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1
Divisors
Let X be an irred variety
Def. A prime divisor is an irred codim 1 subvar of X
Def. The group Div(X) of divisors on X is the free
abelian group on the set of prime divisors.
Less formally, a divisor is a finite formal sum
D = n1D1 + . . . nk Dk , ni ∈ Z, Di prime divisors
with the obvious addition
The identity element is the divisor 0
After removing any terms with ni = 0, the support
Supp(D) of D is the codimension 1 subvariety D1 ∪ . . . ∪
Dk ⊂ X
A prime divisor can be identified with a divisor as above
with k = 1 and n1 = 1
Def. A divisor D = niDi is effective if all ni ≥ 0
Sometimes write D ≥ 0 to indicate that D is effective
P
Clearly any divisor can be written as a difference D =
D0 − D00 of two effective divisors
P
If D = niDi put
D0 =
X
niDi, D00 =
i|ni >0
X
i|ni <0
2
(−ni) Di
If 0 6= f ∈ k[X], we have seen that Z(f ) ⊂ X defines a
codimension 1 subvariety
Let Z(f ) = Z1 ∪ . . . ∪ Zk be the decomp into irred
components, all codim 1 (i.e. prime divisors)
f can vanish along Zi to higher order
Will define order νZi (f ) ∈ Z
This leads to the definition of the divisor of zeros of f
(f )0 =
X
i
νZi (f ) · Zi
Definition extends to f ∈ k(X)
Also have divisor of poles
1
= 
f 0

