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Abelian Varieties
Spring 2012, taught by Xinwen Zhu.
Contents
1 Overview of Results
2
2 Basic Results
4
2.1
Some Corollaries of the Theorem of the Cube . . . . . . . . . . . . . . . . . . . . . .
3 Cohomology of Sheaves
5
6
4 More Results on Line Bundles
10
5 Group Schemes
13
6 The Picard Functor and Dual Variety
16
6.1
Hopf Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
6.2
Polarizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
7 Cartier Duality
22
8 Étale and Local Group Schemes
25
9 The Tate Module
29
10 The Isogeny Category
30
11 The Endomorphism Algebra
33
12 The Rosati Involution
39
12.1 Albert’s Classification of Division Algebras With Involution . . . . . . . . . . . . . .
40
13 Abelian Varieties Over Finite Fields
41
14 Good Reduction of Abelian Varieties
45
14.1 Complex Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
46
1
Overview of Results
Theorem 1.1. Let Y = V /Λ for Λ ⊆ V = Cn a lattice. Then Y can be embedded as a projective
variety if and only if there exists H : V × V → C a positive definite Hermitian form such that
=H|Λ×Λ is integral.
Lemma 1.2. Let V be a complex vector space and VR the underlying real space. Then there is a
bijection between hermitian forms H on V and skew-symmetric forms ω on VR satisfying ω(ix, iy) =
ω(x, y).
Proof. The mapping is H 7→ =H.
Assuming the theorem, we can embed the Jacobian J(X) of an algebraic curve X if we construct
a symplectic pairing H1 (X, Z) × H1 (X, Z) → Z such that the corresponding form on H 0 (X, ΩX )∗
is positive definite. (This pairing is given by the intersection pairing). We will later see that the
addition law on J(X) is a morphism of projective varieties.
An abelian variety over C is a projective variety with a group law. That is, the multiplication
and inversion maps are morphisms of algebraic varieties.
For another example, if F/Q is a totally real extension of degree g and E/F is an imaginary
quadratic extension, then E is a CM field. Hom(E, C) has 2g elements an a natural complex conjugation involution, denoted by ι.
A CM type is a choice of subset Φ ⊆ Hom(E, C) such that Φ has exactly g elements and
∼
Φ ∪ ι(Φ) = Hom(E, C). Such a Φ gives a map E ⊗Q R −
→ CΦ by e ⊗ r 7→ (rϕ(e))ϕ∈Φ .
Theorem 1.3. CΦ /OE is an abelian variety.
Sketch. Choose α totally imaginary in E, such that =ϕ(α) > 0 for every ϕ ∈ Φ. Then we have a
pairing E × E → Q given by (x, y) 7→ trE/Q (αxy ∗ ). α can also be chosen such that OE × OE is
mapped into Z.
For a simpler example, C/Λ is an abelian variety for any Λ ⊆ C. Without loss of generality, let
1
Λ = h1, τ i for τ ∈ H, and then consider H : C × C → C by (z, w) 7→ =τ
zw so that H is positive
definite with =(H(1, τ )) = −1.
Proposition 1.4. Let X be a connected, compact, complex Lie group. Then X is of the form V /Λ.
(In particular, every abelian variety over C is of the form V /Λ.)
Proof. Let V be the Lie algebra of X, and consider the adjoint representation Ad : X → GL(V ). Ad
is a holomorphic map. But X is compact and GL(V ) is an open subset of a complex vector space.
So Ad must be constant, therefore constant identity. We get that ghg −1 = h, so X is commutative.
Now since X is commutative, exp : V → X is a group homomorphism, and in fact a covering
space. So X ∼
= V /Γ for Γ discrete. As X is compact, Γ must be a lattice.
Corollary 1.5. Let X be an abelian variety over C of dimension g. Then:
• X is commutative and divisible. In particular, nX : X → X given by multiplication by n is a
surjective group homomorphism, and ker nX = n1 Λ/Λ ∼
= (Z/nZ)2g .
2
• If X, Y are two abelian varieties, let Hom(X, Y ) be the group of homomorphisms (as complex
Lie groups) from X to Y . Then Hom(X, Y ) → Hom(H1 (X), H1 (Y )) = Hom(ΛX , ΛY ) is
injective. In fact,
Hom(X, Y ) = {f : VX → VY : f (ΛX ) ⊆ ΛY }.
(1.1)
Now fr an arbitrary field k, an abelian variety over k is a smooth, complete variety, together
with 0 ∈ X(k) and morphisms m : X × X → X and s : X → X obeying the group laws.
Let X, Y be two abelian varieties. Then a morphism f : X → Y is a morphism of algebraic
varieties such that this diagram commutes:
m
X ×X
X
f
f ×f
m
Y ×Y
(1.2)
Y
The set of morphisms from X to Y is denoted by Hom(X, Y ). We get a category AVk , of abelian
varieties over k.
Theorem 1.6 (to be proven later). X(k) is commutative and divisible, and for nX : X → X
multiplication by n,
(
(Z/nZ)2g
ker nx =
(Z/pm Z)i
p = char(k) - n
n = pm (for some 0 ≤ i ≤ g).
(1.3)
Theorem 1.7 (Mordell-Weil). If k is a number field, then X(k) is finitely generated.
These theorems are useful for studying the structure of X(k) and X(k). We can also ask about
the structure of Hom(X, Y ) and its matrix representation.
Theorem 1.8. Hom(X, Y ) is a finitely generated free abelian group.
Remark. Divisibility implies it’s torsion free. We already know that it’s finitely generated for k = C.
Choose a prime ` 6= p = char(k). Let X[`n ] = ker `nX ∼
= (Z/`Z)2g , which is acted on by G =
`
X
Gal(k/k). Also we have X[`n+1 ] −−
→ X[`n ].
The `-adic Tate module of X, T` (X), is defined as limn X[`n ]. This is a free Z` module of rank
←−
2g, with a continuous action by G.
Theorem 1.9. For f : X → Y a morphism, then f : X[`n ] → Y [`n ], so we get a map T` (f ) :
T` (X) → T` (Y ). Then the map
Z` ⊗ Hom(X, Y ) → HomZ` (T` (X), T` (Y ))
is injective. In fact, the image lies in HomZ` [G] (T` (X), T` (Y )).
3
(1.4)
Conjecture 1.10 (Tate). If k is finitely generated over its prime field, then
Z` ⊗ Hom(X, Y ) → HomZ` [G] (T` (X), T` (Y ))
(1.5)
is an isomorphism.
This is known for k = Fq (Tate) and for k a number field (Faltings).
We would also like to understand Pic X:
Theorem 1.11. There is a short exact sequence
0 → Pic0 X → Pic X → N S(X) → 0
(1.6)
b and N S(X) is a
where Pic0 X has a natural abelian variety structure, usually denoted by X,
finitely generated free abelian group.
Theorem 1.12. Let L be an ample line bundle on X. Then L3 is very ample.
Finally, we can consider the moduli space of abelian varieties.
2
Basic Results
Theorem 2.1. X is commutative.
Proposition 2.2 (Rigidity Lemma). Let X be a complete variety and Y, Z varieties. Let f : X ×
Y → Z be a morphism such that f |X×y0 is a constant z0 ∈ Z. Then there exists g : Y → Z such
that f = g ◦ pr2 .
Corollary 2.3. Let X, Y be abelian varieties and f : X → Y be a morphism of algebraic varieties.
Then f (m) = m(h(x), a) for some h ∈ Hom(X, Y ) and a ∈ Y .
Proof. Without loss of generality, we may assume f (0) = 0 (we can then translate afterward). We
need to show that f is a homomorphism. Define g : X × X → Y by
g(x1 , x2 ) = f (x1 · x2 )f (x2 )−1 f (x1 )−1 .
(2.1)
Then g(x1 , 0) = 0 and g(0, x2 ) = 0. By the rigidity lemma, g is identically zero.
This implies that X is commutative by applying to the inversion map.
Proof of Rigidity Lemma. Let U be an open affine neighborhood of z0 in Z. Then f −1 (U ) contains
X × {y0 } and is open in X × Y . So W = X × Y \ f −1 (U ) is closed in X × Y . As X is complete,
pr2 (W ) is closed in Y and does not contain y0 . So V = Y \ pr2 (W ) is a nonempty open subset of
Y . Now for v ∈ V . f (X × {v}) ⊆ U . As X × {v} is complete and U is affine, f (X × {v}) must be a
constant map.
Now let x0 ∈ X and define g : Y → Z by g(y) = f (x0 , y). We have f = g ◦ pr2 on X × V . By
separatedness, we conclude that f = g ◦ pr2 .
4
If A is an abelian variety, we now know its operation is commutative, so we will now use additive
notation. nA , multiplication by n, is a morphism A → A.
Lemma 2.4. If char(k) = p - n, then nA is surjective.
Sketch. (dnA )0 : T0 A → T0 A is multiplication by n. So if p - n, then (dnA )0 is an isomorphism. This
implies that nA is smooth, so in particular surjective.
Theorem 2.5 (Theorem of the Cube). Let X, Y be complete varieties and Z any variety. Let
x0 ∈ X, y0 ∈ Y, z0 ∈ Z. Let L be a line bundle on X × Y × Z. If L|{x0 }×Y ×Z , L|X×{y0 }×Z , and
L|X×Y ×{z0 } are trivial, then L is trivial.
Remark. Let Pk+ be the category of pointed complete varieties over k (that is, X with x ∈ X(k)).
Let T : Pk+ → Ab be a contravariant functor. Given X0 , . . . , Xn , we have maps
ci × · · · × Xn
πi : X0 × · · · × Xn → X0 × · · · × X
(2.2)
ci × · · · × Xn → X0 × · · · × Xn .
σi : X0 × · · · × X
(2.3)
and
Now we can define
αn =
X
T (πi )
(2.4)
i
βn =
Y
T (σi ).
(2.5)
i
Lemma 2.6. T (X0 × · · · Xn ) = im αn ⊕ ker βn .
The functor T is of order n (linear for n = 1, quadratic for n = 2) if βn is injective, or equivalently
if α is surjective. Then the theorem of the cube implies Pic : Pk+ → Ab is quadratic, as β2 is injective.
As another example, let A be an abelian variety. Then HomP + (−, A) : Pk+ → Ab is linear,
k
because of the rigidity lemma.
Exercise: if k = C, then H 2 (−; Z) : PC+ → Ab is quadratic.
2.1
Some Corollaries of the Theorem of the Cube
Corollary 2.7. Let X, Y, Z be complete. Then every line bundle on X × Y × Z is of the form
p∗12 L3 ⊗ p∗23 L1 ⊗ p∗31 L2 .
Proof. β2 injective implies α2 surjective by Lemma 2.6.
Corollary 2.8. Let A be an abelian variety and X some variety. Let f, g, h : X → A be three
morphisms. Then for any line bundle L on A,
(f + g + h)∗ L ∼
= (f + g)∗ L ⊗ (g + h)∗ L ⊗ (f + h)∗ L ⊗ f ∗ L−1 ⊗ g ∗ L−1 ⊗ h∗ L−1 .
5
(2.6)
Proof. It’s enough to consider the universal case X = A × A × A and f, g, h the projections, via
f ×g×h
X −−−−→ A × A × A. Then let pij : A × A × A → A × A and pi : A × A × A → A be the projections,
m : A × A → A be multiplication, and n : A × A × A → A be multiplication of three elements. We
want to show that
n∗ L = (m ◦ p12 )∗ L ⊗ (m ◦ p23 )∗ L ⊗ (m ◦ p13 )∗ L ⊗ p∗1 L−1 ⊗ p∗2 L−1 ⊗ p∗3 L−1 .
(2.7)
By the theorem of the cube, it’s enough to show this holds when restricted to 0 × A × A.
Now the left hand side is (m ◦ p23 )∗ L while the right hand side is
p∗2 L ⊗ (m ◦ p23 )∗ L ⊗ p∗3 L−1 ⊗ 0 ⊗ p∗2 L−1 ⊗ p∗3 L−1 ,
(2.8)
so we see that equality holds.
2
2
n +n
n −n
Corollary 2.9. Let L be a line bundle on A. Then n∗A L ∼
= L 2 ⊗ (−1)∗A L 2 .
Proof. Take f = nA , g = 1A , and h = (−1)A . Then by Corollary 2.8,
n∗A L = (n + 1)∗A L ⊗ 0 ⊗ (n − 1)∗A L ⊗ n∗A L−1 ⊗ L−1 ⊗ (−1)∗A L−1 .
(2.9)
i h
i
i h
h
(n + 1)∗A L ⊗ n∗A L−1 = n∗A L ⊗ (n − 1)∗A L−1 ⊗ L ⊗ (−1)∗A L .
(2.10)
So
Now we obtain n∗A L⊗(n−1)∗A L−1 = Ln ⊗(−1)∗A Ln−1 , and then n∗A L = L
n2 +n
2
⊗(−1)∗A L
n2 −n
2
.
2
Corollary 2.10. If L is symmetric (that is, (−1)∗A L ∼
= Ln .
= L), then n∗A L ∼
For x ∈ A(k 0 ), let TX : Ak0 → Ak0 be translation y 7→ x + y.
Corollary 2.11 (Theorem of the Square). For any x, y ∈ A and L a line bundle on A, we have
∗ L ⊗ L = T ∗ L ⊗ T ∗ L.
Tx+y
y
x
Proof. Take f : A → A to be f (A) = x, g : A → A to be g(A) = y, and h = 1A .
Let L be a line bundle on A. Define φL : A(k 0 ) → Pic(Ak0 ) by x 7→ Tx∗ L ⊗ L−1 . Then φL is a
group homomorphism.
