] treated as a polynomial in Af [x] for some f ∈ A). This map again sends principal divisors
to principal divisors. In the case of X × Pn , we thus get a map Cl PnK → 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 corresponds
to P where P ∈ A1K is a closed point. By the same reasoning (with induction), a prime divisor of
X × 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 ones
corresponding to hypersurfaces of PnK .
Theorem. It thus follows that Cl X × An = Cl X and Cl X × Pn = Cl X ⊕ Z.
II.6.1. X ×Pn is indeed Noetherian, separated, and integral (see TIL:7/20/2016), and since X ×An
is 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 be
a prime divisor, and note that X × Pn \ X × H ' X × An . This gives us a exact sequence:

Cl Pn ' Z → Cl X × Pn → Cl X × An → 0

where the first map sends [H] to [X × H], which is the map π2∗ . this map is injective since if
[X × H] = 0, then for f ∈ K(X × Pn ) such that div f = X × H, f restricted to Spec A × Ui for
x
any i 6= 0 and for any Spec A ⊂ X is just x0 /xi ∈ A[ xji s], but this is impossible. Moreover, note
∼
that composing Cl X × An → Cl X and π1∗ , we get a map σ that is a section of the second map in
the 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(Pnk ), 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 a
hyperplane, we get a map ϕ : Cl Pn → Cl X that is well-defined.
(c): We need show that νYi (f i ) = µpi (S/I(X) + (f )) where pi is the homogeneous prime in
S := k[x]• corresponding to Yi . Combining the two facts in TIl 7/22/2016, we have that if p0i is
the prime corresponding to Yi ∩ Ui in the coordinate ring A(X ∩ Ui ), then νYi (f i ) = νp0i (f i ) =
µp0i (A(X ∩ Ui )/I(X ∩ Ui ) + (f )) = µpi (S/IX) + (f )), as desired.
Now, forP deg(D.X) = deg D · deg X, it suffices to show for D = V a hypersurface. But
deg(V.X) = i i(X, V ; Yi ) deg Yi = (deg V )(deg X) 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 deg D = 0 from the equation in part (c). Hence, we get a well-defined
map Cl X → Z. Composing with the map ϕ in part (b), and using part (c), we get a commutative
diagram:
ϕ
Cl Pnk −−−−→ Cl X
 
'ydeg
 deg
y
× deg X
Z −−−−−→ Z
And 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 ) and
S(X) = k[x, y]• /I(V )• . Thus, on Vi = V ∩ Uxi , we have that π −1 (Vi ) = Spec k[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 TIL
8/26/2016 about An -bundles over X in regards to class groups.
(b): [V ] ∈ Cl X is given by y = 0 (continuing with S(X) = k[x, y]• /I(V )• ). If ` is a linear form
in 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 sequence
Cl V (y) → Cl(X \ P ) → Cl(X \ P ) → 0. Under the isomorphism in (a), the first map becomes
Z ,→ Cl V, 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 → Cl V → Cl X → 0

(c) Follows immediately from part (b) and II.6.2 as S(Y ) without grading is the same as
A(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 Cl X  Cl(Spec OP ) and a section map Cl(Spec OP ) ,→ Cl X. Now,
if p ⊂ A(X) of height 1 is not contained in m, then we claim that it is actually principal (see TIL
8/19/2016). I use Fulton’s intersection theory here... Is there a way to avoid this???
II.6.4. As f is square-free, (z 2 − f ) is irreducible in k(x)[z], so that K := k(x)[z]/(z 2 − f ) is
an Galois extension of k(x) of degree 2. It is also the quotient field of A = k[x, z]/(z 2 − f ). We
show that A is the integral closure of k[x] in K (and hence is integrally closed in K). Consider
α = g + hz ∈ K where g, h ∈ k(x) that is integral over k[x]. Note that α satisfies the monic
polynomial q(t) := t2 − 2gt + (g 2 − h2 f ) ∈ k(x)[t], so the minimal monic polynomial p(t) of α (over
k[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 of
degree 2, then p(t)|q(t) implies that p(t) = q(t). This implies that 2g ∈ A hence g ∈ k[x], and

25
moreover, g 2 − h2 f ∈ A so that g 2 − h2 f ∈ k[x] so that h2 f ∈ k[x], implying h ∈ k[x] also since f
is square-free.
Lemma. A short lemma if you wish to avoid the degree 1 and 2 argument: Let A be integrally
closed 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 polynomial
q(t) ∈ K[t] of α over K (after scaling it to monic). Proof: Clearly, q(t)|p(t) in K[t] by minimality
of q(t). But if p(t) factors into two monic polynomials f, g, then both f, g has coefficients that are
integral over A, and hence in A as A is integrally closed.
II.6.5. (b) Let R := Spec k[x0 , . . . , xr ]/(−x0 x1 + x22 + · · · + x2r ) with r 6= 3 and X := Spec R. Let
H := V (x1 ). Since x1 is a nonzerdivisor in R, we have dim R/x1 = dim R − 1, and the minimal
primes of R/x1 have height 1 over R, hence the irreducible components of X ∩ H are to prime
divisors of X, and for r 6= 3 we actually have X ∩ H has one irreducible component. This only
preserves 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 Cl X \ H = 0. Hence, we have a
surjective Z  Cl X 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]/(z 2 ), whose unique minimal prime (z) has z 2 valued to 2 so that
we get [X ∩ V (x1 )] = 2[V (y, z)]. However, when r ≥ 4, we get instead Spec k[x0 , x2 , . . . , xr ]/(x22 +
· · · + x2r ), and (x22 + · · · + x2r ) is prime so that we just get [X ∩ V (x1 )] = [V (x22 + · · · + x2r )]. That
is, if r = 2, we have 2[X ∩s H] = [X ∩ H] = 0. However, if r ≥ 4, then [X ∩s H] = [X ∩ H] so that
the map Z  Cl X 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 irreducible
components Z1 , . . . , Zr , then Z⊕r → Cl X → Cl X \ 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, pullback
of structure sheaf is the structure sheaf.
(b): Let f : X → Y be a finite map of nonsingular curves. Denote by fD ∗ : Cl Y → Cl X the
∗
map just to distinguish it from f : Pic Y → Pic X. We need show that
Cl Y −−−−→ CL X
 
 
y y
Pic Y −−−−→ pic X

is commutative. As both horizontal maps are group homomorphisms, it suffices to check for [Q] ∈
Cl Y where Q ∈ Y is a point. As Q is locally principal, take a affine cover of Y where [Q] is
principal in each. On affines V not containing Q, the diagram is clearly commutative. On U 3 Q,
let’s say U = Spec B and f −1 (U ) = Spec A, and h ∈ B exactly carves out Q (i.e. B/h ' κ(Q)).
We have f # : B → A, and denoting h0 := f # (h). Note that Spec A/h0 gives the divisor fD
∗ (Q) on
−1 0 −1
X. Now, we also see that the pullback is A ⊗B B · h = A · (h ) , 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, then
their alternating sum (as an element in G0 (X)) is zero.

