1Divisors
III Positivity in Algebraic Geometry
1.4 Computations of class groups
To explicitly compute the class group, the following proposition is useful:
Proposition.
Let
X
be an integral scheme, regular in codimension 1. If
Z ⊆ X
is an integral closed subscheme of codimension 1, then we have an exact sequence
Z → Cl(X) → Cl(X \ Z) → 0,
where n ∈ Z is mapped to [nZ].
Proof.
The map
Cl
(
X
)
→ Cl
(
X \ Z
) is given by restriction. If
S
is a Weil divisor
on
X \ Z
, then
¯
S ⊆ X
maps to
S
under the restriction map. So this map is
surjective.
Also, that [
nZ
]
|
X\Z
is trivial. So the composition of the first two maps
vanishes. To check exactness, suppose
D
is a Weil divisor on
X
, principal on
X \ Z
. Then
D|
X\Z
=
dim
(
f
)
|
X\Z
for some
f ∈ K
(
X
). Then
D − div
(
f
) is just
supported along Z. So it must be of the form nZ.
If we remove something of codimension at least two, then something even
simpler happens.
Proposition.
If
Z ⊆ X
has codimension
≥
2, then
Cl
(
X
)
→ Cl
(
X \ Z
) is an
isomorphism.
The proof is the same as above, except no divisor can be supported on Z.
Example. Cl(A
2
\ {0}) = Cl(A
2
).
In general, to use the above propositions to compute the class group, a good
strategy is to remove closed subschemes as above until we reach something we
understand well. One thing we understand well is affine schemes:
Proposition.
If
A
is a Noetherian ring, regular in codimension 1, then
A
is a
UFD iff A is normal and Cl(Spec A) = 0
Proof.
If
A
is a UFD, then it is normal, and every prime ideal of height 1 is
principally generated. So if
D ∈ Spec A
is Weil and prime, then
D
=
V
(
f
) for
some f, and hence (f) = I
D
.
Conversely, if
A
is normal and
Cl
(
Spec A
) = 0, then every Weil divisor is
principal. So if
I
is a height 1 prime ideal, then
V
(
I
) =
D
for some Weil divisor
D
. Then
D
is principal. So
I
= (
f
) for some
f
. So
A
is a Krull Noetherian
integral domain with principally generated height 1 prime ideals. So it is a
UFD.
Example. Cl(A
2
) = 0.
Example. Consider P
n
. We have an exact sequence
Z → Cl(P
n
) = Cl(P
n
\ P
n−1
) = Cl(A
n
) → 0.
So
Z → Cl
(
P
n
) is surjective. So
Cl
(
P
n
) is generated by a hyperplane. Moreover,
since any principal divisor has degree 0, it follows that
nH 6
= 0 for all
n >
0.
This H corresponds to O(1).
Example.
Let
X
=
V
(
xy − z
2
)
⊆ A
3
. Let
Z
=
V
(
x, z
). We claim that
Cl(X)
∼
=
Z/2Z, and is generated by [Z].
We compute
K[X \ Z] =
K[x, x
−1
, y, z]
(xy − z)
∼
=
K[x, x
−1
, t, z]
(t − z)
= K[x, x
−1
, z],
where t = xy, and this is a UFD. So Cl(X \ Z) = 0.
We now want to compute the kernel of the map Z → Cl(X). We have
O
X,Z
=
K[x, y, z]
xy − z
2
(x,z)
=
K[x, y, y
−1
, z]
x − z
2
(x,y)
= K[y, y
−1
, z]
(z)
.
Unsurprisingly, this is a DVR, and crucially, the uniformizer is
z
. So we know
that
div
(
x
) = 2
Z
. So we know that 2
Z ⊆ ker
(
Z → Cl
(
X
)). There is only one
thing to check, which is that (x, y) is not principal. To see this, consider
T
(0)
X = A
3
, T
(0)
(Z ⊆ X) = A
1
.
But if
Z
were principal, then
I
Z,0
= (
f
) for some
f ∈ O
X,0
. But then
T
0
(
Z
)
⊆
T
0
X
will be
ker
d
f
. But then
ker
d
f
should have dimension at least 2, which is
a contradiction.
Proposition. Let X be Noetherian and regular in codimension one. Then
Cl(X) = Cl(X × A
1
).
Proof. We have a projection map
pr
∗
1
: Cl(X) → Cl(X × A
1
)
[D
i
] 7→ [D
i
× A
1
]
It is an exercise to show that is injective. ‘
To show surjectivity, first note that we can the previous exact sequence and
the 4-lemma to assume X is affine.
We consider what happens when we localize a prime divisor
D
at the generic
point of
X
. Explicitly, suppose
I
D
is the ideal of
D
in
K
[
X × A
1
], and let
I
0
D
be the ideal of K(X)[t] generated b I
D
under the inclusion
K[X × A
1
] = K[X][t] ⊆ K(X)[t].
If
I
0
D
= 1, then
I
D
contains some function
f ∈ K
(
X
). Then
D ⊆ V
(
f
)
as a subvariety of
X × A
1
. So
D
is an irreducible component of
V
(
f
), and in
particular is of the form D
0
× A
1
.
If not, then
I
0
D
= (
f
) for some
f ∈ K
(
X
)[
t
], since
K
(
X
)[
t
] is a PID. Then
divf
is a principal divisor of
X × A
1
whose localization at the generic point is
D
.
Thus,
divf
is
D
plus some other divisors of the form
D
0
× A
1
. So
D
is linearly
equivalent to a sum of divisors of the form D
0
× A
1
.
Exercise. We have Cl(X × P
n
) = Z ⊕ Cl(X).