3
Cohomology of Sheaves
Let X be a scheme, QCoh(X) be the abelian category of quasi-coherent sheaves on X, and Coh(X)
be the subcategory of coherent sheaves. Given f : X → Y , we haev the functor f∗ : QCoh(X) →
QCoh(Y ). The derived functors of f∗ are Ri f∗ : QCoh(X) → QCoh(Y ), together with δ i : Ri f∗ F 00 →
Ri+1 f∗ F 0 for each exact 0 → F 0 → F → F 00 → 0, such that:
1. R0 f∗ = f∗ .
6
2. For each exact 0 → F 0 → F → F 00 → 0, we have the long exact sequence
· · · → Ri f∗ F 0 → Ri f∗ F → Ri f∗ F 00 → Ri+1 f∗ F 0 → · · ·
(3.1)
3. The sequence of (2) is natural: given a commutative diagram with two exact rows, the resulting
diagram from taking long exact sequences is commutative.
If Y = Spec R, then Ri f∗ X can be written as H i (X, F). Now assume Y = Spec R and X is
separated. Let U = {Ui } be a cover of X by affine opens. Then H i (X, F) ∼
= Ȟ i (U, F).
Corollary 3.1 (Künneth Formula). Let X, Y be separated schemes over a field k, F ∈ QCoh(X),
and G ∈ QCoh(Y ). Then
H n (X × Y, F G) =
M
H i (X, F) ⊗k H j (Y, G).
(3.2)
i+j=n
Theorem 3.2. Let f : X → Y be a proper morphism of noetherian schemes, and F ∈ Coh(X).
Then Ri f∗ F ∈ Coh(X). In particular, if Y = Spec k, then H i (X, F) is finite dimensional.
Theorem 3.3. Let f : X → Y be as above, and let F ∈ Coh(X) be flat over Y . That is, for every
x ∈ X, Fx is a flat OY,f (x) -module. Then there is a finite complex
0 → K0 → · · · → Kn → 0
(3.3)
of locally free Y -modules of finite rank over Y , such that for every u : S → Y , and for v :
X ×Y S → X and g : X ×Y S → S the projections, we have Ri g∗ (v ∗ F) = H i (u∗ K • ).
As an example, if L is a line bundle over X ×k T , then taking f to be the projection X ×k T → T ,
then L is flat over T .
Let X/k be proper and F ∈ Coh(X). Then let
χ(F) =
X
(−1)i dimk H i (X, F).
(3.4)
i
Corollary 3.4. Let f : X → Y and F be as in the theorem. Then χy (F) = χ(Fy ) is locally constant
on Y , where Fy is F|Xy .
Proof. Let 0 → K 0 → · · · → K n → 0 be as in the theorem. We have the diagram
Xy
X
f
Spec k(y)
(3.5)
Y
Then H i (Xy , Fy ) = H i (K • ⊗Oy k(y)) so
χ(Fy ) =
X
X
(−1)i H i (K • ⊗Oy k(y)) =
(−1)i dim(K i ⊗Oy k(y))
i
i
and each dimension is locally constant.
7
(3.6)
Corollary 3.5. Let f : X → Y and F be as before. Then y 7→ hi (y, F) = dimk(y) H i (Xy , Fy ) is
upper semicontinuous.
Proof. We have
hi (y, F) = dim H i (K • ⊗ k(y))
(3.7)
i
i−1
= dim(ker(d ⊗ k(y))) − dim(im(d
i
⊗ k(y)))
i
(3.8)
i−1
= dim(K ⊗ k(y)) − dim(im(d ⊗ k(y))) − dim(im(d
⊗ k(y))).
(3.9)
After the above corollary, it’s enough to show that for each n, {y ∈ Y : dim(im(di ⊗ k(y))) < n}
is closed. This is clear since locally di is a matrix, and having low rank is equivalent to the vanishing
of minors.
Corollary 3.6. Let f : X → Y and F be as above, and assume that Y is connected and reduced.
Then the following are equivalent:
• hi (y, F) is locally constant.
∼
• Ri f∗ F is locally free of finite rank and Ri f∗ F ⊗ k(y) −
→ H i (Xy , Fy ).
Theorem 3.7 (Seesaw Theorem). For k algebraically closed, X/k complete, and T /k any variety,
let L be a line bundle on X × T . Then:
1. T1 = {t ∈ T : L|X×{t} is trivial} is closed in T .
2. There exists some line bundle M on T1 such that LX×T1 ∼
= p∗2 M.
Proof.
1. Let X be a complete variety. We claim that a line bundle L is trivial if and only if
Γ(X, L) and Γ(X, L−1 ) are nonzero. The “only if” implication is clear. For the “if” implication,
nonzero maps s : OX → L and t : OX → L∨ yield t∨ ◦ s : OX → OX nonzero. Since
Γ(X, OX ) = k, s is an isomorphism.
Now T1 can be expressed as
n
o n
o
t ∈ T : H 0 (X × {t}, L|X×{t} ) 6= 0 ∩ t ∈ T : H 0 (X × {t}, L∨ |X×{t} 6= 0
(3.10)
which is closed.
2. Consider p2 as a map X × T1 → T1 . Since dim H 0 (X × {t}, L|X×{t} ) = 1, (p2 )∗ L is locally
∼
freee of rank 1. Let M = (p2 )∗ L, then p∗2 M −
→ L.
If k is not algebraically closed, then Γ(X, OX ) is the field k 0 of algebraic elements of K(X) and
then there exists M on (T1 )k0 such that L = p∗2 M for p2 : X ×k0 (T1 )k0 → (T1 )k0 .
Proof of the Theorem of the Cube. Suppose X, Y /k are complete, Z/k is any variety, and L is a line
bundle on X × Y × Z such that L|X×Y ×{z0 } , L|X×{y0 }×Z , and L|{x0 }×Y ×Z are trivial. We want to
show that L is trivial.
8
Lemma 3.8. For every x0 , x1 on X, there exists an irreducible curve C containing x0 and x1 .
(The lemma will be proven later.)
Now it’s enough to show that L|{x}×Y ×{z} is trivial for all (x, z) ∈ X × Z, for then L = p∗13 M.
But L|X×{y0 }×Z is trivial, which would make M trivial.
To show that L|{x}×Y ×{z} is trivial for all (x, z), by the lemma, we can choose C through x0
and x. So we can replace X with C. In addition, we can replace C with its normalizations, so we
can assume that X is a smooth projective curve.
Now let g be the genus of X. There exists a divisor E on X of degree g such that
H 0 (X, ΩX (−E)) = 0 =⇒ H 1 (X, O(E))) = 0.
(3.11)
Let M = p∗1 O(E) ⊗ L. Then W = supp(R1 (p23 )∗ M) is disjoint from Y × {z0 }, since
H 1 (X × {y} × {z0 }, M|X×{y}×{z0 } ) = 0,
{z
}
|
(3.12)
O(E)
and then we can use upper semicontinuity.
Now the projection of W to the Z factor is a closed subset, which does not contain z0 . As a
result, there exists Z 0 ⊆ Z open containing z0 , such that R1 (p23 )∗ M|Y ×Z 0 = 0. It’s enough to prove
the theorem for X ×Y ×Z 0 , so we can replace Z by Z 0 , therefore we can assume that R1 (p23 )∗ M = 0.
Therefore (p23 )∗ M is locally free of rank
χ(M|X×{y}×{z} ) = χ(M|X×{y0 }×{z0 } ) = χ(O(E)) = 1
(3.13)
on Y × Z.
Let N = (p23 )∗ M, and let {Ui } be an open cover on Y × Z over which N |Ui is trivial, and
∼
choose αi : OUi −
→ N |Ui . Now αi (1) ∈ Γ(Ui , N ) = Γ(X × Ui , M). Let Di be the set of zeros of
αi (1). On Ui ∩ Uj , we have αi (1) = fij αj (1) for some fij ∈ O(Ui ∩ Uj )× , so Di |Ui ∩Uj = Dj |Ui ∩Uj .
So we can define D ⊆ X × Y × Z such that D|X×Ui ∼
= Di .
Now L is trivial if and only if M = p∗1 O(E) if and only if D = E × Y × Z. Let p ∈ X
with p 6∈ supp(E). Then {p} × Y × {z0 } 6⊆ D ∩ ({p} × Y × Z), since M|X×{y}×{z0 } = O(E).
The projection of D ∩ ({p} × Y × Z) to the Z factor is a closed subset T not containing z0 , so
D ∩ ({p} × Y × Z) = {p} × Y × T as D has codimension 1. But D ∩ ({p} × Y × Z) does not contain
{p} × Y × {z0 }, so T = ∅, giving D ∩ ({p} × Y × Z) = ∅. We conclude that D = E × Y × Z.
Proof of lemma. Assume dim X ≥ 2. By Chow’s Lemma, we can assume that X is projective. Let
f
e be the blow-up of X (X
e −
e is also projective, so embeds into PN for
X
→ X) at {x0 , x1 }. Then X
e is irreducible of
some N . By Bertini. we can find a general hyperplane section H such that H ∩ X
−1
−1
e to X via
codimension 1. As dim f (xi ) ≥ 1, we have H ∩ f (xi ) 6= ∅, so we can project H ∩ X
f , giving us a proper subvariety of X through x0 and x1 .
Over C, there is a simpler proof of the theorem of the cube. Use the exponential sequence
×
0 → Z → OX OX
→1
and the fact that H 1 (X, OX ) is linear and H 2 (X, Z) is quadratic.
9
(3.14)
4
More Results on Line Bundles
∗ L ⊗ L ∼ T ∗ L ⊗ T ∗ L, so we can define a homomorphism
Recall that for L ∈ Pic A, we have Tx+y
= x
y
∗
φL : A(k) → Pic Ak by x 7→ Tx L ⊗ L−1 . Also φL1 ⊗ φL2 = φL1 ⊗L2 , so we get φ : Pic A →
Hom(A(k), Pic Ak ).
Let Pic0 A = ker φ. Concretely, L ∈ Pic0 A if and only if Tx∗ L = L for every x. As for every
L ∈ Pic A, φL has image contained in Pic0 Ak , we get an exact sequence
φ
0 → Pic0 A → Pic A −
→ Hom(A(k), Pic0 Ak ).
(4.1)
Lemma 4.1. L ∈ Pic0 A if and only if m∗ L ∼
= p∗1 L ⊗ p∗2 L.
Proof. If m∗ L ∼
= p∗1 L ⊗ p∗2 L = L L, then restricting this equality to {x} × A, we obtain Tx∗ L = L.
This holds for every x, so L ∈ Pic0 A.
Conversely, if L ∈ Pic0 A, then letting M = m∗ L ⊗ p∗1 L−1 ⊗ p∗2 L−1 , M|A×{x} and M|{x}×A are
trivial for every x. By Seesaw, M is trivial.
For any L ∈ Pic A, define K(L) = {x ∈ A(k) : Tx∗ L = L}, so K(L) = A if and only if L ∈ Pic0 A.
Lemma 4.2. K(L) is closed in Ak .
Proof. For M = m∗ L ⊗ p∗2 L−1 , K(L) = {x : M|A×{x} is trivial}, which is closed.
Theorem 4.3. Let L = O(D), where D is an effective divisor. Then the following are equivalent:
1. L is ample
2. K(L) is finite
3. H(D) = {x ∈ A : x + D = D} is finite
4. |2D| is basepoint free and defines a finite morphism A → |2D|∗ . (Note: |2D| is always basepoint
free.)
Corollary 4.4. Every abelian variety is projective.
Proof of corollary. Pick an affine open nieghborhood U of 0 in A, and D the divisor given by A \ U .
Then H(D) ⊆ U . But also H(D) is closed: we have m : H(D) × D → D. Taking the closure,
H(D) × D maps into D, therefore D. We get H(D) = H(D). As A is complete, H(D) is finite.
Proof of theorem. (1) =⇒ (2) : Suppose K(L) is not finite. Then B = K(L)0 is an abelian variety
of positive dimension. By definition, L|B ∈ Pic0 B, so m∗ L|B ∼
= p∗1 L|B ⊗ p∗2 L|B . But considering the
1×(−1)
map B −−−−→ B × B, we get that OB ∼
= L ⊗ (−1)∗ L is ample, a contradiction.
(2) =⇒ (3) is trivial. For (3) =⇒ (4), we have (x + y + D) + D ∼ (x + D) + (y + D) for every
x, y. In particular, for every x, (x + D) + (−x + D) ∈ |2D|.
Now we show that |2D| has no basepoints. For every u, there exists x such that x 6∈ u − D and
x 6∈ −u + D. So u 6∈ x + D and u 6∈ −x + D, implying u is not a basepoint.
10
So we need to show that ϕ : A → |2D|∗ is finite. Since both A and |2D|∗ are proper, it suffices
to show that ϕ has finite fibers. Suppose this were not the case. Then ϕ maps some curve C ⊆ A to
a point p. Let E ∈ |2D|. Then either E ∩ C = C or E ∩ C = ∅. For generic E, c 6⊆ E, so we must
have E ∩ C = ∅.
Lemma 4.5. Let C ⊆ A be an irreducible curve, and E a divisor such that C ∩ E = ∅. Then for
every x, y ∈ C, x − y + E = E.
The lemma will give us a contradiction, since if C ∩ D = ∅, then H(D) contains C − C.
Proof of lemma. Let L be the line bundle on A given by O(E). Then L|C = OC . Now m∗ L is a line
bundle on C × A, for m : C × A → A multiplication. Now for x ∈ A, we have
χ(T ∗ L|C ) = χ(m∗ L|C×{x} ) = χ(m∗ L|C×{0} ) = χ(OC ).
(4.2)
Therefore deg Tx∗ L|C = deg OC = 0. So either x + E ⊇ C or (x + E) ⊇ C = ∅. So fix x, y ∈ C
and z ∈ E. Then (x − z) + E contains x ∈ C, so (x − z) + E contains all of X, therefore y. So
x − y + E contains z.