26
 (b): If 0 → F 0 → F → F 00 → 0 is an SES, then 0 → Fξ0 → Fξ → Fξ00 → 0 is an SES
of K-modules so that rk F 0 + rk F 00 = rk F . Hence, the map G0 (X) → Z is well-defined. For
⊕n
surjectivity, note that OX = n ∀n ≥ 0.
(a): If R is a PID, and X = Spec R, then the rank map G0 (X) → Z is in fact an isomorphism.
Suppose M is torsion R-module (i.e. rk Mf = 0), then there exists a resolution 0 → Rm → Rn →
M → 0 of M (as R is a PID) and rk M = 0 forces m = n, so that as an element of G0 (X), we have
f = O n − O n = 0.
M X X
(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 finitely
generated 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. Since
∼
tensoring is right exact, we have ϕk : A ⊗ k → A ⊗ k so that ϕ(1) ∈ / m. Hence, the map ϕ is
multiplication 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 that
span(tj1 , . . . , tjn ) = V (otherwise, dim V > n + 1 and thus a contradiction). WLOG let jk = k, and
let L = V (y0 , . . . , yn ) ⊂ Pm . Since {t0 , . . . , tn } generate V ⊂ Γ(X, L ), we have that ψ −1 (Pm \ L) =
ψ π
X and we can the composition with the projection π : Pm \ L → Pn to get ψe : X → Pm \ L → Pn .
Now, each tj can be expressed as a linear combination of si ’s, and moreover, by construction
there 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 line
bundle 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 − dim V − 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 linear

systems V, W , with V ⊂ W and m ≤ n, then there exists a linear projection π : Pnk − L → Pm k
which when composed with an automorphism ι of Pm k 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 be
such that ϕ∗ (OPm (1)) ' OPn (d), and give Pm coordinates x0 , . . . , xm . Consider the linear system
V := span(ϕ∗ x0 , . . . , ϕ∗ xm ) ⊂ Γ(Pn , ϕ∗ (O(1))). If dim V ≤ n, then V cannot be base-point-free
unless d = 0, and hence ϕ(Pn ) = pt. Thus, to have ϕ(Pn ) 6= pt, we need dim V ≥ n + 1, and
so n ≤ m necessarily. FSOC, if dim ϕ(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 that
ϕ∗ 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 such X admits a ample invertible sheaf L , then there exists an immersion i : X ,→ PnA
by L n .

27
 (b): Pic X = 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 also
globally 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 OX n+1
 L and OX m+1
 M give maps ι : X → PnA
ι
and ϕ : X → PnA where ι is an immersion factoring through X → X ,→ PnA (the first map is
open embedding and the second map is closed embedding). As PnA is separated, the graph map
mn+m+n
Γϕ : X → X × Pm n m
A is a closed embedding. Hence, X → PA × PA ' PA is a locally closed
embedding, and X Noetherian means it is the same as an immersion. Now, pulling back O(1) on
Pmn+m+n exactly gives us L ⊗ M , and thus L ⊗ M is very ample.

(e): Follows from (d) as L m 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 g where S = k[x]• and
I = Γ• (IX/Pnk ). Now, we have (S/I)n ' Γn (OX ) for all n >> 0 by Ex.II.5.9.
(b): We have a map deg : Cl X → Z, and if D is torsion, then deg D = 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 , p21 , p22 , p0 p1 , p1 p2 , p2 p0 ) for a point [p0 : p1 : p2 ] ∈ P2k . Then,
the linear system V is ker p~(2), so the linear system has rank 4 (dimension 5). Now, WLOG let
P = [0 : 0 : 1]. Then, the map ϕ : P2k \ P → P4 is given by sections (x2 , y 2 , 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; e locally on Uz , we get X e ∩ 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 Proj
is A2k ×k P1k . Uy , Ux parts doesn’t change at all, so we have X e = 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 e : Xe → P4 where
[0 : 0 : 1] × [u : v] 7→ [0 : 0 : 0 : u : v] (on Uz , the map is [x : y : 1] 7→ [x2 : y 2 : xy : x : y], and which
if WLOG x 6= 0 if [x : y 2 /x : y : 1 : y/x], so that since v/u = y/x, we obtained the map as above
as (x, y) limits to (0, 0)). To check that ϕ e is in fact a closed embedding on Uz (it is already on Ux
and Uy ), we note that on Uz , we have ϕ(x, e y, z, u, v) = [ux, vy, vx = uy, u, v], and one checks easily
that 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 of
vector spaces V → W to be injective it is sufficient to show that the induced dual map W ∗ → V ∗
δ∗
is surjective. We will thus show that Homκ(B) (ΩB/k ⊗ κ(B), κ(B)) → Homκ(B) (m/m2 , κ(B)) is

28
surjective. Note that Homκ(B) (ΩB/k ⊗ κ(B), κ(B)) ' Derk (B, κ(B)). The map δ ∗ sends D ∈
0 0 0
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 e h such
∗
that δ (h) = h. Well, Leibniz rule implies that giving a derivation B → κ(B) is the same as
e
giving 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.
e
(b): Note that as B is a localization of finite type k-algebra, let’s say B = Ap , we have
dim B + tr degk κ(B) = dim Ap + tr degk κ(p) = dim Ap + dim A/p = dim A = tr degk K where K is
the 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 dim B + tr degk κ(B), then m/m2 has κ(B)-dimension dim B so that B is
regular. If B is regular, then the proof is exactly the same as Proposition II.8.8 (noting that as k
is perfect, tr degk K = 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 dim X, since rkK(X) (ΩX/k ⊗ K(X)) = dim X always, from part (a) we see that we only
need ΩX/k |x := ΩX/k ⊗ κ(x) to have κ(x)-rank = dim X. Let’s now work affine locally, and say
X = Spec A(X) where A(X) := k[x1 , . . . , xn ]/I so that we have
J
I/I 2 → 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:

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

So that we exactly need the rank of J(x) to be n − dim X. But its rank is always ≤ n − dim X,
since otherwise, we would have rk ΩX/k |x < dim X. Hence, that rank ≤ n − dim X − 1 is a closed
condition implies that rankS= n − dim X is an open condition.
Generalization: If X = i Zi is not integral, for x ∈ Zi not in any other irreducible components,
replace dim X with dim Zi and the same argument pulls through. If X = Spec k[x1 , . . . , xn ]/I, then
n − dim Zi = 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 ∈ X
closed point, consider the map ϕx : V → E |x := Ex ⊗ κ(x). ϕx is surjective, and as a k-vector space
ker ϕx thus has dimension s − r[κ(x) : k]. Thus W ⊂ X ×k |V | defined by W := {(x, s) : sx ∈ mx Ex }
is an algebraic subset whose dimension is ≤ sup{dim X + 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 exists
s∈ / mx Ex ∀x ∈ X closed. As closed points are dense in finite type k-schemes, this
/ p2 (W ), i.e. s ∈
also means that s doesn’t vanish on any x ∈ X (not necessarily closed). Defining a map OX → E
by 1 7→ s, over any x ∈ X, we have sOX,x = OX,x so that