(f )∞

and the divisor of f
(f ) = (f )0 − (f )∞
Such a divisor is called a principal divisor
The set of all principal divisors forms a subgroup P (X) ⊂
Div(X)
The divisor class group is the quotient Cl(X) = Div(X)/P (X)
3
Examples. 1. X = An
The prime divisors are in 1-1 corr with irred polys F (x1, . . . , xn)
up to scalar multiple
P
Let D = niZi be an arbitrary effective divisor, ni > 0
Write Zi = Z(Fi), Fi = Fi(x1, . . . , xn) irred, and put
FD (x1, . . . , xn) =
Y
i
Fini
Then νZi (FD ) = ni
(FD )0 = D, (FD )∞ = 0, (FD ) = D
Thus every effective divisor on An is principal
Let D ∈ Div(An) and write as D = D0 − D00, Di ≥ 0
f = FD0 /FD00 ∈ k(An), (f ) = D
This shows that all divisors are principal
Div(An) = P (An)
Cl(An) = 0
4
2. X = Pn
Any prime divisor Z ⊂ Pn is associated to an irred homog poly FZ (x0, . . . , xn), unique up to scalar mult
Let D = niZi be a divisor, with Zi associated to an
irred homog poly Fi of degree di = deg(Zi)
We would like to conclude as in the case of An that
P
f=
Y
i
Fini
satisfies (f ) = D
Big problem: f need not be a well defined rational function on Pn
P
f is a ratl fcn iff i nidi = 0, so that f is homog of deg 0
Claim: Cl(Pn) ' Z, the isomorphism corresponding to
the degrees of prime divisors.
To see this, define
φ : Div(Pn) → Z,
X
niZi 7→
X
nideg(Zi)
φ is surjective
The above calculation shows that P (Pn) ' ker(φ)
Thus Cl(Pn) = Div(Pn)/ ker φ ' Z
5
3. Cl(Pn1 × · · · × Pnk ) ' Zk
The isomorphism is induced by
Z(F ) 7→ (d1, . . . , dk ) ∈ Zk
where Z(F ) is a prime div corr to an irred poly which is
homogeneous separately, in each of the k sets of homog
coords, of respective degrees d1, . . . , dk
Let X be an irred alg variety, nonsing in codim 1
Let D ⊂ X be a prime divisor
Let k(X)∗ = k(X) − 0, a multiplicative group
Lemma. There exists a group homomorphism
νD : k(X)∗ → Z
satisfying
• νD (f1f2) = νD (f1) + νD (f2) (rephrasing νD homom)
• νD (f1 + f2) ≥ min(νD (f1), νD (f2)) if f1 + f2 6= 0
Pf (sketch). Find open U ⊂ X s.t.
• U ∩ D is nonempty
• U is affine and nonsing
• ID ⊂ k[U ] is principal, ID = (g) for some 0 6= g ∈
k[U ]
6
At this point proof is entirely similar to proof that smooth
curves have local param
k
Claim: ∩∞
k=1 ID = 0
This can be proven by algebra in the local ring OD of D,
the ring of germs of expressions
g
,
g ∈ k[U ], h ∈ k[U ], h|U ∩D 6≡ 0
h
where U is a nonempty open in X intersecting D
See text for details
Each IDk+1 is a proper subset of IDk
Given f ∈ k[U ] have unique k s.t. f ∈ IDk − IDk+1
Put νD (f ) = k
If f ∈ k(X), write f = g/h, g, h ∈ k[U ]
Put νD (f ) = νD (g) − νD (h)
Can show that the definition is well-defined:
Independent of choice of U and representation of f as
g/h
7
Lemma. Given 0 6= f ∈ k(X), there are only finitely
many irred divisors D s.t. νD (f ) 6= 0
Assuming the lemma, we define the divisor of f
div(f ) = (f ) =
X
νD (f ) · D
D
(f ) is called a principal divisor
Let P (X) ⊂ Div(X) denote the set of principal divisors
It follows immediately from the corresponding properties
of νD that
• (f1f2) = (f1) + (f2)
•
f1
f2
= (f1) − (f2)
• (f ) = 0 if f ∈ k ∗
• (f ) ≥ 0 if f ∈ k[X]
It follows from the second bullet point above that P (X)
is a subgroup of Div(X)
8
If νD (f ) = k > 0, say that f has a zero of order k along
D
If νD (f ) = −k < 0, say that f has a pole of order k
along D
Can define the divisor of zeros
div(f )0 = (f )0 =
X
νD (f ) · D
D|νD (f )>0
and the divisor of poles
div(f )∞ = (f )∞ = −
X
D|νD (f )<0
Clearly (f )0 ≥ 0 and (f )∞ ≥ 0
(f ) = (f )0 − (f )∞
9
νD (f ) · D
Let X be smooth and irreducible
Recall that if f ∈ k[X], then (f ) ≥ 0
Lemma. If f ∈ k(X)∗ and (f ) ≥ 0, then f ∈ k[X]. If
in addition X is projective, then f ∈ k
Pf. Suppose f is not reg at p ∈ X
Write f = g/h with g, h ∈ Op but f 6∈ Op
Since Op is a UFD can assume g, h have no common
factor
Let r ∈ Op be an irred elt which is a factor of h but not