(4) =⇒ (1): We have ϕ : A → PN = |2D|∗ . We need to show that for coherent F, ϕ∗ OA ⊗
Γ(F ⊗ Ln ) → ϕ∗ (F ⊗ Ln ) is surjective for some n. But we know this since OPN ⊗ Γ(ϕ∗ F ⊗ O(n)) →
ϕ∗ F ⊗ O(n) is surjective, because O(1) is ample.
Corollary 4.6. nA : A → A is surjective.
Proof. This is equivalent to showing that ker nA is finite. Choose some ample L. Then n∗A L =
n2 +n
L 2 ⊗ (−1)∗ L
point.
n2 −n
2
is also ample. But also n∗A L|(ker nA )0 is trivial. Therefore (ker nA )0 is a single
Recall that if f : X → Y is dominant and dim X = dim Y , then deg f = [K(X) : K(Y )].
Theorem 4.7.
• deg nA = n2g , where g = dim A.
• nA is separable if and only if p - n.
• The inseparable degree of pA is at least pg .
Let A, B be abelian varieties. A homomorphism α : A → B is called an isogeny if α is surjective
and ker α is finite.
Theorem 4.8. deg nA = n2g .
For X a complete variety of dimension g and L a line bundle on X, let F be a coherent sheaf on
X. Then PL (F, n) = χ(F ⊗ Ln ) is a polynomial of degree at most g. The degree of F with respect
to L, dL (F), is g! times the ng term of PL (F, n). We write dL for dL (OX ).
Proposition 4.9.
1. Let F be a coherent sheaf on X with generic rank r. Then dL (F) = r · dL .
2. Let f : X → Y be a dominant morphism of complete varieties of the same dimension, and L
a line bundle on Y . Then df ∗ L = (deg f )dL .
11
Proof of theorem given proposition. Let L be ample on A. We can assume that L ∼
= (−1)∗ L by
2
tensoring with (−1)∗ L if necessary. Then n∗A L ∼
= Ln so (deg nA )dL = n2g dL , and dL is positive as
L is ample.
Proof of proposition. We’ll prove this under the additional assumptions that X is smooth and f is
finite.
−1
1. Let U ⊆ X be open such that F|U ∼
.
= OUr and D = X \ U be a divisor. Let M = O(D) = ID
There exists σ ∈ Γ(X, M) such that the zero set of σ is D. For any m ∈ Γ(U, F), there exists
N such that m ⊗ σ N extends to a section of F ⊗ MN . We can choose a single N to work for
r → F ⊗ MN , whose cokernel is suported on
every basis element of F|U . This gives a map OX
D. We claim that this map is injective, since it is after restricting to U .
After tensoring by M−N , we get an exact sequence 0 → (IN D )r → F → τ 0 → 0 with τ 0
supported on D. Hence dL (F) = dL (τ 0 ) + r · dL (IN D ). Now dL (τ 0 ) = 0 since τ 0 is supported
on a set of smaller dimension, and by a similar argument using ON D , we get dL (IN D ) = dL ,
so dL (F) = r · dL .
2. Since f : X → Y is finite, for sufficiently large n,
H i (X, f ∗ Ln ) ∼
= H i (Y, f∗ f ∗ Ln ) ∼
= H i (Y, f∗ OX ⊗ Ln ),
(4.3)
so χ(f ∗ Ln ) = χ(f∗ OX ⊗ Ln ). We get df ∗ L = dL (f ∗ OX ) = (deg f )dL .
Given L/K finite, we can find L ⊃ F ⊃ K with F/K separable and L/F purely inseparable.
We can use this to define the separable and purely inseparable degrees of a morphism.
Theorem 4.10.
1. If p - n, then nA is separable.
2. The inseparable degree of pA is at least pg .
Proof.
1. Assume p - n. To show that nA is separable, it suffices to show that nA is smooth at
0. That is, dnA : T0 A → T0 A is surjective. But this is clear, since it is multiplication by n.
2. dpA is the zero map, so p∗A : ΩA/k → ΩA/k is zero. In particular, for f ∈ K(A),
p∗A (df ) = 0 =⇒ d(p∗A f ) = 0 =⇒ p∗A f ∈ k · K(A)p .
(4.4)
So p∗A : K(A) → K(A) factors through k ·K(A)p . As K(A) is purely inseparable over k ·K(A)g
of degree pg , we get a bound for the inseparable dogree of pA .
Corollary 4.11.
(
(Z/nZ)2g
A[n] =
(Z/pm Z)i
p-n
n = pm for some 0 ≤ i ≤ g.
12
(4.5)
Proof. #A[n] = n−1
A (0) = degs nA . This uniquely determines A[`] for ` prime not equal to p (and
some information for ` = p). Now we use the exact sequence
0 → A[`] → A[`k+1 ] → A[`k ] → 0
(4.6)
and induction on k.
5
Group Schemes
Fix a (locally) noetherian scheme S. A group scheme G → S is a group object in the category of
schemes over S.
For C any category, we can define Cb = {F : C op → Set}. There is a natural h : C → Cb given
by h(X) = hX where hX (Y ) = HomC (Y, X). The Yoneda lemma implies h is fully faithful. The
essential image consists of objects which are called representable functors.
A functor G : C op → Grp is called a group functor. A group object in C is a triple (G, X, α) where
b That is, one assigns a group
G is a group functor, and α is a natural isomorphism G → hX in C.
structure on each set hX (Y ), and for any Z → Y , hX (Y ) → hX (Z) is a group homomorphism.
Assume that finite products exist in C. (In particular, final objects exist.) Then giving a group
object (G, X, α) is equivalent to giving an object X of C and morphisms m : X × X → X, s :
X → X, and e : ∗ → X satisfying the group laws. In this case, we have an induced multiplication
hX (Y ) × hX (Y ) → hX (Y ) given by m ◦ −. Conversely, given the original definition, we can take the
identity element of the group hX (X × X) to be multiplication.
Now we can interpret a group scheme G either as giving G ×S G → G, G → G, and S → G
satisfying the group laws, or for every T → S, a functorial group structure G(T ). (This is given by
G(T ) = HomS (T, G).) Also, for any S 0 → S, GS 0 = G ×S S 0 has a natural structure as a group
scheme over S 0 .
We can define right translation as follows: given T → S and g ∈ G(T ), first define
1×g
m
Rg : GT ∼
= GT ×T T −−→ GT ×T GT −→ GT .
(5.1)
Then Rg (T ) : GT (T ) = G(T ) → G(T ) is the normal right translation of G(T ) by g.
Let G → S be a group scheme and H ⊆ G an open (respectively closed) S-subscheme. We say
that H is an open (respectively closed) group subscheme of G if the inclusion H → G is compatible
with multiplication.
Let G, H be two group schemes over S and f : G → H a morphism. f is a morphism of group
schemes is m ◦ (f × f ) = (f × f ) ◦ m.
Associated to f , we have
ker f
G
f
S
e
13
H
(5.2)
ker f is a group scheme over S. If e is a closed embedding, then ker f is a closed group subscheme
of G.
Here are some examples:
• S = Spec k and A any abelian variety over k.
• The additive group Ga is Spec OS [t], having m : Ga ×S Ga induced by OS [t] → OS [t] ⊗ OS [t]
with t 7→ 1 ⊗ t + t ⊗ 1, s : Ga → Ga induced by t 7→ −t, and e : S → T induced by t 7→ 0. We
have Ga (T ) = Γ(T, OT ) with usual addition.
• The multiplicative group Gm is Spec OS [t, t−1 ], with multiplication, inversion, and identity
induced by t 7→ t ⊗ t, t 7→ t−1 , t 7→ 1. We have Gm (T ) = Γ(T, OT× ) with multiplication.
• We have then nth power map n : Gm → Gm , either by n(T ) : Gm (T ) → Gm (T ) given by
f 7→ f n , or by specifying n∗ : OS [t, t−1 ] → OS [t, t−1 ] by t 7→ tn . Then we can define Gm [n] =
ker n = Spec OS [t]/(tn − 1), also denoted by µn . We have µn (T ) = {f ∈ Γ(T, OT× ) : f n = 1}.
• An example of a group functor: let X → S be a scheme. We can define PicX/S : Sch/S → Grp
by
T 7→ {isomorphism classes of line bundles on XT }
/{isomorphism classes of line bundles on T }.
(5.3)
(5.4)
If PicX/S is representable, then the corresponding scheme is called the Picard scheme of X/S.
From now on, we assume that G → S is locally of finite type. Then we have the sheaf of
differentials ΩG/S such that for any quasi-coherent M, HomOG (ΩG/S , M) = DerOS (OG , M). In
particular, HomOG (ΩG/S , G) = DerOS (OG , OG ) are called vector fields on G.
Let D = DerOS (OG , OG ) be a vector field. For any T → S, if u is the associated map GT → G,
then u∗ ΩG/S ∼
= ΩGT /T . We obtain u∗ (Hom(ΩG/S , OS )) → Hom(ΩGT /T , OT ) so D induces a vector
field on GT , also denoted by D.
We say that D is a right invariant vector field if for every T → S, g ∈ G(T ), and f ∈ OG , we
have D(Rg∗ (f ⊗ 1)) = Rg∗ D(f ). We denote the set of right invariant vector fields on G by Lie(G). In
fact, Lie(G) is a sheaf of OS -modules.
Lemma 5.1.
• Let D1 , D2 ∈ Lie(G). Then [D1 , D2 ] = D1 D2 − D2 D1 ∈ Lie(G).
• If pOS = 0, then for D ∈ Lie(G), Dp ∈ Lie(G).
Corollary 5.2. Lie(G) is a restricted Lie algebra.
For simplicity, assume S = Spec k and G/k a group scheme. Then Lie(G) is the set of left
invariant vector fields on G; that is, derivations D : OG → OG such that DL∗x = L∗x D. Then Lie(G)
is a restricted Lie algebra, also called a p-Lie algebra.
A restrited Lie algebra is a k-space g with [−, −] : g × g → g and (−)(p) : g → g such that [−, −]
is k-bilinear, [x, x] = 0, [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0, (λx)(p) = λp x(p) , and ad x(p) = (ad x)(p) .
(Here (ad X)(Y ) = [X, Y ].)
14
Finally, (x + y)(p) = x(p) + y (p) + Fp (ad x, ad y)y for some universal non-commutative polynomial
Fp , depending only on p, such that Fp (0, 0) = 0. In particular, if [x, y] = 0, then (x + y)(p) =
x(p) + y (p) . We say that g is abelian if [x, y] = 0 for every x, y. In this case, (−)(p) is p-semilinear.
Let Te G be the tangent space of G at e; that is, (me /m2e )∨ . This is also expressible as the
x
set of x : Spec Λ → G such that Spec k ,→ Spec Λ −
→ G is e, for Λ = k[]/2 . In turn, this is
ker(G(Λ) → G(k)).
Proposition 5.3. The map Lie(G) → Te G given by D 7→ D|e (where D|e is given by f 7→ D(f )
(mod m2e ) for f ∈ me ) is an isomorphism.
Lemma 5.4. For X/k a scheme, then to give a derivation D : OX → OX , it is equivalent to give
e : OX ⊗k Λ → OX ⊗k Λ an automorphism of Λ-algebras such that D
e (mod ) = 1.
D
This is given by f 7→ f + Df .
e = X × Spec Λ, we give D
e :X
e →X
e over Spec Λ such that the restriction
In other words, for X
to X → X over Spec k is the identity. If X = G is a group scheme, then D is left invariant if and
only if
e×G
e
G
e
1×D
e×G
e
G
m
e
G
m
e
D
(5.5)
e
G
Proof of proposition. The inverse map is given as follows: for x ∈ Te G = ker(G(Λ) → G(k)) ⊆ G(Λ),
e→G
e given by right multiplication is left invariant, so D ∈ Lie(G).
Rx : G
Let Λ0 = Λ⊗k Λ = k[1 , 2 ]/(21 , 22 ). We have maps p1 , p2 , p3 : Spec Λ0 → Spec Λ for 7→ 1 , 2 , 1 2
respectively.
f1 )p∗ (D
f2 )p∗ (D
f1 )−1 p∗ (D
f2 )−1 =
Lemma 5.5. Let D1 , D2 ∈ Lie(G) and D3 = [D1 , D2 ]. Then p1 ∗ (D
2
1
2
f3 ).
p∗3 (D
Corollary 5.6. If G is commutative, then g = Lie(G) is abelian.
f1 = Rx , D
f2 = Ry , and Rx Ry Rx−1 Ry−1 = 1.
For x, y ∈ Te G, we have D
To prove the lemma, cover G by open affines and reduce to the affine case.
Now suppose the group scheme G/k is locally of finite type, and G0 is the connected component
containing e.
Lemma 5.7.
1. G0 is an open and closed group subscheme.
2. G0 is geometrically irreducible and of finite type.
Sketch of proof.
1. Connected components are always closed, and they are open in the locally
noetherian case. Also
15
G0 × G0
G0
(5.6)
m
G×G
G
takes (e, e) to e and G0 × G0 is connected, so the map factors through G0 × G0 → G0 .
2. It is a fact that any connected space containing a rational point is geometrically connected.
To see that G0 is geometrically irreducible, base change to k and consider G0red , which is also
connected. We have
G0red × G0red
G0red
(5.7)
G0
G×G
so G0red is a group scheme over k, and is in fact smooth (it’s a fact that it has a nonsingular
point, and then translate). So G0red is connected and smooth, so is irreducible, implying that
G0 is irreducible.
m
Also for any U ⊆ G0red open, U × U −→ G0red is surjective.
Theorem 5.8. If char(k) = 0, then G0 is smooth (and therefore G is smooth).
Sketch. We can replace k with k and check that G0 is nonsingular. If suffices to check this at e.
Let dx1 , . . . , dxn form a basis for me /m2e , then δ1 . . . , δn ∈ Te G the dual basis gives rise to
D1 , . . . , Dn ∈ Lie(G).
bG,e be the complete local ring of G at e, and lift dxi to xi , giving a map
Let A = O
k[[x1 , . . . , xn ]] → A,
f →7
X 1
D(α) (f ).