0 → OX,x → Ex → Ex0 → 0

is a SES of free modules. Thus, we get the desired SES of locally free sheaves.
s E
Alternate Soln See TIL 10/6/2016. The main task is to interpret V → E |x as OX

29
II.8.3. (a): A bit of CA review: if A → B, A → A0 and B 0 = B ⊗A A0 , then ΩB 0 /A0 ' ΩB/A ⊗A A0 via
d(b⊗a0 ) ↔ db⊗a0 . Now, in our case, this implies that ΩX×S Y /X ' p∗2 ΩY /S and ΩX×S Y /Y ' p∗1 ΩX/S .
We thus get a commutative diagram

p∗2 ΩY /S
'

 &
p∗1 ΩX/S / ΩX× Y /S / ΩX× Y /X /0
S S

'
& 
ΩX×S Y /Y

so that the horizontal line is in fact a split short exact sequence. Thus, ΩX×S Y /S ' p∗1 ΩX/S ⊕p∗2 ΩY /S .
(b): Let X r , 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∗1 ωX ⊗ p∗2 ωY (check affine locally for pedantic details).
V
(c): Recall that PX×Y = PX PY (with X × Y embedded via Segre embedding), and hence
pa (X × Y ) = pa (X)pb (Y ) + (−1)s pa (X) + (−1)r pa (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, as
ωX ' 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 that
if A is a UFD, and I ⊂ A a (f )-primary ideal for f ∈ A irreducible, then I = (f m ) for some m;
if m is the least such that f m ∈ I, then any g ∈ I factors to f k g1 · · · g` , and as no power of gi ’s
are in I (∵ f does not divide them) f k ∈ I so that k ≥ m. Second, if X ⊂ An a closed subscheme
such that I(X)fj is principal for some open cover {D(fj )}j of An , then Ass(I(X)fj ) = ∅ or all
associated primes of I(X)fi ’s are codimension 1 by the unmixedness theorem, so that all associated
primes of I(X) is codimension 1. Thus,Q as k[x1 , . . . , xn ] is a UFD, this implies that the primary
decomposition of I(X) is i (gini ) = ( i gini ) for irreducible gi ’s so that I(X) is in fact principal.
T
Thus, if H is a hypersurface in Pnk , then H ∩Uk = V (g (k) ) so that I(H) = g for some g homogeneous
polynomial in S.
Thus, if Y = H1 ∩ · · · ∩ Hr as schemes, then Y = V (f1 ) ∩ · · · ∩ V (fr ) = V (f1 , . . . , fr ) for some
homogeneous polynomials f1 , . . . , fr , so that I(Y ) = (f1 , . . . , fr ).
(b): Let X = Spec S/I be the affine cone of Y . Its codimension in An+1 is still codimX Y . It
is indeed a (locally) complete intersection, so that A(X) is Cohen-Macaulay. To check that X is
regular in codimension 1, note that the cone point P is of codimension ≥2 since dim Y ≥ 1, and as
X \ {P } is a A1 -bundle over Y , we have that A(X)p ' OY,y [t] for some y ∈ Y for any codimension
1 points p in X.
(c): Follows immediately from II.5.14(d). The surjection implies that OY (Y ) is of k-rank at
most 1, so that Y is connected.
(d): Suppose X ⊂ Pnk is codimension r ≤ n − 2 nonsingular subvariety. Consider the d-uple
(n)−1
embedding of νd : Pnk ,→ Pkd . By Bertini’s theorem, there exists a hyperplane section H such
that H 6⊃ νd (X) and H ∩ νd (X) is nonsingular variety, so that X ∩ νd−1 H is a nonsingular variety
of codimension r − 1. Use this to induct and make the desired complete intersection.
(e): Use II.8.20 repeatedly.

30
 (f): We have ωX = OX (d − n − 1). If d < n + 1, then Γ(X, ωX ) = 0 (as a global section
would give an effective divisor of negative degree). If d ≥ n + 1, apply − ⊗ OP (d − n − 1) to
the SES 0 → Op (−d) → OP → OX → 0
(which stays SES by part (c)) to get that 0

h (ωX ) =
0 0 d−1 z+n
h (OP (d − n − 1)) − h (OP (−n − 1)) = n (plug in z = d − n − 1 into PP (z) = n ). 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−1 e−1 d+e−1


n + n − n . 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
ϕ
0 → ZU → Z → ZY → 0, we get 0 → Z → Z → Z ⊕ Z → H 1 (ZU ) → H 1 (Z) → · · · , and the image
of 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 ) is
well-defined, as if s ∈ ΓU ∩Y (U, F ), then ϕ(U )(s)P = ϕP (sP ) = ϕP (0) = 0. We have shown that
HY0 (−) : 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.
α β
To see that this is left exact, let 0 → F 0 → F → F 00 → 0 be a SES of sheaves on X. That
Fα is injective and Fβ ◦ Fα = 0 is clear. Now, if s ∈ ΓY (X, F ) maps to 0 in ΓY (X, F 00 ), then by
exactness sp ∈ Fp0 ∀p ∈ X, so that s ∈ F 0 (U ) in particular, and its support is still in Y .
(b): We need show that ΓY (X, F ) → ΓY (X, F 00 ) is surjective. Given s00 ∈ ΓY (X, F 00 ), con-
sidering as an element of F 00 (X), there exists s ∈ F (X) such that s 7→ s00 (by II.1.16). Now, as
s|U ∈ F (U ) maps to s00 |U = 0 ∈ F 00 (U ), we have s|U ∈ F 0 (U ). Extend s|U to t ∈ F 0 (X) (possible
since F 0 is flasque), then s − t has support in Y since s|U − t|U = 0, and s − t still maps to s00 since
t 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 → HY0 (F ) → F → j∗ (F |U ) is exact, and moreover,
res
if F is flasque, the last map on any open V is F (V )  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 HYi (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 actually
true?. Hence, no I i has a global section supported in Y , and thus HYi (X, FU ) = 0 for all i as
desired.