of g
Have nbhd U of p s.t. r is reg on U and Z(r) is irred
Put D = Z(r) ⊂ X
Then D is an irred divisor and νD (f ) < 0, contradiction
If in addition X is proj, then k[X] = k
Cor. If X is smooth, irred, and proj, f, g ∈ k(X)∗, then
(f ) = (g) iff f and g are scalar mults of each other
Pf. Apply lemma to f /g and note (f /g) = (f ) − (g) = 0
10
Linear equivalence of divisors is a simple but fundamental notion
Def. Divisors D1 and D2 are linearly equivalent, denoted D1 ∼ D2, if D1 = D2 + (f ) for some principal
divisor (f ).
It is easy to see that the relation of linear equivalence is
an equivalence relation, as the name suggests:
Reflexivity: D ∼ D for any D, as D = D + (1)
Symmetry: if D1 = D2 + (f ), then D2 = D1 + (1/f )
Transitivity: if D1 = D2 + (f ) and D2 = D3 + (g)
Then D3 + (gf ) = D3 + (g) + (f ) = D2 + (f ) = D1
11
The notion of a Cartier divisor provides a general class
of divisors which behave much like divisors on a smooth
variety
First, describe Cartier divisors for smooth varieties X
If D ⊂ X is a prime div, can find an open cover X = ∪Ui
s.t. D has a local eqn fi on Ui, fi ∈ k[Ui]∗
In Ui ∩ Uj , we have (fi) = (fj ) = D ∩ Ui ∩ Uj
Therefore fi/fj is regular and nowhere zero, by earlier
lemma
Of course we could have used a different cover Vj and
local eqns gj for D
However, {(Ui, fi)} is related to {Vj , gj )}:
fi/gj is reg and nowhere vanishing on Ui ∩ Vj
12
We can now give a similar description for arbitrary diviP
sors D = nα Dα
Let fiα by local equations for Dα in Ui
Q
Put fi = α (fiα )nα ∈ k(X)∗
Then fi/fj is regular and nowhere vanishing in Ui ∩ Uj
Def. A Cartier (or locally principal) divisor on an irred
X (not nec smooth) is defined by the following data with
an equivalence reln
• An open cover X = ∪Ui
• 0 6= fi ∈ k(X)∗ for each i
• In Ui ∩ Uj , fi/fj is regular and nowhere vanishing
The equiv reln is defined by saying that
{(Ui, fi)} and {(Vj , gj )} define the same Cartier div if
fi/gj is reg and nowhere vanishing in each Ui ∩ Vj
13
Given a Cartier divisor rep by {(Ui, fi)}, can define an
ordinary divisor as follows
Given a prime div D ⊂ X, find a Ui intersecting D
Put nD = νD (fi)
This is independent of i: if Ui and Uj both intersect D,
then have
νD (fi) = νD (fi/fj ) + νD (fj )
and νD (fi/fj ) = 0 since fi/fj is reg and nowhere vanishing
Then to {(Ui, fi)} can associate the divisor
X
nD · D (a finite sum)
Easy to check that an equivalent {(Vj , gj )} gives the same
divisor
Thus, we can think of (equivalence classes of) Cartier
divisors as being ordinary divisors, with the special property of being locally principal
14
Examples. 1. Consider the hyperplane D = Z(x0) ⊂ Pn
Take the standard open cover Ui = Ani
In each Ui, D is defined by the rational fcn fi = x0/xi
fi ∈ k[Ui] and (fi) = D ∩ Ui
Note that fi/fj = xj /xi, which is reg and nowhere van
in Ui ∩ Uj
2. D ⊂ Pn an hypersurface defined by F (x0, . . . , xn) =
0, F homog of deg d
In this case can take fi = F/xdi for each Ui
3. Let X be the quadric cone xy = z 2, which is irred and
reg in codim 1 (the origin is the only singular pt)
Let L ⊂ X be the line x = z = 0, so that L is an irred
divisor
It turns out that L is not Cartier
Discuss instead of proving
If we try to define L by z = 0, get the divisor L + L0,
where L0 is the line y = z = 0
If we try to define L by x = 0, get z 2 = 0
2L is Cartier, but not L
15
One of the useful things about Cartier divisors is that
they can be pulled back by certain reg maps φ : X → Y
Suppose D is a Cartier divisor in Y and φ(X) is not
contained in the support of D
Will show that D can be pulled back to a divisor φ∗(D)
on X, much like the pullback of a reg fcn
Let {(Vi, fi)} be a representative of D
Then the sets Ui = φ−1(Vi) are open subsets of X
{Ui} forms an open cover of X
The assumption on φ and D implies that if Ui is nonempty,
then
φ∗(fi) ∈ k(Ui)∗ = k(X)∗