α!
α
(It is an exercise to show that the two maps are inverses.) Here for α = (m1 , . . . , mn ),
1
m1 !···mn ! .
6
(5.8)
1
α!
=
The Picard Functor and Dual Variety
Let X/k be a projective variety and x ∈ X(k). Then let
PicX/k (T ) = {isomorphism classes of line bundles on X × T }/{those of the form p∗ L}
for p : X × T → T .
16
(6.1)
Lemma 6.1. PicX/k (T ) ∼
= {(L, α) : L a line bundle on X × T , α : L|{x}×T ∼
= OT }/ ∼.
Theorem 6.2.
• PicX/k is represented by a (group) scheme, locally of finite type over k.
• Pic0X/k is quasi-projective, and is projective if X is smooth.
As an example, we have
PicX/k (k) = {(L, α) : L a line bundle on X, α : L ∼
= k} = Hom(Spec k, PicX/k ).
(6.2)
Also PicX/k (T ) = Hom(T, PicX/k ) for any T .
In the case T = PicX/k , we have 1 ∈ Hom(PicX/k , PicX/k ), which gives us a universal pair
(Puniv , αuniv ) ∈ PicX/k (PicX/k ). Now for any (L, α) ∈ PicX/k (T ), the corresponding morphism
can be composed with 1, so if ϕ : T → PicX/k is the corresponding morphism, then (L, α) =
ϕ∗ (Puniv , αuniv ).
For any λ ∈ PicX/k (k 0 ), Puniv |X×{λ} is a line bundle on Xk0 , which represents isomorphism
classes of line bundles on Xk0 .
Let M, N be two line bundles on X. We say that M, N are algebraically equivalent if there
exist T1 , . . . , Tn connected schemes of finite type over k, {si , ti } geometric points of Ti , and Li line
bundles on X × Ti such that
L1 |X×{s1 } ∼
= Mk
Li |X×{t } ∼
= Li+1 |X×{s }
i
Ln |X×{tn }
i
∼
= Nk .
(6.3)
(6.4)
(6.5)
Lemma 6.3. Let L be a line bundle on X and λ ∈ PicX/k (k) be the corresponding point. Then
λ ∈ Pic0X/k (k) if and only if L and OX are algebraically equivalent.
Proof. Suppose {λ, 0} ∈ Pic0X/k (k). Then there exists a curve C containing λ and 0, contained in
(Pic0X/k )red , therefore PicX/k . C → PicX/k gives a line bundle P on X × C such that PX×{0} ∼
= OX
∼
and PX×{λ} = L. So L and OX are algebraically equivalent (and just one connecting bundle is
needed).
Conversely, suppose L ∼ OX algebraically. Choose T1 , . . . , Tn , {si , ti }, and Li according to the
definition. We can shrink the Ti if necessary, so that we can equip the Li with a trivialization αi
on {x} × Ti . We get (Li , αi ) : Ti → PicX/k . The image lies in a single connected component, so by
reverse induction, the image is always contained in Pic0X/k .
b = Pic0 . Then A
b is
Now consider X = A, an abelian variety. We have 0 ∈ A(k). Define A
A/k
b is smooth, so A
b is an abelian
a connected projective group scheme over k. We will prove that A
variety, called the dual abelian variety of A. But recall that we have also defined Pic0 A = ker φ :
Pic A → Hom(A(k), Pic(Ak )).
b
Theorem 6.4. Pic0 A = A(k).
17
b → A × A.
b Let
Proof. For (⊇), let P = Puniv |A×Abred . Then we have maps m, p1 , p2 : A × A × A
∗
∗
−1
∗
−1
M = m P ⊗ p1 P ⊗ p2 P . It suffices to show that M is trivial, which follows from the theorem
of the cube. For (⊆), we need:
Lemma 6.5. Let L ∈ Pic0 A and L =
6 OA . Then H i (A, L) = 0 for every i.
Proof. First, show that H 0 is zero. Suppose not. Then L = O(D) for effective D. Then
OX = (1 ⊗ (−1))∗ m∗ L = (1 ⊗ (−1))∗ (p∗1 L ⊗ p∗2 L) ∈ L ⊗ (−1)∗ L
0×1
(6.6)
m
so 0 = D + (−1)D, implying D = 0 and L = OA . For k > 0, consider A −−→ A × A −→ A and
pull back, then use Kunneth for A × A.
Theorem 6.6. Let L be an ample line bundle. Then the map φL : A(k) → Pic0 Ak given by
x 7→ Tx∗ L ⊗ L−1 is surjective.
Proof. We can assume k = k. Suppose M ∈ Pic0 A is not of the form Tx∗ L ⊗ L−1 . Then consider
N = m∗ L ⊗ p∗1 L−1 ⊗ p∗2 (L−1 ⊗ M−1 ) on A × A. We have spectral sequences
H i (A, Rj (p1 )∗ N ) =⇒ H • (A × A, N ), H i (A, Rj (p2 )∗ N ) =⇒ H • (A, N |{x}×A ).
(6.7)
To determine the beginning of the first spectral sequence, consider H • (A, N |{x}×A ). We have
N |{x}×A = Tx∗ L ⊗ L−1 ⊗ M−1 . This is nontrivial and in Pic0 , therefore H • (A, N |{x}×A ) = 0. By
applying a flat base change, Rj (p1 )∗ N = 0. We get H • (A × A, N ) = 0.
On the other hand, we can also use the second spectral sequence. Consider H • (A, N |A×{x} ).
N |A×{x} = Ty∗ L ⊗ L−1 is trivial at the points of K(L), which is finite for L ample. We get
H • (A, N |A×{y} ) = 0 for y ∈
/ K(L). We get Rj (p2 )∗ N |A\K(L) = 0, so that Rj (p2 )∗ N is a coherent sheaf on A, set-theoretically supported on K(L). Hence
M
H i (A, Rj (p2 )∗ N ) =
i,j
M
Γ(A, Rj (p2 )∗ N )
(6.8)
Γ(A, Rj (p2 )∗ N ).
(6.9)
j
and the spectral sequence degenerates. We obtain
0=
M
H j (A × A, N ) =
j
M
j
So Rj (p2 )∗ N = 0 for every j. Therefore, for every y, H • (A, N |A×{y} ) = 0, a contradiction for
y = e.
Fix Me Where does this proof end? It must end at or before here. (1)
b = Pic0 is smooth.
Theorem 6.7. A
A/k
b
Proof. φL : A(k) → A(k)
given by x 7→ Tx∗ L ⊗ L−1 can be extended to S-points. For any k-scheme
0
b
S and x : S → A, φL (x) = Tx∗ LS ⊗ L−1
S ; this is then represented by φL : A → PicA/k = A. But:
18
Corollary 6.8. If L is ample, then φL is surjective with finite kernel.
For ker φL = K(L).
b Therefore to prove that A
b is smooth, it’s enough to show that dim T0 A
b≤
So g = dim A = dim A.
g. We have
b = ker(A(Λ)
b
b
T0 A
→ A(k))
(6.10)
= ker(PicA/k (Λ) → PicA/k (k))
(6.11)
∼
= {(L, α) : L a line bundle on A × A, α : L|{0} −
→ O such that β ∗ L|A,→A×Λ ∼
= OA }.
(6.12)
×
Now isomorphism classes of line bundles on X is given by H 1 (X, OX
). We have an exact sequence
0
OA
f 7→1+f
×
OA×Λ
×
OA
1
(6.13)
×
×
) → 0.
) → H 1 (A, OA
0 → H 1 (A, OA ) → H 1 (A × Λ, OA×Λ
(6.14)
giving the exact
b = H 1 (A, OA ), so it remains to show that dim H 1 (A, OA ) ≤ g.
We obtain T0 A
L
Now for any complete variety X/k with H 0 (X, OX ) = k, consider HX = i H i (X, OX ). Then
HX is a graded commutative k-algebra. For the case X = A an abelian variety, HA is a commutative
∼
coalgebra by m∗ : HA → HA×A ∼
→ HA . We have
= HA ⊗ HA . And we also have s∗ = (−1)∗ : HA −
m∗ (h) = 1 ⊗ h + h ⊗ 1 +
i−1
X
h0j ⊗ h00i−j
(6.15)
j=1
for h ∈ H i .
and cocommutaClaim. (∆∗X , m∗ , s∗ ) makes HA a finite dimensional positive graded commutative
P
tive Hopf algebra over k, such that H 0 = k and m∗ (h) = 1 ⊗ h + h ⊗ 1 + h0 ⊗ h00 .
6.1
Hopf Algebras
Let H/k be a k-vector space, together with m : H ⊗ H → H, ∆ : H → H ⊗ H, s : H → H,
: H → k, and δ : k → H. We say that H is a Hopf algebra over k if:
• m, δ make H an algebra over k.
• ∆, make H a coalgebra over k.
• ∆ is an algebra homomorphism, and m is a coalgebra homomorphism.
19
• s is an anti-algebra and anti-coalgebra homomorphism with
Fix Me Is this correct? (2)
H ⊗H
1⊗s,s⊗1
H ⊗H
m
◦∆
k
δ
(6.16)
H
Lemma 6.9. G/k 7→ H = Γ(G, OG ) is an equivalence of categories from affine group schemes over
k to commutative Hopf algebras.
The inverse map is k[G] →7 G.
We can also consider graded Hopf algebras. Here H is a graded vector space and m, δ, ∆, , s preserve gradings. Graded commutative means that ab = (−1)deg a deg b ba, and graded cocommutative
is similar.
Examples of graded Hopf algebras:
1. For Ga = k[t], we have the Hopf algebra k[t], with ∆(t) = 1 ⊗ t + t ⊗ 1. If we put deg t = 2n,
then k[t] is graded commutative and cocommutative.
n
2. αpn = Spec k[t]/tp is done similarly, for p = char(k).
3. Spec k[t]/t2 , but this time will be graded commutative and cocommutative if deg t is odd. This
is denoted by Spec Λk [t].
4. If H1 , H2 are two graded commutative and cocommutative bialgebras, then so is H1 ⊗ H2 .
Theorem 6.10 (Borel). Let k be a perfect field and H a positive graded commutative and cocommutative k-bialgebra. If H0 = k (H is connected) and dim Hm < ∞, then H ∼
= H1 ⊗ · · · ⊗ Hn with
Hi given in (1)-(3).
Recall that HA =
L
H i (A, OA ) is a graded commutative and cocommutative k-bialgebra.
Corollary 6.11. dim H 1 (A, OA ) = g, and Λ• H 1 (A, OA ) → HA is an isomorphism.
Proof. We can assume that k = k.mWe know
that H n (A, OA ) = 0 for every n > g, so we must
p 1
p mn
•
have HA = Λ (V ) ⊗ k[t1 , . . . , tn ]/(t1 , . . . , tn ) with V a graded vector space concentrated in odd
degrees. Since Λtop V1 is 1-dimensional for V1 the 1-dimensional part of V , we have dim V1 ≤ g. (This
already shows that dim H 1 (A, OA ) ≤ g.)
b is smooth, dim V1 = dim H 1 (A, OA ) = g. Now suppose x ∈ Vd for d > 1 or x = ti .
Now that A
Then deg(x ⊗ Λtop V1 ) > g, a contradiction. So HA = Λ• (V1 ).
∼
Corollary 6.12. Λi H i (A, OA ) −
→ H i (A, OA ), so in particualar hi = gi .
b is smooth.
Corollary 6.13. A
20
6.2
Polarizations
b be as before. If L is ample, ker φL is finite, so φL is an
Let L be a line bundle and φL : A → A
isogeny. Also by the theorem of the square, φL⊗L0 = φL for L0 ∈ Pic0 A.
b such that λ⊗k = φL for some ample
A polarization of an abelian variety is an isogeny λ : A → A
line bundle L. λ is called a principal polarization if it is an isomorphism (equivalently, deg λ = 1).
Let X be a complete curve over k with x0 ∈ X(k). Then PicX/k is representable. Let J(X) = Pic0X/k ,
called the Jacobian of X. Then J(X)(k) = {L on Xk : deg L = 0}, and T0 J(X) = H 1 (X, OX ).
For every d, we have the Abel-Jacobi map AJ d : X d → J(X) by (x1 , . . . , xd ) 7→ O(x1 + · · · +
xd − dx0 ). Or, given xi : S → X, we have Γxi = graph(xi ) ,→ X × S a Cartier divisor, giving a line
bundle O(Γxi ) on X × S. We can then apply the same map on symmetric powers.
If d > 2g − 2, then the fibers of AJ d are projective spaces of degree d − g. In fact, (AJ d )−1 (L) =
P(Γ(X, L(dx0 ))).
In particular, dim J(X) ≥ g. As T0 J(X) = H 1 (X, OX ), we find that J(X) is smooth.
Theorem 6.14. J(X) admits a canonical principal polarization.
Let Θ be the scheme-theoretic image of AJ g−1 , which is {L ∈ J(X) : Γ(L(g − 1)x0 ) 6= 0}.
Claim. Θ is an ample divisor, and φΘ is a principal polarization.
As a special case, consider X = E an elliptic curve, and take x0 = 0. Then we claim that
[ is an isomorphism.
J(X) ∼
= E. For Θ = {O(x − x0 ) : Γ(x − x0 ) 6= 0} = {x0 }. So φΘ : J(E) → J(E)
Proof of claim. Let AJ 1 : X → J(X). Then f : (X, x0 ) → (Y, y0 ) induces f ∗ Pic0Y /k → Pic0X/k , so
\ → J(X). It is enough to show that τ ◦ φΘ = −1, for then φΘ has trivial
we get τ = (AJ 1 )∗ : J(X)
kernel, and Θ effective with φΘ of finite kernel will imply that Θ is ample.