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-coherent
sheaf G on Xred , we have H i (Xred , G ) = H i (X, j∗ G ) which is zero for all i > 0 as X is affine
and Noetherian-ness implies that j∗ G is quasi-coherent. Now, suppose Xred is affine, and let
F ∈ QCohX and N the sheaf of nilpotents. Note that we have a filtration F ⊃ N ·F ⊃ · · · (which
is 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 all
vanish.
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 as
X is reduced). If one of Y, Z is contained in another, then we are done, so suppose not, so that
IY 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 SES
0 → IY F ⊗ OX /IZ → F → F ⊗ OX /IY → 0 (as F · OX /IY IZ = F OX /0 ' F ). By LES, we
have that higher cohomologies of F all vanish, and hence X is affine.
∼
III.3.3. Let X = Spec A. We have shown that ∼: A-Mod → QCohX is an equivalence√of categories.
And moreover, for Y = V (a), we have m ∈ ΓY (M
f) ⇐⇒ V (Ann m) ⊂ Y ⇐⇒ a ⊂ Ann m ⇐⇒
∞
m ∈ (0 :M a ) = Γa (M ), so that as functors we have ΓY ◦ ∼= Γa . This makes all (a,b,c) parts
clear.
III.3.4. (a): If deptha M ≥ 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 finitely
generated, then deptha M = 0 implies that a is contained in some associated prime (of which there
are finitely many as M is f.g.). This associated prime is Ann m for some 0 6= m ∈ M , so that
m ∈ 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
f
begin at n = 1). Consider the SES 0 → M → M → M/f → 0 where f ∈ a is a nonzerodivisor of M .
Then deptha (M/f ) = deptha M − 1, so that deptha M ≥ n ⇐⇒ deptha (M/f ) ≥ n − 1. Moreover,
f
we have the LES from the SES: · · · → Hai (M ) → Hai (M ) → Hai (M/f ) → Hai+1 (M ) → · · · . Hence,
we have Hai (M ) = 0 ∀i < n ⇐⇒ Hai−1 (M/f ) = 0 ∀i < n: The =⇒ direction is easy; for ⇐=
direction, note that by Exer.III.3.3.(c), for i < n − 1, Hai (M ) is zero when localized at p 6⊃ a, and
for p ⊃ a, Nakayama implies that Hai (M ) is zero, and for i = n − 1, note that ×f cannot be an
injective map as the support of every element of Hai (M ) is contained in V (a) ⊂ V (f ) (so some
power 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 have
a LES by Exer.III.2.3.(e): 0 → HP0 (U, OU ) → H 0 (U, OU ) → H 0 (U − P, OU ) → HP1 (U, OU ) → · · · .
Now, depth OP ≥ 2 ⇐⇒ HP0 = HP1 = 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 . Then
U := {Ui := f −1 (Vi )} is an affine cover X such that C • (V, f∗ F ) = C • (U, F ), and hence by III.4.5
we have H i (X, F ) ' H i (Y, f∗ F ) ∀i.
III.3.4.2. (a): Let V := Spec B ⊂ Y and U := Spec A = f −1 (Spec B) ⊂ X. Then K(A) is a
finite extension of K(B) (∵algebraic+f.g.), and thus if [K(A) : K(B)] = r, then shrinking V if
necessary (thus shrinking U ), we have that B r ' A as B-modules. Letting M := OU , we see that
M is coherent since X is separated (and U is affine). Moreover, define a map α : OYr → f∗ (OU ) by
selecting r sections in Γ(Y, f∗ (OU )) = OU (U ) = A that span A as a B-module as given by B r ' A.
(b): Just follow the hint, and for isomorphism restrict to an affine (since quasi-coherent) and
use 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, we
need only show that each irreducible components of Y is affine, so assume Y is integral. But then
f 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 subset
of Y , then f : f −1 (Z) → Z is again a finite surjective morphism where f −1 (Z) is affine since it is
closed subscheme of X. Hence, we reduce to showing that Y is affine given that every closed subset
of Y is affine.
Now, let F be any coherent sheaf on Y , and let β : f∗ G → F r be as in part (b). Via
Cech 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 → ker β → f∗ G → F r → coker β → 0. As f∗ G and F r are both
coherent (since f is finite), ker β and coker β are both coherent, and hence their supports, let’s say
Y1 , Y2 , are both closed. Thus, we have ker β = (j1 )∗ (j1 )∗ (ker β) and similarly for coker β. Hence,
as Y1 and Y2 are affine (by Noetherian hypothesis), the higher cohomologies of ker β and coker β
vanish. Now, splitting the sequence above into 0 → ker β → f∗ G → im β → 0 and 0 → im β →
F r → coker β → 0, and applying LES, we obtain H i (Y, f∗ G ) ' H i (Y, im β) ' 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. The
kernel is thus k[x, y], and the image (as k-vector space) is all monomials xα y β where at least one
of α, β is non-negative. Hence, H 0 (X, OX ) = k[x, y] and H 1 (X, OX ) = k{xi yi : i, j < 0}.
III.4.4. (a): For each p, we define λ ep : C p (A, F ) → C p (B, F ) by sending α ∈ F (Ui ,...,i ) 7→
Q 0 p

J α| U J
where the J’s are (j 0 , . . . , j p ) such that λ(jk ) = ik . As λ is an index map (preserving the
well-ordering), one checks easily that λ ’s form a map of chain complexes C (A, F ) → C (B, F ).
e p • •

In fact, in the same way, we have a map λ e• : 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 • and
•

ϕB : C • (B, F ) → I • induced from the identity map on F . Moreover, ϕB is homotopic to ϕA ◦ λ e• ,

and hence the maps Ȟ (A, F ) → Ȟ (B, F ) are compatibles with maps Ȟ (A, F ) → H (X, F ).
i i i i
∼
(c): Following the hint and using LES, we need show that limA D• (A) → limA C • (A, F ). Well,
−→ −→
the injectivity always holds, and for surjectivity, use Exer.II.1.3(a).
III.4.5. Define a map Pic X → H 1 (X, OX ∗ ) as follows: Given L , take any trivializing cover
∼
A = {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) in
C 1 (A, OX
∗ ). That the transition maps satisfy the cocycle condition implies that L (A) ∈ ker d :
1
C → C 2 . Now, if L (A) ∈ im d0 , then L ' OX (can construct an explicit isomorphism by
1

34
 ×fi
OX (Ui ) → OX (Ui ) where fi ∈ OX ∗ (U )). We thus have an element of Ȟ 1 (A, O ∗ ). Lastly, the
i X
∼
isomorphism limmf A Ȟ (A, OX ) → H 1 (X, OX
1 ∗ ∗ ) implies that we can consider L (A) as an element of
−→
H 1 (X, OX∗ ) and that the map Pic X → H 1 (X, O ∗ ) is indepndent of the choice of the cover A. To
X
see that this is a group homomorphism, given L , M , taking common refinement of each trivializing
cover.
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, but
direct limit commutes with cohomology.
(b): Let X be open subscheme of a projective scheme Y ⊂ Pnk . Suppose H i (X, E ) = 0 ∀i > n
for any locally free coherent sheaf E . Now, let F be a coherent sheaf on X; by Exer.II.5.15
there exists a coherent sheaf F 0 on Y such that F 0 |X = F . By Serre’s theorem II.5.18 we
have OYr  F 0 ⊗ OY (m) for some r, m. Tensoring by OY (−m) (which is exact as OY (−m) is
locally free), we obtain a SES 0 → K 0 → E 0 → F 0 → 0, where E 0 is locally free of rank r, and
K 0 is coherent, as E 0 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 that
H i (X, F ) ' H i+1 (X, K ) ∀i > n. Grothendieck vanishing implies that H d+1 (X, K ) = 0, and
hence H d (X, F ) = 0 where d = dim X. But this implies that any coherent sheaf of X has
vanishing dth cohomology, so applying this to K we have that H d−1 (X, F ) = 0. Continuing this
on, 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.
n
p (e): Y ⊂ Pk is a set-theoretic intersection of codim Sr r if Y = V (f1 , . . . , fr ) (that is, I(Y
+
) =
(f1 , . . . , fr ) in the coordinate ring). Thus, X \ Y = i=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 H 2 (X \ Y, OX\Y ) ' HY3 (X, OX ). Setting P = Y1 ∩ Y2 ,
and suppressing some notations, we have by Mayer-Vietoris the following sequence:

→ HP3 → HY31 ⊕ HY32 → HY3 → HP4 → HY41 ⊕ HY42 → · · ·

Now, as depth of P is 4, HP3 = 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 ) ' HYi+1
i
∀` > 0. This
` 3 4 4
implies that HYi = 0 ∀` ≥ 3. Thus, we now have that HY ' HP . But HP 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 taking
the derived LES of Γ(V, ·) from this SES for every V , and noting that I is injective and F has all
vanishing higher cohomologies, we have that H i (V, G |V ) = 0 ∀i > 0 as well. Now, the rest of the
proof 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.
S
Lemma. Let A be a Noetherian ring of finite dimension, and f ∈ A such that f ∈ / p∈Min(A) p.
Then, dim A/f = dim A − 1. Proof: Topologically, Spec A is the union of irreducible components

35
V1 ∪ · · · ∪ Vr , and Spec A/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∗ F
is 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, we
induct on the dimension of Supp F (where dim= −1 means the √ support∼ is empty). Note that
it is a closed subscheme carved out by the sheaf of ideals I := Ann M ; for each Ui , we have
q √
I (Ui ) = AnnS(xi ) M(x0 ) = Ann M (x0 ) . Now, base case is the the support being empty, in which
case we have a zero sheaf, so that χ(M f(n)) = 0 ∀n ∈ Z. Now, suppose Supp M f 6= ∅. Then as an S-
module, the irrelevant ideal (x0 , . . . , xn ) is not
Sa minimal homogeneous prime of M , so that there is
a 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
f f
x0
x
0 → K → Mi →i Mi → Mi /( xx0i )Mi → 0 (for i 6= 0; the i = 0 case is trivial). As xx0i is a not in any
minimal prime of Mi := M(xi ) , we have that Supp Mi /( xx0i )Mi ⊂ V (Ann M + ( xx0i )) has dimension 1
less than Supp Mi by the Lemma above. The same holds for Supp K. 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, and
hence χ(F (n)) is a polynomial.
(b): By III.5.2, we have χ(F (n)) = dimk H0 (X, F (n)) for n >> 0. But by definition of
M := Γ∗ (F ), we have H 0 (X, F (n)) = Mn .
III.5.3.(a): pa (X) = (−1)r (χ(OX ) − 1) = −(−1)r + rj=0 (−1)r+j dimk H j (X, OX ) which is
P
Pr r+j dim H j (X, O ) =
Pr−1 r−i
j=1 (−1) k X i=0 H (X, OX )
(b): If X = Proj S/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 a
coherent 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 Cech
complex, 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 = Proj S/(f1 , . . . , fr−q ). Then S is Cohen-Macaulay, so dim C(Y ) = q +1 = r+1−(r−q)
implies that (f1 , . . . , fr−q ) is a regular sequence. Hence, Y 0 = Proj S/(f1 , . . . , fr−q−1 ) is also a
complete intersection of dimension q + 1. Now, we have 0 → OY 0 (− deg fr−q ) → OY 0 → OY → 0.
Now, the statement follows from part (c) below.
(b): k = H 0 (X, OX )  H 0 (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 a
nonzero (nonzero follows from base-changing to k and using the lemma above) k-algebra, we have
that 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 0 of dimension q 0 > 1, and let Y be as in (a) (of dimension q = q 0 − 1). Then
again, we have the SES 0 → OY 0 (−d) → OY 0 → OY → 0 where −d = deg fr−q , tensoring by OY 0 (n)
and taking the LES, we have H i (Y, OY (n)) ' H i+1 (Y 0 , OY 0 (n − d)) = 0 ∀0 < i < q 0 − 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 (X, F ) ⊗k H j (Y, G ).
` i
L

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 have
0 → 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 U0 ) ' k[v/u][x, y]• (only x, y has weight 1) and
k
Γ∗ (OQ |P1 ×k U1 ) ' k[u/v][x, y]• . Hence, as OY (Y ) is again nonzero, we have Y is connected.
k
(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 (exists
by Bertini’s theorem). Taking i−1 , we get a curve that is connected (from (b1)) and regular at each
point (and hence irreducible as well).
(b3): If |a − b| ≤ 1 as well, then by part (a), we have that H 0 (Q, OQ (n))  H 0 (Y, OY (n)) ∀n so
that Y is projectively normal (Y is already normal as it is nonsingular). Now, if |a − b| > 1, then
by 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+r r 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
a0 b0 x = a0 b0 z = a0 b1 b0
mind the map × 7→ . Thus, the family of lines P1 × are given
a1 b1 y = a1 b0 w = a1 b1 b1
 
a
by VQ (b1 x − b0 z, b1 y − b0 w) as [b0 : b1 ] varies, and likewise we have 0 × P1 corresponding to
a1
VQ (a1 x − a0 y, a1 z − a0 w). Thus, (x, y), (z, w) are in the same family of lines, and (x, z), (y, w) are
in another family of lines.
Now, let [sn : sn−1 t : stn−1 : tn ] define the rational nth curve C on Q. We claim that its
type is (1, n − 1). Well, let Y = VP3 (wn−2 y − z n−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 − z n−1 , x − yz) = VUw (y − z n−1 , x − z n ),
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 − z n−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 a
general line in each family. The intersection of C with [z = w = 0] (i.e. P1 × [1 : 0]) has degree
n − 2; consider k[sn , sn−1 t, stn−1 , tn ]/(stn−1 , tn ) and count monomials. Likewise, intersection of C
with [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 any

A-module N . Note that M is faithfully flat iff M ⊗A N = 0 =⇒ N = 0 for any N cyclic
module iff M/mM 6= 0 for any m ⊂ A maximal. If ϕ : A → B is faithfully flat map of rings,
then Spec B → Spec A is surjective. In fact, ϕ is strongly injective in the sense that for
I ⊂ A we have I ec = 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, then
A = B; well, if b ∈ B, then b = aa21 , and so a2 b = a1 ∈ A, and by faithfully flatness we have
a2 B ∩ A = (a2 )A so that a2 b = a2 a0 for some a0 ∈ A, and thus a2 b = a2 a0 = a1 ⇒ a0 = aa21 = b.
As a result, surjective birational map of varieties is flat iff it is an isomorphism. This shows
that 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 ∈ Spec B. Proof: Let I ⊂ A be an
ideal. Consider I ⊗A M → M . Note that this map is injective as a map of A-modules iff it is
injective as a map of B-modules (set-wise, they are the same map!). Now, for any maximal
ideal m ⊂ B and p := f −1 (m) ⊂ A, note that (I ⊗A M ) ⊗B Bm ' (I ⊗A M ) ⊗B Bm ⊗A Ap '
Ip ⊗A Mm ' 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 canonically
an S −1 A-module; put in another way, if M is a S −1 A-module, then A M ⊗A S −1 A ' 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 TorR
1 (R/J, R/I) = I ∩ J/IJ = 0 for any J, but setting J = I we
have I/I 2 = 0. Hence, I ⊗ A/I = 0 =⇒ Supp(I) ∩ Supp(A/I) = ∅, but Spec A =
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-Down
property holds for the map Spec B → Spec A; given p = ϕ−1 (q) and p ⊃ p0 , there exists q0 ⊂ B
such that ϕ−1 (q0 ) = p0 . Proof: Base change to A/p0 (i.e. A/p0 → B/p0 B) so that we can assume
A domain and p0 = 0. Let q0 be a minimal prime of B contained in q. Now, ϕ−1 (q0 ) = 0 since
0 6= a ∈ A is a nonzero divisor and remains nonzerodivisor in B as A → B is flat. But a minimal
prime consists of zero divisors.
The lemma above implies that f (U ) is stable under generization. Combined with the fact the
f (U ) is also constructible, we have that f (U ) is open.