Thus {Ui, φ∗(fi)} is a Cartier divisor, denoted φ∗(D)
Note that φ∗(fi)/φ∗(fj ) = φ∗(fi/fj ) is reg and nowhere
vanishing in Ui ∩ Uj , since fi/fj has the corresponding
property in Vi ∩ Vj
Straightforward to check that equivalent Cartier divisors
define equiv pullbacks
16
Examples. 1. Let f ∈ k(X) − k be a nonconstant ratl
Can identify f with a ratl map F = (1, f ) : X −− → P1
F is nonconstant since f 6∈ k
Therefore F has dense image and pullbacks of Cartier
divisors are defined
Let U ⊂ X be the largest open set on which F is regular
F : U → P1, X − U ⊂ X has codim ≥ 2
So Div(U ) ' Div(X)
Induced by map of prime divs taking D ⊂ U to D ⊂ X
0 = (1, 0) ∈ P1 is a Cartier divisor
So F ∗(0) is a Cartier divisor in U0
In fact, F ∗(0) = (f )0 ∩ U as a divisor
To see this, write 0 as the Cartier divisor
x1
{(A10, x = ), (A11, 1)}
x0
Then F ∗(0) is the Cartier divisor
{(F −1(A10), F ∗(x)), (F −1(A11), F ∗(1)))}
= {(U0, f ), (U1, 1)},
where U0 is the set on which f is reg
U1 is the set where 1/f is reg
Clearly the divisor associated to this Cartier divisor is
(f )0 ∩ U
The corresponding divisor in X is (f )0
17
2. Similarly, letting ∞ = (0, 1) ∈ P1, have ∞ ∈ P1 is a
Cartier divisor, and F ∗(∞) = (f )∞
Note that (f )0 and (f )∞ are fibers of f
(f )0 ∼ (f )∞, linearly equiv
Can think of linear equivalence as the relation obtained
by identifying fibers of f : X → P1 as a point moves
linearly along the projective line P1
3. Consider φ : P1 → P2, φ(x0, x1) = (x0, x1, 0)
Identifies P1 with the line L ⊂ P2 given by x2 = 0
φ−1(L) = P1, not a divisor
So divisors can’t always be pulled back by maps that
aren’t dense
But any other line can be pulled back, and φ∗(L0) is the
divisor of the point L ∩ L0
Cartier divisors form a multiplicative group:
{(Ui, fi)} · {(Vj , gj )} := {(Ui ∩ Vj , figj )
This is just a refinement of the addition of ordinary divisors
If D is the divisor corresponding to the Cartier divisor
{(Ui, fi)}
D0 is the divisor corresponding to the Cartier divisor
{(Vj , gj )}
Then the divisor corr to the product is just D + D0
18
Lemma. If X and Y are smooth, and φ : X → Y is
regular with dense image, then there is a group homomorphism φ∗ : Div(Y ) → Div(X)
Pf. Since Y is smooth, all divisors are Cartier
Since φ(X) has dense image, it can’t be contained in the
support of any divisor
Thus the pullback map is defined
Then compute, with notation as above
φ∗(D+D0) = φ∗({(Ui∩Vj , figj )}) = {(φ−1(Ui∩Vj ), φ∗(figj ))}
= {(φ∗(Ui ∩ Vj ), φ∗(fi)φ∗(gj ))) = φ∗(D) + φ∗(D0)
Lemma. φ∗(P (Y )) ⊂ P (X)
Pf. Compute
φ∗((f )) = (φ∗(f ))
Details omitted
By passing to the quotients we have proved
Prop. If X and Y are smooth, and φ : X → Y is regular
with dense image, then there is a group homomorphism
φ∗ : Cl(Y ) → Cl(X)
It turns out that this prop is true even without the assumption that φ is dense!
19
In other words, while we can’t necessarily pull back divisors, we can pull back divisor classes
This is true for arbitrary irred varieties which are reg in
codim 1, if we restrict to Cartier divisors
Lemma. Let X be nonsing and D a div on X. Let
x1, . . . , xn be any finite set of pts on X. Then there
exists a div D0 on X with D0 ∼ D and xi 6∈ Supp(D0)
for each i
Pf. See text
The idea is that there are sufficiently many rational functions whose zeros or poles can be used to cancel prime
divisors contained in Supp(D) which contain any xi
Assuming the lemma, we prove
Thm. Let X, Y be nonsing and φ : X → Y reg. Then
have a group homom
φ∗ : Cl(Y ) → Cl(X)
Remark. Note that it has not been assumed that φ(X)
is dense in Y
20
Pf. Let y ∈ φ(X) ⊂ Y
Represent a divisor class on Y by D ∈ Div(Y )
By lemma, find D0 s.t. D0 ∼ D and y 6∈ supp(D0)
φ(X) is not contained in Supp(D0)
So D0 can be pulled back
φ∗[D] := [φ∗(D0)],
where [D] denotes the divisor class of a div D
If D00 were chosen instead, D00 = D0 + (f ), then