Let S be a noetherian scheme, L a line bundle on X × S, and π : X × S → S. For s ∈ S, consider
Λtop (H 0 (X, L|X×s )) ⊗ Λtop (H 1 (X, L|X×s ))−1 ; this is a 1-dimensional space. Choose K 0 → K 1
locally free of finite rank, such that Ri π∗ L ∼
= H i (K • ). Let detL = Λtop K 0 ⊗ (Λtop K 1 )−1 . This is a
line bundle on S, and independent of the choice of (K 0 → K 1 ). detL is called the determinant line
bundle for L. Its formulation commutes with any base change.
p2
As an example, take S = X and X × X −→ X. detO(−∆) = OX and detO(∆) = ωx .
Let P be the universal bundle on X × J(X), and M = P ⊗ p∗1 O((g − 1)x0 ). Let Ma = M|X×a
and Pa = P|X×a for a ∈ J(X). Then Ma ∼
= Pa ⊗ O((g − 1)x0 ).
We have χ(Ma ) = 0. Consider K 0 → K 1 on J(X) representing Ri (p2 )∗ M. For every a,
dim K 0 |a − dim K 1 |a = χ(Ma ) = 0, so rank(K 0 ) = rank(K 1 ). We get that Λtop K 0 ,→ Λtop K 1
is injective, so Λtop K 0 ⊗ (Λtop K 1 )−1 → OJ(X) is injective.
To show injectivity of ψ : Λtop K 0 → Λtop K 1 , consider
21
{a ∈ J(X) : ψ|a = 0} = {a ∈ J(X) : Ka0 → Ka1 is not an isomorphism}
(6.18)
0
(6.19)
= {a ∈ J(X) : h (Ma ) > 0}
= {a ∈ J(X) : h (Pa ⊗ O((g − 1)x0 )) > 0}
= im(AJ
g−1
:X
(6.17)
0
g−1
→ J(X))
(6.20)
=Θ
(6.21)
of dimension at most g − 1. In particular, it cannot be all of J(X). So ψ is not identically zero,
therefore injective.
So now Λtop K 0 ⊗ (Λtop K 1 )−1 ,→ OJ(X) is injective. Its quotient has support Θ, so is OnΘ for
∼
some n ≥ 1. We get det−1
M = O(nΘ) for some n ≥ 1.
Claim. τ ◦ φdet−1 = −1 : J(X) → J(X). (Then we’re done, since φdet−1 = nφΘ . This also implies
M
M
n = 1, so det−1
M = O(Θ).)
The claim is equivalent to showing that τ (φdet−1 (a)) = −a for every a ∈ J(X). That is,
M
−1
∗
τ (Ta∗ (det−1
M ) ⊗ detM ) = −a, or Ta (detM ) ⊗ detM |AJ 1 (X) = −a.
Now Ta∗ detM |AJ 1 (X) = detM |AJ 1 (X)−a = det(M|AJ 1 (X)−a ) . And M|AJ 1 (X)−a = P ⊗ p∗1 O((g −
1)x0 )|AJ 1 −a , and P|X×AJ 1 (X) = O(∆ − ({x0 } × X)).
Let N = M|X×(AJ 1 (X)−a) = O(∆ + (g − 2({x0 } × X)) ⊗ p∗1 (Pa ). Now we just need:
Lemma 6.15. Let L be a line bundle on X, and J = O(∆) ⊗ p∗1 L. Then detJ = detO(∆) ⊗L.
The lemma implies the claim because then
Ta∗ detM |X×AJ 1 (X) = detN = detO(∆+(g−2)({x0 }×X)) ⊗Pa = detM |X×AJ 1 (X) ⊗ Pa .
(6.22)
Proof. By induction, it’s enough to prove that detO(∆)⊗p∗1 L(x) = detO(∆)⊗p∗1 L ⊗O(x). On X × X,
we have
0 → O(∆) ⊗ p∗1 L → O(∆) ⊗ p∗1 L(x) → (ιX )∗ O(X) → 0.
(6.23)
Now these are flat over the base, so we can take determinants.
7
Cartier Duality
Let G be a finite commutative group scheme over k and H = k[G] = Γ(G, OG ), so H is a finite
dimensional commutative and cocommutative Hopf algebra over k. Then let H ∗ = Homk (H, k)
with comultiplication m∗ and multiplication ∆∗ . H ∗ is also a finite dimensional commutative and
b = Spec H ∗ , called the Cartier dual to G, a finite commutative
cocommutative Hopf algebra. Let G
b is a duality functor: D2 = 1.
group scheme over k. Then D : G → G
Let G1 , G2 be two commutative group schemes over S. Define Hom(G1 , G2 ) : Sch/S → Ab by
(T → S) 7→ HomT −grp sch (G1 ×S T, G2 ×S T ).
22
b∼
Proposition 7.1. G
= Hom(G, Gm ).
∼
b
Proof. Let R be a k-algebra. We need to show that G(R)
= Hom(G ⊗k R, Gm ⊗k R). To show this,
first we have
∗
b
, R) = HR = H ⊗k R. (7.1)
G(R)
= Homk−alg (H ∗ , R) = HomR−alg (H ∗ ⊗k R, R) ⊆ Homk−linear (HR
| {z }
∗
HR
∗ , R). Then we have
Let ϕ ∈ Homk−linear (HR
∗
ϕ ∈ HomR−alg (HR
, R) ⇐⇒ ϕ(a · b) = ϕ(a) · ϕ(b), ϕ(1) = 1
⇐⇒
ϕ(∆∗R (a
⊗ b)) = ϕ(a) · ϕ(b)
(7.2)
(7.3)
⇐⇒ (∆R (ϕ))(a ⊗ b) = ϕ(a) · ϕ(b) = (ϕ ⊗ ϕ)(a ⊗ b)
(7.4)
⇐⇒ ∆R (ϕ) = ϕ ⊗ ϕ.
(7.5)
And ϕ(1) = 1 ⇐⇒ R (ϕ) = 1 (done similarly).
∗ , R) ⇐⇒ ∆ (ϕ) = ϕ ⊗ ϕ and (ϕ) = 1 if and only if ∆ (ϕ) = ϕ ⊗ ϕ and
So ϕ ∈ HomR−alg (HR
R
R
R
ϕ is invertible in HR . These ϕ are classified by HomR−Hopf (R[t, t−1 ], HR ) = Hom(GR , (Gm )R ).
As an example, for G = Z/nZ,
b
G(R)
= Hom(GR , (Gm )R ) = {f ∈ R× : f n = 1} = µn (R).
(7.6)
\ = µn and µ
So Z/nZ
cn = Z/nZ. It is an exercise to show that α
cp = αp .
b → A
b is also an
Theorem 7.2. Let f : A → B be an isogeny of abelian varieties. Then fb : B
b
[
isogeny, and ker f = ker f .
∼
b = {L on A : m∗ L =
Here is an informal discussion (which can be made rigorous): Recall that A
×
L L}. For L an invertible sheaf on X, L = Tot(L) is an affine scheme over X, then L = L \{0}
is a Gm -torsor. For f : X → Y and L on Y , Tot(f ∗ L)× = X ×Y L× .
b consists of the Gm torsors L× → A such that
So A
L× × L×
L×
(7.7)
m
A×A
A
In turn, this consists of the commutative group schemes L× with an exact sequence
1 → Gm → L× → A → 1.
(7.8)
This equals Ext1 (A, Gm ). Now for 0 → K → A → B → 0, since Hom(complete, affine) = 0, we
get
23
0 → Hom(K, Gm ) → Ext1 (B, Gm ) → Ext1 (A, Gm ) → 0
(7.9)
b →B
b→A
b → 0.
as Ext1 (K, Gm ) = 0 for K finite. We get 0 → K
Proof of theorem. ker fb(S) consists of (L, α) for L a line bundle on B × S and α : L|0×S ∼
= OS such
that f ∗ (L, α) ∼
= (OA×S , ). This is turn is the set of isomorphism classes of L on B × S such that
f ∗L ∼
= OA×S .
Now we’ll use the theory of fppf descent: if f : X0 → Y is faithfully flat and locally of finite
presentation, then for F ∈ QCoh(Y ), letting G = f ∗ F ∈ QCoh(X0 ), we have
X1 ×X0 X1
X2
∼
X0 ×Y X0 ×Y X0
X0 ×Y X0
X1
X0
Y.
(7.10)
For θ :
p∗1 G
∼
= p∗2 G, we have p∗23 (θ)p∗12 (θ) = p∗13 (θ). Now define
n
o
Desc(X0 , Y ) = (G, θ) : G ∈ QCoh(X0 ), θ : p∗1 G ∼
= p∗2 G : θ|∆ = 1, p∗12 (θ)p∗23 (θ) = p∗13 (θ) .
(7.11)
Then f ∗ : QCoh(Y ) → QCoh(X0 ) factors through the forgetful functor
Desc(X0 , Y ) → QCoh(X0 ),
(7.12)
giving f ∗ : QCoh(Y ) → Desc(X0 , Y ).
Theorem 7.3 (Grothendieck). If f is fppf, then f ∗ : QCoh(Y ) → Desc(X0 , Y ) is an equivalence.
Now let G = ker f , Y = B × S, and X0 = A × S, with map f × 1. Then X1 ∼
= A × S × G and
∼
X2 = A × S × G × G with the corresponding diagram
p1
A×S×G×G
1A×S ×m
p
A×S×G
A×S
f ×1
B × S.
(7.13)
m
p2
∼
Now ker fb(S) = {(OA×S , θ) ∈ Desc(A × S, B × S)}, where θ : p∗ OA×S −
→ m∗ OA×S is an
isomorphism OA×S×G → OA×S×G . So θ corresponds to f ∈ Γ(A × S × G, OA×S×G )× = Γ(S ×
G, OS×G )× . θ|∆ = 1 implies (f ) = 1 for : OG → k, and the cocycle condition implies ∆(f ) = f ⊗f .
But now
b
{f ∈ Γ(S × G, OS×G )× : (f ) = 1, ∆(f ) = f ⊗ f } = Hom(GS , (Gm )S ) = G(S).
b finite.)
(fb is an isogeny since G finite implies G
24
(7.14)
Corollary 7.4. If f is an isogeny, then deg f = deg fb.
Proof. deg f = dimk k[ker f ].
Let A, B be two abelian varieties over k of the same dimension. A line bundle Q on A × B is
b with
called a divisorial correspondence if Q|0×B ∼
= OA . Q induces κQ : B → A
= OB and Q|A×0 ∼
∼
b
κQ (0) = 0, so κQ is a homomorphism. Now for σ : B × A = A × B, we also get κσ∗ Q : A → B.
ˆ
b κP = 1 by definition of A.
b Let κ = κσ∗ P : A → Â.
For P the Poincaré line bundle on A × A,
Proposition 7.5. Let L be a line bundle on A. Then ϕL = ϕ
bL ◦ κ.
Proof. We have (1×ϕL )∗ P ∼
= m∗ L⊗p∗1 L−1 ⊗p∗2 L−1 by Seesaw. Then (1×ϕL )∗ P|{x}×A ∼
= Tx∗ L⊗L−1
implying ϕ
bL (κ(x)) = ϕL (x).
Corollary 7.6. κ is an isomorphism.
(Apply the proposition to L ample to get that κ is an isogeny, and deg ϕL = deg ϕ
bL implies
deg κ = 1.)
b That is, λ = λ
b ◦ κ.
b is called symmetric if λ = λ.
λ:A→A
Proposition 7.7. For f : A → B and L a line bundle on B, ϕL ◦ f = fb ◦ ϕf ∗ L .
8
Étale and Local Group Schemes
Let G be a finite group scheme over K. We say that G is local if G = G0 , and GQis étale if [G] is
an étale k-algebra. (A of finite type over k is étale if ΩA/k = 0. Equivalently, A = Li for Li finite
separable field extensions of k.)
Lemma 8.1. Let k s be the separable closure of k. Then the category of finite étale k-algebras is
equivalent to the category of finite Gal(k s /k)-sets.
Proof. Represent a k-algebra as a scheme X. Then Gal(k s /k) acts on X(k s ). And if Gal(k s /k) acts
on T , then consider
!Gal(ks /k)
Y
k
s
.
(8.1)
orbits of T
Corollary 8.2. The category of étale k-groups is equivalent to the category of finite groups with
Gal(k s /k)-action.
As an example, take µn for n not divisible by char(k). Then µn (k s ) ∼
= Z/nZ and is acted on
naturally by Gal(k s /k) as the nth roots of unity. We get a cyclotomic character χ : Gal(k s /k) →
(Z/nZ)× , defined by σ(ζ) = ζ χ(σ) where ζ is a primitive nth root of unity.
Let X be a k-scheme of finite type over k.
25
Proposition 8.3. There exists a finite étale k-scheme π0 (X), together with q : X → π0 (X) which
is universal in the following sense: if q 0 : X → Y with Y finite étale, the there exists a unique
f : π0 (X) → Y with q 0 = f ◦ q. In addition, q is faithfully flat, and the fibers of q are connected
components of X.
Sketch of proof. Define π0 (X) by π0 (X)(k s ) = π0 (Xks ) Gal(k s /k) and define q : X → π0 (X) by
Xks → Spec k s .
Corollary 8.4. If G is a k-group scheme of finite type, then π0 (G) is an étale k-group, and q is a
homomorphism.
Proposition 8.5.
1. Every finite k-group G fits into an exact sequence 1 → G0 → G →
π0 (G) → 1 for G0 local and π0 (G) étale.
2. If k is perfect, then this exact sequence splits canonically.
Sketch of (2). If k is perfect, then a fiber product of reduced schemes is reduced, so Gred is a kgroup. But a reduced k-group is smooth, therefore étale. Then claim that Gred ,→ G → π0 (G) is an
isomorphism, which can be checked over k s .
b is
For G a commutative finite group scheme, we say that G is étale-local if G is étale and G
local, and define étale-étale, local-étale, and local-local similarly.
Corollary 8.6. If k is perfect, then G = Gét,ét × Gét,loc × Gloc,ét × Gloc,loc .