38
III.9.2. We have the twisted cubic X parametrized as [s3 , s2 t, st2 , t3 ], and P = [0 : 0 : 1 : 0]. We
do 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 hx2 − y 3 , a3 x − z 3 , a2 y − z 2 , ay 2 − xz, ax − yzi,

and setting a = 0, we get hx2 − y 3 , z 2 , xz, yzi. The underlying set is indeed the cuspidal cubic
V (x2 − y 3 ), and when x 6= 0 we have z = 0 from xz = 0 so that it is reduced everywhere except at
the cusp (0, 0, 0) on which we have z 2 = 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 by
S := 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 a
nonzero 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 2
everywhere except over the origin (fuzz-size 3). Thus, again, x⊗z−y⊗w becomes zero but is nonzero
in (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] = Y
indeed, and moreover, as (z, w) is annihilator of z or w, and (z, w) is a minimal prime of X, if
p = 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 a
unit in every affine piece, in particular f ∈ OX (Ui ) for all i for any cover {Ui } of X, we have
f ∈ OX (X) = k.

• Recall that we had L (D)|Ui = OX · fi−1 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 define
L (D) as: Γ(U, L (D)) = {f ∈ K(X)× : div f |U + D|U ≥ 0} ∪ {0}. On a cover {Ui } where
D 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 on
this however: For f ∈ Γ(X, L (D)) ⊂ K(X), the notation div f can be ambiguous—f can be
considered either as a rational function or as a section of a line bundle. To distinguish these
two situations, we write div f if considering f as a rational function, and (s)0 if considering
s 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 where
Y is a proper k-scheme. Then ϕ extends uniquely to ϕ e: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 of
degree [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 fe : X → P1k where fe(0) = div f + and fe(∞) = div f − . As a result a nonsingular
curve 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 nonconstant
global section, say f , which we can consider as an element of K(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 function
f ∈ K(X) that is regular on X \ {P1 , . . . , Pr−1 } and poles at P1 , . . . , Pr−1 . Let nr be the order of f
at 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 vanish
on any of P1 , . . . , Pr−1 . As g is nonconstant, we also have that g has a pole of order at least one
at Pr . Now, take f g nr +1 as the function that is regular on X \ {P1 , . . . , Pr } with poles exactly on
P1 , . . . , P r .
IV.1.3. By (I.6.10), X is quasiprojective. Let’s say X ,→ X. As D := X \ X is a finite collection
of 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 → A1k ,→ P1k . And this map extends
uniquely to fe : 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| ≤ deg D is equivalent to l(K − D) ≤ g, which always holds as l(K) = g
and 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, then
l(P ) = 2 for any P ∈ X in fact. Conversely, if l(P ) ≥ 2, then there is a nonconstant rational
function f ∈ Γ(X, L (P )) ⊂ K(X) such that f has a pole of order 1 at P . This induces a degree 1
map 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 that
l(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 function
in 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 deg D = 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 0 + Q0 , with {P, Q} distinct from {P 0 , Q0 } (∵ if P = P 0 , say, then
Q − Q0 ∼ 0, so that g ≥ 1 forces Q = Q0 , implying s = ct for some c ∈ k × ). Thus, we have
V (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
P pulls back to a line in Q that intersects X with degree 2. Hence, ϕ : X → P1 is a degree 2 map.
1

IV.1.8. (a): Note that for a point P on a 1-dimensional integral k-scheme X, we have that
P ∈ Xreg iff OX,P is integrally closed. Now, as Xreg an open subset of X, if f : X
e → X is the
normalization of X (which is a finite map), then OX → f∗ OXe is an isomorphism at stalks P ∈ X
for all but finitely many points. Hence, we have an SES
X
0 → OX → f∗ OXe → OeP /OP → 0
P ∈X
P
Now, since Euler characteristic is additive and f is affine, we have χ(OX ) + P ∈X δP = χ(OXe ) so
that 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 X
is 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 the
node, note that dimk k[t]/k[t2 − 1, t3 − t] = 1.
I.1.9. (a): As supp D ∈ Xreg , the divisor D is locally principal, and hence defines a Cartier divisor
D on X so that we can consider the line bundle L (D). Note that χ(L (D)) = deg D + 1 − pa
holds when D = 0. Now, we show that the equation holds for D iff it holds for D + P for any
P ∈ Xreg . Consider the SES 0 → L (−P ) → OX → κ(P ) → 0 and tensor by L (D + P ) to get
0 → 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 via
some line bundle O(1) (which corresponds to a very ample Cartier divisor). Given a line bundle
L , note that L (n) globally generated for some n by (II.5.17). Moreover, by Exer.II.7.5(d) we have
that 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 supp D ∈ Xreg .
Well, take a hyperplane L ∈ O(1) (that does not meet any of singular points of X) and pull it back
to 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, by
Hurwitz formula, we have that deg R = 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 with
ramification 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, assume
αi 6= 0. As αi ’s are distinct, h is square-free, so that K = k(x)[z]/(z 2 − h) is a degree 2 finite Galois
extension of k(x). As K is separable degree 1 over k, we can consider the projective nonsingular
curve Y := CK whose function field is K. K ⊃ k(x) then gives a map f : Y → P1 of degree 2. On
one chart of this map, we have f −1 (U0 ) → Spec k[x] via k[x] ,→ k[x, z]/(z 2 −ph), as k[x, z]/(z 2 − 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 (where
√
z 0 := x−3 z/ α1 · · · α6 )

k[x−1 , z 0 ]/(z 0 2 − i (x−1 − αi−1 )) −−−−→ k[x± , z]/(z 2 − h) ←−−−− k[x, z]/(z 2 − h)
Q
x x x
  
  
k[x−1 ] −−−−→ k[x± ] ←−−−− k[x]
The leftmost is justified as
k(x)[z] k(x−1 )[z 0 ] k(x−1 )[z 0 ]
= =
(z 2 − h) ((x6 α1 · · · α6 )(z 0 2 − e
h)) (z 0 2 − e
h)

h = i (x−1 − αi−1 ). So on the other affine patch, the map f is exactly ramified over the
Q
where e
points αi as well. Thus, f : Y → P1 is a degree 2 map with deg R = 6, so that g(Y ) = 2 by the
Hurwitz formula. Now, the map f is given by a divisor D on Y . Then deg D = 2 and l(D) ≥ 2
implies (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 Aut Pnk ' P GLn+1 (k). Proof: Let
∼
f : Pnk → Pnk be an isomorphism. Then f ∗ : Pic Pn = Z → Pic Pn = 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 aij xj , we note that
P
V (s0 , . . . , sn+1 ) = ∅ iff A = (aij ) is an invertible matrix, thus we get an element of P GLn+1 (k). It
is clear that this gives an isomorphism of groups Aut Pn ' P GLn+1 (k). Note that the map given
by A where A(e1 , . . . en+1 ) = (s0 , . . . , sn+1 ) is actually just A[~
p].
     