φ∗(D00) = φ∗(D0+(f )) = φ∗(D0)+φ∗((f )) = φ∗(D0)+(φ∗(f )) ∼ φ∗(D0)
φ∗ : Cl(Y ) → Cl(X) is well-defined
Straightforward to check it’s a group homomorphism
21
Example. Return to φ : P1 → P2, φ(x0, x1) = (x0, x1, 0)
Want to understand φ∗ : Cl(P2) → Cl(P1)
Pick a point in the image of φ, e.g. p = (0, 1, 0)
Let D ∈ Div(P2)
Let L0 be the line x1 = 0
If D has degree d, then D ∼ dL0
The support L0 of dL0 does not contain p
So the image of φ is not contained in supp(dL0)
It follows that φ∗(dL0) is defined
Using [ ] to denote linear equivalence classes of divisors
i.e. images of divisors in the corresponding divisor class
groups
φ ([D]) = [φ∗(dL0)]
But φ∗(L0) = (1, 0) =: q ∈ Div(P1)
φ∗(dL0) = d · q
Denote by ψr the isom Cl(Pr ) ' Z
Then ψ2(dL0) = d and ψ1(d · q) = d
Thus φ∗ : Cl(P2) → Cl(P1) is identified with the identity
map Z → Z
22
Returning to L given by x2 = 0, φ−1(L) not a divisor
It is lin equiv to L0 given by x1 = 0
L − L0 = (x2/x1), L ∼ L0
So can use L0 if we just want the divisor class of the
pullback
Now φ−1(L0) = {(1, 0)} = q
Easy to compute that (1, 0) occurs in φ∗(L0) with mult 1
Another important application of these ideas is a hyperplane section
Let X ⊂ Pn be a smooth irred proj var
Let L be any linear form on Pn which does not vanish
identically on X
The hyperplane H = Z(L) is a (prime) divisor
The image of the inclusion ι : X ,→ Pn is not contained
in H
So ι∗(H) ∈ Div(X) is defined
Any such ι∗(H) defines a hyperplane section divisor
Given two such hyperplanes H, H 0, have H ∼ H 0 and
ι∗(H) ∼ ι∗(H 0)
This leads to a well defined hyperplane section class
[ι∗(H)] ∈ Cl(X)
23
More generally, consider a homogeneous form F on Pn
of degree d which does not vanish identically on X
Let Y ∈ Div(Pn) be the divisor associated to the vanishing of F with multiplicities, a hypersurface divisor
Since the image X of ι is not contained in the supp of Y ,
ι∗(Y ) ∈ Div(X) is defined
In terms of divisor classes, this is nothing dramatically
new
Y ∼ dL since Y = dL + (F/Ld)
[ι∗(Y )] = [ι∗(dL)] = d[ι∗(L)]
Thus the degree d hypersurface section class is simply d
times the hypersurface section class
Note that the hyperplane section class is not intrinsic to
X, but depends on the projective embedding X ,→ Pn
An important special case is X = Pn itself
The hyperplane class H ∈ Cl(Pn) ' Z is a generator of
this group
We will see later how to start with a divisor class satisfying certain properties and produce a projective embedding!
24
Remark. For varieties X which are regular in codim 1
but not necessarily smooth, divisors need not be Cartier
divisors
Let’s see what this says about divisor classes
Denote by CaDiv(X) ⊂ Div(X) the subgroup of Cartier
divisor
Principal divisors are always Cartier:
(f ) is obviously represented by the Cartier divisor (X, f )
P (X) ⊂ CaDiv(X) ⊂ Div(X)
This induces an inclusion of abelian groups
Div(X)
CaDiv(X)
,→
= Cl(X)
P (X)
P (X)
Def. The Picard group of X is the group
Pic(X) :=
CaDiv(X)
P (X)
We have just observed that Pic(X) is a subgroup of
Cl(X)
25
Examples. 1. If X is smooth, then Pic(X) = Cl(X)
2. If X ⊂ P3 is the projective quadric cone x1x2 = x23
then Pic(X) is a proper subgroup of Cl(X)
Cl(X)/Pic(X) ' Z2
For ι : X → P3 the inclusion have
ι∗ : Z ' Cl(P3) → Cl(X)
taking the hyperplane class of P3 to the hyperplane class
H ∈ Cl(X)
ι∗ injective
Choosing the hyperplane to pass through (1, 0, 0, 0), the
hyp section is either two lines or one line L with mult 2
L is not in the image of ι∗, and generates the Z2 quotient
26
Example. The Veronese embedding
ν3 : P1 → P3, ν3(s, t) = (s3, s2t, st2, t3)
whose image is the twisted cubic curve
Describe all hyperplane section divisors
General hyperplane defined by linear form L = ax0 +
bx1 + cx2 + dx3
φ∗(L) is the divisor on P1 consisting of the 3 roots (including mult) of as3 + bs2t + cst2 + dt3
Rewrite ν3 in terms of rational fcns rather than cubic
forms