Here are some basic examples: µn for p - n is étale-étale. Z/pZ is étale-local, µp is local-étale,
and αp is local-local.
If k = k, then Gét,ét is simply a finite abelian group with étale dual, so is a product of constant
cyclic group schemes of order not divisible by p. Similarly we have
Gét,loc ∼
=
Y
Z/pm Z,
Gloc,ét ∼
=
Y
µp m .
(8.2)
Also, if char(k) = 0, then G = Gét,ét .
A local group is said to be of height one if xp = 0 for every x ∈ m, the maximal ideal given by
e ∈ G.
p
p
Lemma 8.7. Let G be of height one. Then k[G] ∼
= k[x1 , . . . , xn ]/(x1 , . . . , xn ).
Proof. Let xi ∈ m form a basis of m/m2 . Then k[xi ]/(xpi ) → k[G] is surjective since k[G] is a local
ring and xpi = 0. Let Di be the left invariant vector Q
fields on G dual to the xi , so that Di (xj ) = δij
(mod m). Then there cannot be a relation between i xni i for 0 ≤ ni < p, otherwise taking di gives
a smaller relation unless all Di vanish.
Lemma 8.8. For F the Frobenius, F (1) : G → G(1) is a group homomorphism if G is a k-group of
finite type. And if GF = ker F (1) , then k[GF ] = OG,e /mpG,e . In particular, GF is of height one. Also
Lie(GF ) = Lie(G).
26
Suppose char(k) = p > 0. Then for F the Frobenius, we have
F
G
F (1)
G(1)
G
(8.3)
Spec k
Let GF
(1)
be ker F (1) . Then GF
(1)
Spec k
is local of height one, and GF
G(1)
Spec OG,e
F
(1)
,→ G factors through
Spec OG(1) ,e
(8.4)
F (1)
G
G(1)
(1)
so k[GF ] = OG,e /m(p) , for m(p) = {xp : x ∈ m}. This shows that GF
(1)
and Lie(GF ) = Lie(G).
(1)
is local of height one,
Theorem 8.9. G 7→ Lie(G) is a equivalence of categories between height one groups and p-Lie
algebras.
Sketch construction of inverse functor. Given g, form the universal enveloping algebra
U (g) =
M
g⊗n /(x ⊗ y − y ⊗ x − [x, y]).
(8.5)
n≥0
U (g) is a cocommutative Hopf algebra, with ∆(x) = 1 ⊗ x + x ⊗ 1 for every x ∈ g (this can be
extended to U (g)). We have that
{v ∈ U (g) : ∆(v) = 1 ⊗ v + v ⊗ 1} = g.
(8.6)
Now let u(g) = U (g)/(xp − x(p) , x ∈ g). Then u(g) is also a cocommutative Hopf algebra, and
turns out to be finite dimensional (this is clear if G is commutative). So u(g)∗ is a finite dimensional
commutative Hopf algebra; take G = Spec u(g)∗ , and then can show that G is of height one.
Corollary 8.10. If G is of height one and commutative, then pG : G → G is the zero map (G →
e
Spec k ,−
→ G).
For pG is a group homomorphism, and dpG is multiplication by p on Lie(G), therefore zero. As
G is of height one, we obtain pG = 0.
Corollary 8.11. Let G be local. Then there exists n such that nG : G → G is zero.
Proof. Iterate the Frobenius:
27
F (2)
G
F (1)
(1)
We have an ascending chain GF ,→ GF
for each k, we have the exact sequence
1 → GF
As p kills the right functor, p : GF
(k)
tion, pk kills GF .
F (1)
G(1)
(k)
(2)
···
(8.7)
,→ · · · . As G is local, G = GF
→ GF
(k+1)
G(2)
(k+1)
→ GF
→ (GF
(k+1)
(k)
)F
(n)
for some n. But
(1)
factors through GF
(8.8)
(k+1)
→ GF
(k)
. By induc-
Corollary 8.12. Let G be finite commutative. Then there exists n such that nG : G → G is zero.
For Gét is a finite group, killed by its order.
Corollary 8.13. Let f : A → B be an isogeny of abelian varieties, Thn there exists nA : A → A
and an isogeny g : B → A such that g ◦ f = nA .
Proof.
Theorem 8.14. Let F : U → V be fppf and X be a scheme. Then
!
Hom(V, X)
∼
eq Hom(U, X)
Hom(U ×V U, X)
(8.9)
Apply this theorem to f : A → B and X = A. We have A ×B ker f = A ×B A with two
projections to A. Then n : A → A factors through f if and only if the two projections become equal
after composing with n. We can take n killing ker f .
As an example, if char(k) = p > 0, F (1) : A → A(1) has AF ⊆ A[p], so there exists a unique
isogeny V : A(1) → A such that V ◦ F (1) = pA . We can then show that F (1) ◦ V = pA(1) . V is called
the Verschiebung.
Now consider A[n] for n = pm n1 , p - n1 . We have a natural morphism A[pm ] × A[n1 ] → A[n].
Claim. This is an isomorphism.
\
b b], since nc
Lemma 8.15. A[n
A ] = A[nA
A = nA
b.
∗
For nc
A (L) = nA L = L
∗
m L = L L.
n2 +n
2
⊗ (−1)∗ L
n2 −n
2
, which is Ln if (−1)∗ L = L−1 . This happens for
Proof of claim. A[n1 ] is étale-étale. And we have
A[pm ]k = (A[pm ]k )ét,loc × ( )loc,ét × ( )loc,loc
(8.10)
because over k, no nontrivial group of p-power order is étale-étale. So the claim is shown after
base changing to k.
28
We have A[n1 ] = (Z/n1 Z)2g with Gal(k s /k)-action (this will be easy to show). The étale-local
part over k must be a product of cyclic groups of p-power order. By and induction argument as
done earlier, it must be of the form (Z/pm Z)r for some r. Similarly, the local-étale part must be of
the form (µpm )s .
Claim. r is an invariant under isogeny, called the p-rank of A.
Let i = dim k[ker f ]. f : A → B induces f : A[pm ] → B[pm ]. We get pmrA ≤ i · pmrB for every
m, so rA ≤ rB . But there exists an isogeny g : B → A also, so rA = rB .
b we find that s = r. Also r + s ≤ 2g, so r ≤ g.
By taking the isogeny A → A,
9
The Tate Module
Suppose p - n. Then A[n](k) = A[n](k s ) = (Z/nZ)2g , with Gal(k s /k)-action. For n = `m with ` 6= p
prime, we can apply this, and then define the Tate module
T` (A) = lim A[`m ](k)
←−
(9.1)
m
`
with the inverse system given by A[`m ] ←
− A[`m+1 ]. Then T` (A) is a free Z` -module of rank 2g,
with a continuous action of Gal(k s /k). Any f : A → B induces T` (f ) : T` (A) → T` (B) which is a
continuous group homomorphism.
The definition of Tate module makes sense for any commutative group scheme. For example,
T` (Gm ) = lim(µ` ← µ`2 ← µ`3 ← · · · )
←−
(9.2)
which we call Z` (1). The corresponding action is the cyclotomic character χ : Gal(k s /k) → Z×
`
by σ(ξ) = ξ χ(σ) .
If M is a free Z` -module of finite rank with action of Gal(k s /k), then define
(
M ⊗Z` Z` (1)⊗n
M (n) =
M ⊗Z` (Z` (1)∨ )⊗−n
n≥0
n<0
(9.3)
b
Proposition 9.1. (T` (A))∨ (1) ∼
= T` (A).
m ] = A[`
\
b m ]. So
Proof. A[`
b m ](k s ) = Hom(A[`m ]ks , (Gm )ks ) = Hom(A[`m ]ks , (µ`m )ks ) = Hom(A[`m ](k s ), µ`m (k s )).
A[`
(9.4)
Then take inverse limit.
Proposition 9.2. Let f : A → B be an isogeny and N the kernel. Then we have a short exact
sequence of Z` [Gal(k s /k)]-modules given by
T` (f )
0 → T` (A) −−−→ T` (B) → N` (k s ) → 0
where N` (k s ) is the `-Sylow subgroup of N (k s ).
29
(9.5)
Proof. We have
T` (A) = lim A[`n ] = lim Hom(Z/`n Z, A(k)) = Hom(lim Z/`n Z, A(k)) = Hom(Q` /Z` , A(k)). (9.6)
−→
←−
←−
We begin with 0 → N (k) → A(k) → B(k) → 0, and obtain
0 → Hom(Q` /Z` , N (k)) → T` (A) → T` (B) → Ext1 (Q` /Z` , N (k)) → Ext1 (Q` /Z` , A(k)).
(9.7)
As A(k) is divisible, it is an injective abelian group, so Ext1 (Q` /Z` , A(k)) = 0. Also N (k) finite
implies Hom(Q` /Z` , N (k)) = 0. It remains to consider Ext1 (Q` /Z` , N (k)).
First write N = N` × N ` for N` the `-torsion and N ` the torsion relatively prime to `. We have
Ext1 (Q` /Z` , N ` (k)) = 0, so Ext1 (Q` /Z` , N (k)) = Ext1 (Q` /Z` , N` (k)).
Now 0 → Z` → Q` → Q` /Z` → 0 exact gives
0 = Hom(Q` , N` (k)) → Hom(Z` , N` (k)) → Ext1 (Q` /Z` , N` (k)) → Ext1 (Q` , N` (k)) = 0.
(9.8)
So Ext1 (Q` /Z` , N` (k)) ∼
= Hom(Z` , N` (k)) = N` (k), which equals N` (k s ) since N` is étale.
For ` = p, we can still define the p-adic Tate module Tp,ét (A) = lim(A[pm ](k)), a free Zp -module
←−
of rank r. (Here we must use k, not k s . Or use A[pm ]ét instead.)
Let S be a base scheme. A p-divisible group X is a directed system {Xn , ιn : n ≥ 0} where the
Xn are commutative, finite flat over S, X0 = S, and ιn : Xn → Xn+1 a closed embedding such that
p : Xn → Xn factors as ιn+1 ◦ πn , for πn : Xn → Xn−1 faithfully flat.
As an example, A[p∞ ]{A[pn ], ιn } for ιn the natural embedding.
p : X1 → X1 kills X1 , so rank(OX1 ) = ph for some h, called the height of X. For example, A[p∞ ]
has height 2g. By induction, rank(OXn ) = pnh .
10
The Isogeny Category
Let AV0k denote the category of abelian varieties up to isogeny: its objects are abelian varieties over
k, and its morphisms are given by Hom(A, B) ⊗ Q, sometimes written Hom0 (A, B).
Lemma 10.1. Let f be an isogeny. Then f is invertible in AV0k .
Because nA and nB are.
Theorem 10.2 (Poincaré complete reducibility). Let A ⊆ B be an abelian subvariety. Then there
exists C ⊆ B such that A × C → B is an isogeny.
Proof. Pick L ample on B, then we have
30
ι
A
B
ϕi∗ L
b
A
b
ι
(10.1)
ϕL
b
B
Let C = ker(b
ι ◦ ϕα )0red . Then the kernel of A × C → B is A ∩ C = ker ϕi∗ L ample. By a dimension
count, we see that A × C → B is an isogeny.
An abelian variety is called simple if it does not contain any abelian subvariety other than 0, A.
Corollary 10.3. Up to isogeny, any abelian variety is a product of simple varieties. In fact, the
decomposition is unique up to permutation and isogeny.
Remark. In fact, AV0k is a semisimple abelian category.
Remark. If A is simple, then End0 (A) = End(A) ⊗ Q is a division algebra over Q.
Corollary 10.4. For every A, End0 (A) is semisimple and expressible as
End0 (A) =
Y
Mni (Di ),
A∼
i
Y
Ani i ,
Di = End0 (Ai ).
(10.2)
i
Let E be a field, and V a vector space over E. A function f : V → E is called (homogeneous) of
degree n if the restriction of f to any finite-dimensional subspace is a polynomial function of degree
n. Equivalently, for every v1 , v2 ∈ V , f (λ1 v1 + λ2 v2 ) is homogeneous of degree n in λ1 , λ2 .
Define deg : End(A) → Z by deg f if f is an isogeny, 0 otherwise.
Theorem 10.5. There is a unique way to extend deg to a homogeneous polynomial of degree 2g,
deg : End0 (A) → Q.
Proof. Pick an ample line bundle L on A such that χ(L) 6= 0. Then:
Lemma 10.6. deg f =
χ(f ∗ L)
χ(L) .
GIven the lemma, deg : End(A) → Z is homogenous of degree 2g. That is, deg(nf ) = deg(nA ) deg(f ) =
deg f . Then there is a unique way to extend this to deg : End(A) ⊗ Q → Q which is homogeneous of degree 2g. So to show it is a polynomial function, it is enough to show that for every
∗
1 +f2 ) L)
f1 , f2 ∈ End(A), deg(nf1 + f2 ) is a polynomial in n. By the lemma, deg(nf1 + f2 ) = χ((nfχ(L)
so we can show this for χ((nf1 + f2 )∗ L). But recall that for f, g, h : X → A and L on A,
n2g
(f + g + h)∗ L ∼
= (f + g)∗ L ⊗ (f + h)∗ L ⊗ (g + h)∗ L ⊗ f ∗ L−1 ⊗ g ∗ L−1 ⊗ h∗ L−1 .
(10.3)
Let L(n) = (nf1 + f2 )∗ L. Then
−1
∗
∗
⊗−2
L(n+2) ∼
= L⊗2
(n+1) ⊗ (2f1 ) L ⊗ L(n) ⊗ (f1 L)
−1
∗
∗
⊗−2
∼
=⇒ L(n+2) ⊗ L−1
).