a c e
(c): Let P1 = , P2 = , P3 = . As P1 , P2 are distinct, by scaling a, b and c, d
b d f
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 1
cz+d then is an automorphism of Pk 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 nonsingular
plane 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 ] (fx
denotes ∂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 • (S n ) = S[t1 , . . . , tn ] the tensor algebra where ti ’s don’t commute, xi ’s commute, and xi ’s
commute with tj ’s. Consider two k-linear maps D : T → T and ~x∗ : T → T , where D is of degree 1
and ~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 0
otherwise. Then, we have (as long as ` < d)

~x∗ ◦ D = (x1 ∂x1 + · · · + xn ∂xn ) · − and hence ~x∗ D`+1 f = (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 somewhat
weirdly. 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 a
point P ∈ P2k (0 if not in the intersection). Our goal is to give a criterion for mP ≥ m in terms of
Dm f (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, − ac s − cb t) = (p2 s − p1 t)mP (as − bt)0 things. Then we see that mP ≥ m iff ∂s∂g i ∂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:

a b e f
Define  := de + ah − cf − bg. Let I, J ⊂ {1, 2, 3} of size 2, and given a 3 × 3
c d g h
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 [D2 f ] be the Hessian matrix of f . Then mP ≥ 3 if m ≥ 2 already and
([~n2 ]I,J  [D2 f ]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 is
given p 7→ [Df (p)] × [0 : 0 : 1] = [fy (p) : −fx (p)] (the cross product, which is always nonzero for
p ∈ X since L is not tangent to X). In other words, the map is given by ϕ# : k[x, y] → k[x, y, z]/f
via x 7→ fy , y 7→ −fx (as ϕ# (x, y) carves out V (fx , fy , f ), and L not being tangent to X implies
that 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 is
ramified at p iff V (h) meets V (f ) at p with multiplicity > 1 (note that p ∈ V (h) indeed). This is
equivalent to stating that [Dh(p)] = λ[Df (p)] ∃λ ∈ k. Well, we have
  
fy (p)
[Dh(p)] =  Hf (p)  −fx (p)
0

So, if p ∈ L to begin with, [fy (p) : −fx (p) : 0] = p so that the generalized Euler’s formula (see
Lemma 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

fy2 fxx −2fx fy fxy +fx2 fyy , −fx fy fxz +fx2 fyz −fy fz fxx −fx fz fxy , fz fy fxy −fx fz fyy −fy2 fxz +fx fy fyz

(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 arranged
so 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 the
equation 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- can’t; is ⇐= ⇐= ⇐=
tive ⇐=
0 e.g.P − Q D∼0 ⇐= ⇐= ⇐=
on g > 0
1 ... D ∼ Is very ample, so ι : ⇐= ⇐=
∼
P ∃P X → P1 via L (D) '
ι∗ O(1)
2 ... ... Either: (1) b-p-f L (D) makes X the 2- ⇐=
hence hyperelliptic uple embedding of P1
or (2) rational
3 ... ... Either: (1) b-p-f Either: (1) very ample L (D) makes X
hence trigonal, or (2) hence hyperelliptic (in the 3-uple em-
hyperelliptic or ra- fact elliptic) or (2) only bedding of P1
tional b-p-f so rational (maps
to a singular plane cu-
bic) or (3) not b-p-f
hence rational

IV.3.1. If deg D ≥ 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 deg D ≥ 3; if
deg D = 2, note either one of the following: (1) if E ∼ D effective, then dim |D − E| = 0 implying
that dim |D| ≤ 2, which is a contradiction, (2) deg D = 2 implies by Bezout’s Theorem that X ⊂ H
for some linear space H of dimension 2 (just take any three points on X) and hence X is a conic
in P2 and hence rational. Now using R-R, we have that deg D = 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, deg D ≥ 5, as
desired.
IV.3.2. (a): Adjunction formula KX = (X + KP2 )|X = (4H − 3H)|X = H|X = H · X where H is
a 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 very
ample 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 every
degree 2 effective divisor has rank 0.
= i Hi where deg Hi = di . Since ωX = OX ( i di − n − 1) and 2g − 2 > 0 if
T P
IV.3.3. Suppose X P
g > 2, we have that i di − n − 1 := m for some m > 0. If Pn ,→ PN is m-tuple embedding, then
ωX is OPN (1) pulled back to X, so that it is very ample. Now, if g = 2, we know that K is not
very 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 that
dim |D| = deg D = n, then by (Ex.IV.1.5), we conclude that g(X) = 0, and OX (1) ' OP1 (d) under
the 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 0 := ϕ(X) is regular. Then the birational map X 0 99K X extends to a
full morphism X 0 → X, and thus the projection is an isomorphism X ' X 0 . Thus, we have the
following:

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 0 (1). But h0 (OX (1)) ≥ 4, while
h0 (OX 0 (1)) ≤ h0 (OP2 (1)) = 3 since X 0 ⊂ P2 is a complete intersection (hence by (Ex.III.5.5.)
we have H 0 (P2 , OP2 (1))  H 0 (X 0 , OX 0 (1))). This is a contradiction. Hence, we conclude that a
non-planar curve in P3 cannot be embedded in P2 .
(b): As X is the normalization of X 0 , we have that pa (X 0 ) = pa (X) + P δP . Note that
P

deg X = deg X 0 = d as projection is a degree 1 map, and hence g = d−1 − P δP < d−1 as X 0

P

2 2
is singular.
(c): Since X0red ' ϕ(X), from part (a) we have pa (X0red ) = pa (ϕ(X)) > pa (X). However, by
(III.9.10) we have pa (X0 ) = pa (X1 ) = pa (X) so that X0red 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 that
X ⊂ 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: Let
D be the effective divisor giving the embedding X ⊂ Pd , so that dim |D| ≥ d and deg D = d + 1.
Then R-R implies that g ≤ h0 (K − D) + 1. But if D is a special divisor, then dim |D| ≤ 21 deg D
by 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 two
quadrics 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)) =
5 1 0
2 = 10 and h (OP3 (2)) = 0, and moreover h (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 exist
f1 , 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, by
Bezout’s theorem, C := V (f1 , f2 ) is a degree 4 curve containing X, which is also of degree 4, and
hence 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 X e
3
in P . This is because 3 = pa (X) = pa (X) + 1 so that pa (X) = 2, but curves of degree 4 in P are
e e 3

either g = 0, 1, or 3. Now, V (xyz 2 + x4 + y 4 ) ⊂ 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 family
of tangent lines to X in G(1, 3) is of dimension 1, so that planes containing tangent lines is at most
of dimension 2). Now, we show that the family of multisecant lines of X in G(1, 3) is of dimension
1 as well. stuck here...