2
3
t
t
t


1, ,
,
ν3(s, t) = 
s s2 s3
In this form, L pulls back to the ratl fcn
s3 + bs2t + cst2 + dt3
f=
s3
How are the divisors (f ) and φ∗(L) related?
φ∗(L) = 3 · (0, 1) + (f )
φ∗(L) ≥ 0 since it is a hyp section div
27
Now let’s generalize this setup
Let X be a smooth projective variety, D ∈ Div(X) (not
necessarily effective)
Def.
L(D) = {0} ∪ {f ∈ k(X)∗ | D + (f ) ≥ 0}
Example. On P1



L(3 · (0, 1)) = 

3
2
2
3

as + bs t + cst + dt
s3



Note how this set of rational functions arose from divisor
theory, without any mention of a projective embedding!
Remark. If D =
P
niDi and f ∈ k(X)∗, then
f ∈ L(D) iff νDi (f ) ≥ −ni for all i
Returning to the example, let 0 6= f ∈ L(3 · (0, 1))
Then ν(0,1)(f ) ≥ −3 and νp(f ) ≥ 0 for all p 6= (0, 1)
So s3f is a rational homog form of deg 3 with no poles
s3f is therefore a homog poly of degree 3
Therefore f is of the form (as3 + bs2t + cst2 + dt3)/s3 as
claimed
28
We next establish the basic properties of L(D), then show
how L(D) leads to rational maps from X to projective
space
Lemma. L(D) is a k-vector space
Pf. If f ∈ L(D) and c ∈ k, have to show cf ∈ L(D)
If f = 0 or c = 0 this is clear
Otherwise (f ) is defined and (cf ) = (f )
Thus D + (cf ) = D + (f ) ≥ 0 and cf ∈ L(D)
If f1, f2 ∈ L(D) must show f1 + f2 ∈ L(D)
Clear if f1 = 0, f2 = 0, or f1 + f2 = 0
Otherwise (f1), (f2), (f1 + f2) are all defined
P
Writing D = niDi our hypothesis is
νDi (f1) ≥ −ni, νDi (f2) ≥ −ni
νDi (f1 + f2) ≥ min (νDi (f1), νDi (f2)) ≥ −ni
and therefore f1 + f2 ∈ L(D)
Remark. If X is smooth and projective, it is a fact that
dim(L(D)) < ∞
Will prove this for curves later
Put `(D) := dim(L(D))
29
Lemma. If D ∼ D0, then L(D) is isomorphic to L(D0)
as vector spaces
In particular, `(D) only depends on the linear equivalence
class of D
Pf. Write D = D0 + (g), g ∈ k(X)∗
ρ : L(D) → L(D0), ρ(f ) = f g
is an isomorphism
First check that ρ(L(D)) ⊂ L(D0)
Clear that 0 ∈ L(D) goes to 0 ∈ L(D0) so assume f ∈
L(D) − 0
D + (f ) ≥ 0
Must show f g ∈ L(D0) i.e. D0 + (f g) ≥ 0. But
D0 + (f g) = D0 + (g) + (f ) = D + (f ) ≥ 0
ρ is clearly k-linear
ρ−1 is defined by multiplication by g −1
30
Riemann-Roch theorems give information on `(D) (and
sometimes compute it exactly)
Example. Riemann-Roch for Pn: if H ∈ Cl(Pn) is the
hyperplane class, then