(n+1) = (L(n+1) ⊗ L(n) ) ⊗ ((2f1 ) L ⊗ (f1 L)
31
(10.4)
(10.5)
Letting M = (2f1 )∗ L ⊗ (f1∗ L)⊗−2 , we have that for some Q,
−1
⊗(n−1)
∼
=⇒ L(n) = M⊗
L(n) ⊗ L−1
(n−1) = (L(1) ⊗ L(0) ) ⊗M
{z
}
|
n(n−1)
2
⊗ N ⊗n ⊗ Q.
(10.6)
N
Hence χ(L(n) ) is a polynomial in n.
Proof of lemma. First consider the case where f is an isogeny. Let G be a group scheme over k
of finite type, X a scheme over k of finite type, equipped with the trivial action. A G-torsor P
over X is a scheme P with a right G-action together with a G-equivariant π : P → X such that
P ×k G → P ×X P is an isomorphism. (A G-torsor is also called a principal G-bundle.)
f
As an example, if A −
→ B is an isogeny, then it is an N -torsor of N = ker f .
Theorem 10.7. Let G be finite and π : P → X be a G-torsor, with X proper. Then for any coherent
sheaf F, χ(π ∗ F) = deg(π)χ(F).
Here π will be finite, so deg π = dim k[G]. This will prove the isogeny case of the lemma.
Proof. By noetherian induction, we can assume the theorem holds for dimension less than n = dim X
and that X is integral. Let r be the generic rank of F. We can assume that r > 0, otherwise F is
i
supported on a closed proper subset of X. Then there exists U ⊆ X open with F|U ←
− OUr . We
∼
OUr
(i,1)
OUr .
r
OX
r
OX
then have
,−−→ F|U ⊕
As F ⊕
is coherent, we can extend to E ,→ F ⊕
for some
r
coherent E. Now we have 0 → ker → E → F → coker → 0 and 0 → ker → E → OX → coker → 0,
with both kernels and cokernels supported in dimension less than n. So if the theorem holds for any
r , and then for all F.
F, then it holds for OX
Take F = π∗ OP . We want χ(π ∗ π∗ OP ) = deg(π)χ(π∗ OP ). We have π ∗ π∗ OP = OP ⊗ k[G]. So
both sides equal χ(OP ) deg π (the right hand side does since π∗ is flat).
Now to finish the proof of the lemma, it suffices to prove that if f is not an isogeny, then
χ(f ∗ L) = 0.
Proposition 10.8. Let L be a line bundle on an abelian variety A. Then χ(Ln ) is a homogeneous
polynomial of degree g.
g
(Therefore the polynomial must equal dL ng! . We can assume that k = k, as this statement is
independent of base change.)
g
Assuming the proposition, χ(f ∗ Ln ) = df ∗ L ng! . But also χ(f ∗ Ln ) = χ(Rf∗ OA ⊗ Ln ) from the
spectral sequence H q (Rp f∗ f ∗ Ln ) =⇒ H p+q (f ∗ Ln ) and the projection formula. As Rf∗ OA is
supported on B, χ(Rf∗ OA ⊗ Ln ) is a polynomial of degree less than g. Therefore df ∗ L = 0, so
χ(f ∗ Ln ) = 0.
Proof of proposition. Let p(n) = χ(Ln ). Then we want p(m2 n) = m2g p(n), or
2
χ(Lm n ) = m2g χ(Ln ).
32
(10.7)
2
If L is symmetric (L = (−1)∗ L), then we know that m∗ Ln = Lm n , so that
2
χ(Lm n ) = χ(m∗ Ln ) = deg(mA )χ(Ln ) = m2g χ(Ln ).
(10.8)
If L ∈ Pic0 A,
then L is algebraically equivalent to OA , so χ(Ln ) = χ(OA ), which equals 0 since
hi (A, OA ) = gi . Finally, for general L, L is of the form L1 ⊗L2 with L1 symmetric and L2 ∈ Pic0 A.
Then an L is algebraically equivalent to L1 , we have
2
2
n
χ(Lm n ) = χ(Lm
) = m2g χ(Ln1 ) = m2g χ(Ln ).
1
(10.9)
Lemma 10.9. If k = k, then every L can be expressed in the form L1 ⊗ L2 , for L1 symmetric and
L2 ∈ Pic0 A.
−1
∗ −1
Proof. It’s enough to find L2 ∈ Pic0 A such that L2 ⊗ (−1)∗ L−1
2 = L ⊗ (−1) L , for then L ⊗ L2
0
∗
−1
would be symmetric. If L ⊗ (−1) L
∈ Pic A, then we can take L2 to be a square root of
0
∗
−1
L ⊗ (−1) L , as k = k implies Pic A divisible. So now it sufices to prove that L ⊗ (−1)∗ L−1 ∈
Pic0 A.
We want to show that for each x, Tx L⊗Tx (−1)∗ L−1 = L⊗(−1)∗ L−1 , or equivalently Tx L⊗L−1 ∼
=
(−1)∗ (T−x L ⊗ L−1 ). We have T−x L ⊗ L−1 ∈ Pic0 A, so (−1)∗ (T−x L ⊗ L−1 ) = T−x L−1 ⊗ L. Then
(again using that it’s in Pic0 A), it’s invariant by translation by x, implying the result.
g
Remark. We now know that χ(Ln ) = dL ng! . If L is very ample, then L = i∗ O(1) for some i : A ,→ Pn ,
so dL = deg A (as embedded in Pn ), equal to Dg , the self-intersection number for D a hyperplane
g
section. So if L = O(D) is very ample, χ(L) = Dg! .
In general, for every L, we can associate a Chern class c1 (L) ∈ CH 1 (A). (If A is defined over
C, we can take c1 (L) ∈ H 2 (A; Z).) Then c1 (L)g ∈ CH g (A) (or H 2g (A; Z)). This is equipped with a
g
degree map to Z. We will see the general Riemann-Roch formula: χ(L) = c1 (L)
g! .
11
The Endomorphism Algebra
Theorem 11.1. Let A, B be two abelian varieties over k. Then
Hom(A, B) ⊗ Z` → Hom(T` (A), T` (B))
(11.1)
is injective. In particular, Hom(A, B) is a free Z-module of finite rank.
Proof. Let
Q
Ai → A, B →
Q
Bi be isogenies with the Ai and Bi simple. Then we have
Hom(A, B) ⊗ Z`
Hom(T` (A), T` (B))
(11.2)
Q
i,j
Q
Hom(Ai , Bj ) ⊗ Z`
33
i,j
Hom(T` (Ai ), T` (Bj ))
so it’s enough to show that the bottom row is injective. So we can assume that A and B are
simple. If A is not isogenous to B, then we’re done. Otherwise, let B → A be an isogeny, then
Hom(A, B) ,→ End(A) so we can also assume that A = B.
To show that End(A) ⊗ Z` → End(T` (A)) is injective, it’s enough to show that for any finitely
generated M ⊆ End(A), M ⊗ Z` → End(T` (A)) is injective.
Claim. Let QM = {f ∈ End(A) : ∃ n such that nf ∈ M }. Then QM is finitely generated.
Proof. QM = (M ⊗ Q) ∩ End(A) inside End0 A, and M ⊗ Q is a finite-dimensional vector space
over Q. We have deg : M ⊗ Q → Q a polynomial with deg(f ) ∈ Z \ {0} for any f ∈ End(A) \ {0} by
simplicity. Now 0 = {v : | deg v| < 1} so QM is discrete in M ⊗ R, therefore finitely generated.
Because ofPthe claim, we can assume that M = QM . Let f1 , . . . , fr be a Z-basis of M . Suppose
that we had i ai T` (fi ) = 0 for ai ∈ Z` not all zero. We can arrange for not all of the ai to be
0
divisible by `, since End(T` (A)) is free, in particular
a0i ≡ ai
P 0has no `-torsion. Chosse ai ∈ Z with
0
(mod `). Then T` (F )(T` (A)) ⊆ `T` (A) for f = i ai fi . So A[`] ⊆ ker f =⇒ f = `f for some
f 0 : A → A and f 0 ∈ QM = M . But then every a0i will be divisible by `, a contradiction.
φ
b so N S(A) is a free abelian group of rank
Corollary 11.2. N S(A) = Pic A/ Pic0 A ,−
→ Hom(A, A),
2
at most 4g . The rank is called the base number of A and denoted by ρ.
Corollary 11.3. End0 A is a finite dimensional Q-algebra. Hence
End0 A ∼
= Mn1 (D1 ) × · · · × Mnr (Dr )
(11.3)
where the Di are finite dimensional division algebras over Q.
Let D be a finite dimensional division algebra over Q, K the center of D, so D is a central
division algebra over K. Then D ⊗K K is a central simple algebra over K, so of the form Md (K)
for some d.
Let B be a finite dimensional simple algebra over Q. A function N : B → Q is called a norm
form if N is a polynomial and N (ab) = N (a)N (b). A function T : B → Q is called a trace form if
T is linear and T (ab) = T (ba).
Proposition 11.4. Let B be a finite dimensional simple algebra over Q, and K ⊆ B be the center.
0
0
Then there exists a unique norm form NB/K
: B → K and trace form TB/K
: B → K such that any
0
norm form N : B → Q is of the form N = (NKQ NB/K
)i and any trace form T : B → Q is of the
0
0
form T = ϕ ◦ tr0B/K for some linear ϕ : K → Q, and TB/K
(1) = d, where [B : K] = d2 . NB/K
is
0
called the reduced norm form, and TB/K is called the reduced trace.
Proof. If K = Q, base changing to K gives a norm form on B ⊗K K ∼
= Md×d (K), which must be
a power of the determinant in order to be multiplicative. The N ⊗ K = (det)i . This descends to
K because of the Gal(K/K)-action. Also T ⊗ K : Md (K) → K must kill commutators, so factors
through tr : Md×d (K) → K.
In general, B ⊗Q Q = (B ⊗K (K ⊗Q Q)), and
K ⊗Q Q ∼
=
Y
i:K,→Q
34
K
(11.4)
so
B ⊗Q Q ∼
=
Y
Md (K).
(11.5)
(det)ni
(11.6)
i:K,→Q
Now
N ⊗Q=
Y
i:K,→Q
and must be invariant under the Galois action, so all ni are equal. The argument for trace is
similar.
Theorem 11.5. For V` (A) = T` (A) ⊗ Q` , we have
End0 (A) ⊗ Q`
degQ`
Q`
(11.7)
det
End(V` (A))
Q`
As a consequence, if φ ∈ End(A), then P (n) = deg(n − φ) equals charT` φ (n).
Proof. Observe that both degQ` and det are norm forms on End0 (A)⊗Q` . Also, for any f ∈ End(A),
| deg f |` = | det(T` (F ))|` .
Q
Q
These imply the theorem: Write A ∼ Ani i , then End0 (A) = i Mni (Di ). Now we have
deg =
Y
0
(NC(Di )/Q` ND
)vi
i /C(Di )
(11.8)
i
det =
Y
0
(NC(Di )/Q` ND
)wi
i /C(Di )
(11.9)
i
As End0 (A) is dense in End0 (A) ⊗ Q` , the condition on absolute values implies vi = wi .
To show equality in norms, we can assume that φ ∈ End(A). If φ is not an isogeny, then T` ⊗Q is
not an isomorphism, so det(T` φ) = 0 and we’re done. If φ is an isogeny, we have the exact sequence
T φ
`
0 → T` A −−
→ T` A → (ker φ)` → 0.
So | deg φ|` =
1
#(ker φ)`
(11.10)
= | det(T` φ)|` .
Corollary 11.6. If φ ∈ End(A) and P (n) = deg(n − φ), then P (x) ∈ Z[x] and is monic.
Proof. P is integral-valued, therefore in Q[x]. And also charT` φ (x) ∈ Z` [x]. A priori this just implies
p(x) ∈ Z[ p1 ][x].
Since End(A) is finite over Z, there exists q(x) ∈ Z[x] such that q(φ) = 0. Then q(T` φ) = 0.
The roots of q(x) are algebraic integers, therefore the roots of p(x) are algebraic integers. Since
p(x) ∈ Q[x] is monic, we get p(x) ∈ Z[x].
35
We call p(x) = x2g +a1 x2g−1 +· · ·+a2g the characteristic polynomial of φ, a2g = NQ(φ) = deg(φ),
and −a1 = tr φ. If we write End0 (A) ∼
= B1 × · · · × Br with Bi simple, and deg φ = NB0 i /Q (φ)mi ,
P
then tr φ = mi TB0 i /Q (φi ) by expanding p.
We know that B = End0 A → Q is a polynomial of degree 2g. If A is simple, then we have
0
[End0 (A) : K] = d2 and [K : Q] = e. NB/q
is a polynomial of degree de, so we find:
Corollary 11.7. If A is simple, then de|2g.
Proposition 11.8. If char(k) = 0 and A is geometrically simple, then dim(End0 (A)) = d2 e|2g.
Proof. We can assume that k = C. Let A = V /L; then End0 (A) acts faithfully on LQ . We have
2g
= dimEnd0 A LQ ∈ Z.
d2 e
An abelian variety over k is said to be of CM type if there exists F ,→ End0 A with [F : Q] = 2g.
If A is simple and of CM type, then any maximal subfield in the division algebra End0 A has degree
de over Q, so de must Q
equal 2g. In general,Qif A is of CM type, then A ∼ An1 1 with A1 simple of
CM type. For if A ∼ i Ani i , then F ,→
Mni (End0 (Ai )) so F ,→ Mni (End0 (Ai )) for some i.
Any maximal subfield of Mni (End0 (Ai )) is of degree ni di ei over Q, so 2g ≤ ni di ei ≤ ni · 2gi ≤ 2g.
Equality holds, so A ∼ Ani i , and Ai is of CM type.
Ai s potentially of CM type if Ak is of CM type. If char(k) = 0 is A is potentially of CM type,
then Ak ∼ An1 1 for A1 simple and of CM type, then End0 (Ak ) = Mn1 (D) for D = End0 (A1 ). This
acts faithfully on Ln1 , so n1 d21 e1 |2n1 g1 = 2g, but also 2g ≤ n1 d1 e1 , so d1 = 1 and n1 e1 = 2g. Hence
End0 (Ak ) = Mn (K) for K a field of degree 2g
n over Q.