46
IV.3.10. We say that P1 , . . . , Pm ⊂ Pn are in general position Q
if there is no linear space Lm−2 of
dimension m − 2 such that P1 , . . . , Pm ∈ Lm−2 . Now, let Vm ⊂ m n
i=1 P 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 × m
n
matrix.
Qn In particular, when m = n, we have Qn that Vn is irreducible closed subset. If X ⊂ P , then
( i=1 X) ∩ Vn is proper closed subset of i=1 X, 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 deg D = d ≥ 3 and fix an embedding i : X ,→ Pd−1 given by |D|. Since X
is nonsingular, it is normal, and hence to show projective normality is equivalent to showing
H 0 (OPd−1 (n))  H 0 (OX (n)) ∀n ≥ 0. Now, from the SES 0 → IX → OPd−1 → OX → 0, we
need show that H 1 (IX (n)) = 0 ∀n ≥ 0. This holds for n = 0, 1 since I0 , I1 = 0 since i(X) is
nonempty 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 complete
intersections, the canonical divisor is very ample (Ex.IV.3.3), and hence by (IV.5.2) it cannot be
hyperelliptic.
IV.5.3. The same reasoning as (Ex.IV.2.2) shows that the hyperelliptic genus g curves form an
irreducible 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 complete
intersection of

an (irreducible) quadric and a (irreducible) cubic by (IV.5.2.2). For the quadric,
we have 3+22 − 1 = 9 dimensional space, and the subsequence cubic that does that vanish on the
quadric chosen is of dimension 3+3


3 − 4 − 1 = 15. Lastly, P GL(3) has dimension 15, so we have
9 + 15 − 15 = 9.
Now, for the nonhyperelliptic genus 4 to have a unique g31 , the quadric is a rank 2 quadric,
which forms a family of dimension 8—it’s really a (irreducible) hypersurface in (P9 )∗ given by the
determinant of the symmetric matrix.
IV.5.4. We know that genus 4 curve X canonically embedded is a complete intersection of a
quadric 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 g31 ’s coming from the two family of lines (note
that g31 is equivalent to three collinear points with the canonical embedding of X by R-R); these
g31 ’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 g31 ’s is the canonical divisor. In the singular conic case, there is a unique g31
since the lines in the family are all linear equivalent, and by the following lemma, and two times
this g31 also gives the canonical divisor in this case (as the pullback of hyperplane is the double line
on the singular cone).
Lemma. Let X ⊂ Y closed embedding of schemes satisfying (∗)+locally-factorial of (II.6). Suppose
D, D0 ∈ Div X such that D = X ∩ E, D0 = X ∩ E 0 for some codimension-1 integral subschemes

47
E, E 0 of Y . Then E ∼ E 0 on Y implies that D ∼ D0 on X. Proof: By assumption E, E 0 does not
contain X. Let f ∈ K(Y ) be the rational function such that div f = E − E 0 . Then affine locally
f = a/b where b does not vanish on X. Hence, it makes sense to map f to a rational function
f ∈ K(X) by a/b (if A → A/p induces X ,→ Y , then we have Ap → κ(p)). Then f now induces
D ∼ D0 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 g31 ’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 injective
except at over two points in the image where P is a part of three collinear points. Thus, the
closure of the image X 0 is birational to X, and has two nodes, and thus by (Ex.IV.1.8) we have
pa (X 0 ) = p(X) + 2 = 6 and hence deg X 0 = 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 with
multiplicity 2 as well) where P, Q, R are collinear. Hence the projection has a tacnode and is of
degree 5 (note that δP = 2 for tacnode as twice blow-up resolves the singularity). To only have
nodes, we need project from a point not on the singular cone. But then the number of secants
through 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 0 a
birational map. Then the degree of X 0 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 a
closed subscheme of Y . Suppose dim X = dim Y = 1 and deg X = deg Y . Then X = Y . Proof:
Y is ACM, and hence equidimensional, and moreover C(Y ) the affine cone is reduced since it is
2-dimensional CM ring (hence normal). Now, dim X = dim Y implies that X is an irreducible
component of Yred = Y , and thus deg Y = deg X implies that Y has no other components, thus
Y = 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, so
that X lies on two different irreducible quadrics F1 , F2 (irreducible since X is not in a plane), then
Y := V (F1 , F2 ) is a 1-dimensional k-scheme of degree 4 containing X and hence X = Y . But this
would mean that X is a genus 1 curve. Hence, X lies on a unique quadric. If it lies on the singular
quadric, then a projection from a point on the quadric but not on the curve gives a birational map
to X 0 with a tacnode. But then, pa (X e 0 ) + 2 = 0 + 2 = pa (X 0 ), 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 cubic
surface. [s5 , s4 t, st4 , t5 ] lies on a quadric surface, but [s5 , s4 t + s3 t2 , s2 t3 + 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 unique
one since if X is in two quadric surfaces then its degree is at most 4). Now, given an abstract genus
2 curve X, note that any divisor D of degree 5 is very ample since 2g + 1 = 5 here, and has rank
5 − 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 is
the double line. Thus, deg X ∩ 2L must be even but deg X ∩ H = 5 by Bezout. (This faulty proof
shows that doing intersection theory on singular surfaces gets really tricky!)

48
 However, here is an example of genus 2 curve of degree 5 in P3 lying on a singular quadric. Let
S(X) := S/I have a free resolution:
 
x y
 
y z

 
w 2 z2
I2 (M )
0 −−−−→ S(−4)2 −−−−−−−→ S(−3)2 ⊕ S(−2) −−−−→ 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 none
of 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, X
lies in two independent cubics that are necessarily irreducible (since X does not lie in a plane or
a quadric). But then X is the intersection of these cubic surfaces, in which case the genus will be
10. Thus, X of degree 9 and genus 11 does not exist in P3 .

49
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 = 21 H 2 t2 − (H.K)t + pa + 1, and
that 2π − 2 = H.(H + K) = H 2 + H.K.
V.1.3. (a): (Note that we can’t just use adjunction formula since D may be singular). Take χ of the
SES 0 → OX (−D) → OX → OD → 0 to get 1 − χ(OD ) = 1 + χ(OX (−D)) − χ(OX ) = 12 D.(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 the
hyperplane class (restricted to X). By adjunction formula, −2 = C.(C + K) = C.(C + K) =
C.(C + (d − 4)H) = C 2 + d − 4 (as C.H = 1).
(b): Set X = V (f = xd + xz d−1 + y d + ywd−1 ). Note that Sing(f ) = V (dxd−1 + z d−1 , dy d−1 +
wd−1 , xz d−2 , ywd−2 )
= ∅.
V.1.5. (a): (d − 4)2 H 2 = d(d − 4)2 .
(b):

50