n + d

,
`(dH) = 
n
a degree n polynomial in d
To see this, choose H to be the hyperplane x0 = 0
L(dH) is the set of ratl fcn whose only poles are along
H, with mult at most d

F
L(dH) =  d | F is a deg d homog poly
x0



`(dH) is the dim of the space of deg d homog polys
Now, let V ⊂ L(D) be a finite-dimensional subspace
If f ∈ V − 0, then D + (f ) is an effective divisor
Def. A linear system of divisors on X is a set of effective divisors of the form
{D + (f ) | f ∈ V − 0 ⊂ L(D) − 0}
If V = L(D), the resulting linear system is called a complete linear system and denoted by |D|
31
Example. X = P1, D = 3 · (0, 1) as before
For the complete lin system

as3 + bs2t + cst2 + dt3 
L(D) = 



s3




If f ∈ L(D) − 0, then (f ) = D0 − 3 · (0, 1), where D0 is
the effective divisor consisting of the three points (with
mult) where as3 + bs2 + cs2t + dt3 = 0
Thus D + (f ) = D0
The complete linear system is the totality of all divisors
P
P
nipi with ni = 3
This seems independent of the original divisor D (aside
from its degree)
In fact, it is straightforward to see in the general case
that if D ∼ D0
V ⊂ L(D) corresponding to V 0 ⊂ L(D0)
The linear systems associated to V and V 0 are the same
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Finally, consider a linear system determined by D and
V ⊂ L(D)
Let {f0, . . . , fn} be a basis for V
Have rational map
φV : X − −− → Pn, φV = (f0, . . . , fn)
Choose a different basis {g0, . . . , gn} for V
P
Then gi = j aij fj for an invertible (n + 1) × (n + 1)
matrix A
The matrix A determines an automorphism ψA of Pn,
x 7→ Ax
If φ0V : X − − → Pn is the map corresponding to the
basis {gi}
φ0V = ψA ◦ φV
In conclusion, a linear system determines a rational map
to Pn up to automorphism
Questions:
• In terms of V ⊂ L(D), when is φV regular?
• In terms of V ⊂ L(D), when is φV an embedding?
These questions will be answered for curves
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Example. Consider D = 3 · (0, 1) and the complete
linear system determined by V = L(D)
Choose the basis {1, t/s, t2/s2, t3/s3} of L(D)
t t2 t3  3 2

2 3
φD (s, t) = 1, , 2 , 3  = s , s t, st , t
s s s


We have therefore recovered the twisted cubic from the
single divisor 3 · (0, 1)!
Claim: The linear system associated with V ⊂ L(D) is
isomorphic to the projective space P(V ). In particular,
the complete linear system is isomorphic to P`(D)−1
Pf. The second claim follows from the first, since P(L(D)) '
P`(D)−1
Let LV ⊂ Div(X) denote the linear system
Consider the map
φV : V − 0 → Div(X), f 7→ (f ) + D
If φV (f ) = φV (g), then (f ) = (g)
So f and g are scalar multiples of each other
So φV induces an injective map
φ̃V : P(V ) ,→ Div(X)
LV = im(φ̃) by definition
Therefore LV ' P(V )
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