Let A be an abelian variety of dimension g. We say that A is of CM type if End0 (A) contains a
commutative semisimple algebra B of dimension 2g over Q.
Proposition 11.9. If A/k with A simple and char(k) = 0, then dim End0 (A) = d2 e|2g.
Proof. We can assume k = C. Then End0 (A) ,→ End0 (Ak ) LQ , for AC ∼
= V /L.
If char(k) = 0 and A is simple of CM type, then End0 (A) is a field of degree 2g over Q. If A is
CM by a field F , then A ∼ An1 1 for some A1 simple and of CM type.
∼ [
n ] = Hom(A[`n ], µ n ) since ker `n =
b ∼
b n] −
Recall that T` A
→ A[`
= (T` A)∨ (1). To see this, A[`
`
b
A
n
n
n
n
c
b
\
ker ` = ker ` . We obtain a pairing e`n : A[` ] × A[` ] → µ`n .
A
A
To take inverse limit, we need to show
b n]
A[`n ] × A[`
e`n
`×`
`
b n+1 ]
A[`n+1 ] × A[`
b → µn is defined as follows:
en : A[n] × A[n]
∗
∼
φ L = OA }. Define ev : ker φ × ker φb → Gm
∼
β : φ∗ L −
→ OA , then Tx∗ β : Tx∗ φ∗ L → Tx∗ OA
φ∗ L → OA . Then define ev(x, L) = Tx∗ β ◦ β −1 .
µ`n
e`n+1
(11.11)
µ`n+1
∼
b
[φ ←
ker
− ker φb was defined by ker φ(k)
= {L on B :
b
by, for x ∈ ker(φ)(k) and L ∈ ker(φ)(k),
picking
∗
∗
∗
∗
∗
= OA . But Tx φ L = φ Tφ(x) L = φ L. Let Tx∗ β :
36
b
Now let L ∈ A[m](k)
and x ∈ A[nm](k).
Claim. enm (x, L) = em (nx, L).
∼
∼
To prove the claim, choose β : m∗ L −
→ OA so n∗ β : n∗ m∗ L −
→ n ∗ OA ∼
= OA . Then
enm (x, L) = Tx∗ (n∗ β) ◦ (n∗ β)−1 = n∗ (Tnx β ◦ β −1 ) = n∗ em (nx, L) = em (nx, L).
(11.12)
b
By the same argument, if φ : A → B, then e`∞ (T` φ(x), y) = e`∞ (x, T` φ(y))
for x ∈ T` A and
b
y ∈ T` B.
In the case that L = O(D) for some Cartier divisor D, or equivalently given i : L ,→ κA the
n∗ i
β
sheaf of rational functions, then OA ←
− n∗ L ,−−→ n∗ κA ∼
= κA . Then g = n∗ i ◦ β −1 (1) is a rational
∼
function. Then
(g −1 )
=
n−1 D.
In this case, we have en (x, L) =
1×T φ
e
Tx∗ g
g
=
g(z+x)
g(z)
for any z ∈ A.
∞
` L
`
b −−
−→ T` A × T` A
→ Z` (1). Then E L is skew-symmetric.
Theorem 11.10. Let E L : T` A × T` A −−−−
Proof. It’s enough to show that e`∞ (x, φL (x)) = 1. Let L = O(D). Then φL (x) = Tx∗ L ⊗ L−1 =
O(Tx D − D). Let (g −1 ) = n−1 (Tx D − D). Then we want Tx∗ g = g. It’s enough to show that there
exists z such that g(z + x) = g(z).
∗ E) = T
Write x = ny. Then (g −1 ) = T−y (n−1 D) − n−1 D = T−y E − E. So (Tjy
−(j+1)y E − T−jy E.
We get

n−1
Y

j=0
So h(z) =
Qn−1
j=0

∗ −1 
Tjy
g
=
n−1
X
(T−(j+1)y E − T−jy E) = T−x E − E = 0.
(11.13)
j=0
g −1 (z + jy) is constant. In particular, h(z + y) = h(z), so g(z + x) = g(z).
Corollary 11.11. Let λ be a polarization. Then E λ (x, y) = e`∞ (x, T` λ(y)) is symplectic (skewsymmetric and nondegenerate).
b We have T` (A × A)
b = T` A × T` A.
b
Proposition 11.12. Let P be the Poincaré line bundle on A × A.
P
Then E ((x, x
b), (y, yb)) = e`∞ (x, yb) − e`∞ (y, x
b).
Proof. It’s enough to check for E P ((x, 0), (y, 0)) and E P ((x, 0), (0, yb)). The first is zero (also E P ((0, y), (0, yb)) =
1×0
b we have
0. To do this, for i : A ,−−→ A × A,
E P ((x, 0), (y, 0)) = E P (T` i(x), T` i(y)) = E i
∗P
(x, y) = 0
(11.14)
as i∗ P = OA .
∼
\b
b → A
b×B
b via
\
For E P ((x, 0), (0, yb)), we have ϕP : A × A
× A. We first claim A
×B −
→ A
\b ∼ b bb ∼ b
L 7→ (L|A×0 , L|0×B ) (by Seesaw). So then A × A
= A × A = A × A. We claim that under this
∗
−1 ; when restricting to A × 0, we get
isomorphism, ϕP (x, x
b) = (b
x, x). For ϕP (x, x
b) = T(x,b
x) P ⊗ P
P|A×{x} . Therefore, E P ((x, 0), (0, yb)) = e`∞ ((x, 0), (b
y , 0)) = e`∞ (x, yb).
37
b
Theorem 11.13. The following are equivalent, for φ : A → A:
1. φ is symmetric.
2. E φ is skew-symmetric.
3. 2φ = φL for some L.
4. Over k, φ = φL0 for some L0 .
c ), N S(A ) is identified with the
Corollary 11.14. Under φ : Pic(Ak ) → N S(Ak ) ,→ Hom(Ak , A
k
k
symmetric homomorphisms.
Proof. We’ve seen (4) =⇒ (1) and (2). For (3) =⇒ (4): let N = L⊗2 . Then 4φ = φN . In
particular, A[4] ⊆ ker φN . We want N to be A[2]-equivariant. For then by faithfully flat descent on
m
A × A[2]
A
p
2A
A
(11.15)
we have N = 2∗A L0 =⇒ 4φ = φ2∗A L = 4φL0 =⇒ φ = φL0 .
Now we have equivariance if and only if there exists αx : Tx∗ N ∼
= N such that for all x ∈ A[2]
and every y,
Ty∗ Tx∗ N
Ty∗ (αx )
αy
∼
∗ N
Tx+y
Ty∗ N
αx+y
(11.16)
N
This will be shown later.
(2) =⇒ (3): Take L = (1 × φ)∗ P. Then
E L (x, y) = E P (T` (1 × φ)x, T` (1 × φ)y)
P
(11.17)
= E ((x, T` φ(x)), (y, TL (y)))
(11.18)
= e`∞ (x, T` φ(y)) − e`∞ (y, T` φ(x))
(11.19)
φ
φ
= E (x, y) − E (y, x),
(11.20)
which by skew-symmetry is 2E φ (x, y). We conclude that φL = 2φ.
(1) =⇒ (3): Again take L = (1 × φ)∗ P. Then φL = φ(1×φ)∗ P = 1[
× φ ◦ φP ◦ (1 × φ). Applying
[
[
this to x, the right hand side is 1 × φ ◦ φP (x, φ(x)) = 1 × φ(φ(x), x). It remains to show that
b
b
1[
× φ(b
x, x) = (b
x, φ(x)),
which is clear. So we end up with (φ(x), φ(x)),
which by symmetry is
2φ(x).
38
12
The Rosati Involution
b Then we get a map
On End0 A, we have a Rosati involution. Pick a polarization λ : A → A.
0 : End0 A → End0 A by φ0 = λ−1 ◦ φ
b ◦ λ. We easily see that (φ + ψ)0 = φ0 + ψ 0 , (φψ)0 = ψ 0 φ0 , φ00 = φ.
So 0 is an anti-involution of End0 A. Given two polarizations λ1 , λ2 , we can write λ2 = λ1 ◦a for some
a ∈ End0 A. Then φ(λ2 ) = aφ(λ1 ) a−1 , so there is a well-defined conjugacy class of anti-involutions.
T
`
Claim. E λ (T` φx, y) = E λ (x, T` φ0 y). In other words, (End0 A,0 ) −→
(End(V` (A)), ?) for ? the canonλ
ical anti-involution induced by E .
Theorem 12.1. The Rosati involution is positive. That is, let φ ∈ End0 (A), φ 6= 0. Then tr(φφ0 ) >
0.
It’s enough to show the theorem for φ ∈ End(A).
Proof. We can assume that λ = ϕL for L very ample, by taking an appropriate multiple of λ. Let
L = O(H) for some hyperplane section H.
Claim. tr(φφ0 ) =
2g
g−1
H g (H
· φ−1 (H)).
The claim implies the theorem. To show the claim, first,
deg(ϕφ∗ L−1 ⊗Ln ) = deg(nϕL − ϕφ∗ L )
b L φ)
= deg(ϕL · n − φϕ
(12.1)
(12.2)
b L φ)
= deg(ϕLn − ϕL · ϕ−1
φϕ
| L {z }
(12.3)
= deg(ϕL ) deg(n − φ φ)
(12.4)
= deg(ϕL )p(n).
(12.5)
φ0
0
Theorem 12.2. For any L nondegenerate (i.e. such that K(L) is finite), deg(ϕL ) = χ(L)2 .
Given this theorem, p(n) =
Hg
g!
χ(φ∗ L−1 ⊗Ln )2
.
χ(L)2
But by Riemann-Roch, χ(L) =
c1 (L)g
g! ,
which equals
if L is very ample. We get
p(n) =
(H g )2 n2g − 2g(H g )(H g−1 · φ−1 (H)n2g−1 + · · ·
[(nH − φ−1 (H))g ]2
=
[H g ]2
(H g )2
implying tr(φ0 φ) =
2g
g−1
H g (H
(12.6)
· φ−1 (H)).
We’ll prove a weaker version of the above theorem, namely that
deg(ϕφ∗ L−1 ⊗Ln )
χ(φ∗ L−1 ⊗ Ln )2
=
.
deg ϕL
χ(L)2
(12.7)
To prove this, let M(L) = (1 × ϕL )∗ P = m∗ L ⊗ p∗1 L−1 ⊗ p∗2 L−1 . Then χ(M(L)) = deg(ϕL )χ(P).
But also Ri (p1 )∗ (M(L)) is supported on K(L). Now H i (A × A, M(L)) = Γ(A, Ri (p1 )∗ (M(L)) as
K(L) finite implies the Leray spectral sequence degenerates. But also, by the projection formula,
39
Ri (p1 )∗ (M(L)) = Ri (p1 )∗ (m∗ L ⊗ p∗2 L−1 ⊗ p∗1 L−1 ) = Ri (p1 )∗ (m∗ L ⊗ p∗2 L−1 ) ⊗ L−1 .
(12.8)
We get H i (A × A, M(L)) = Γ(A, Ri (p1 )∗ (m∗ L ⊗ p2 L−1 )), so χ(M(L)) = χ(m∗ L ⊗ p∗2 L). But
m×p2
A × A −−−→ A × A is an isomorphism, and m∗ L ⊗ p∗2 L = (m × p2 )∗ (L L−1 ). So
χ(m∗ L ⊗ p∗2 L−1 ) = χ(L L−1 ) = χ(L)χ(L−1 ) = (−1)g χ(L)2 .
(12.9)
We end up with deg ϕL χ(P) = (−1)g χ(L)2 .
This is all we need for now. The get the full theorem, we also need to show that χ(P) = (−1)g .
If A is simple, D = End0 A is a finite dimensional division algebra with anti-involution 0 : D → D,
such that tr0D/Q is positive. These algebras have been classified by Albert.
Lemma 12.3. Let D be an Albert division algebra, K the center of D, and K0 = {a ∈ K : a0 = a}.
Then K0 is totally real, and either K = K0 or K is a totally imaginary quadratic extension of K0 .
Proof. Let {σi : K0 ,→ R : i = 1, . . . , r1 } and {σr1 +j : K0 ,→ C : j = 1, . . . , r2 } run over the real and
∼
complex embeddings of K0 , so [K0 : Q] = r1 + 2r2 and K0 ⊗Q R −
→ Rr1 × Cr2 via the σi . We know
2
0
that q(x) = trK0 /Q (x ) = trK0 /Q (xx ) > 0. So qR : K0 ⊗ R is positive semidefinite by continuity.
Also q is nondegenerate, so qR is nondegenerate. So qR is positive definite. This implies r2 must be
0. So K0 is totally real.
√
If K = K0 , we’re done. Otherwise, K =QK( α) for some α ∈ K0 . We want to show that
σi (α) < 0 for every i. We have K ⊗Q R = σi (K ⊗K0 R). Now K ⊗K0 R ∼
= C or R × R. But
0
0
trK/Q (xx ) > 0 excludes the case of R × R and having switch factors. So K ⊗K0 R ∼
= C implying
σi (α) < 0.
Let e0 = [K0 : Q], e = [K : Q] (so e equals e0 or 2e0 ), and d2 = [D : K]. Recall that K0 is
totally real, and either K = K0 or K is an imaginary quadratic extension of K0 .
12.1
Albert’s Classification of Division Algebras With Involution
Type I: D = K = K0 .
Type II: K = K0 , D is a quaternion algebra over K, x0 = x∗ = tr0D/K x − x the standard
Q
involution, and D ⊗Q R ∼
= i:K,→R H.
a quaternion algebra over K. There is an isomorphism D ⊗Q R ∼
=
Q Type III: K = K0 , D
0
T , . . . , AT ).
M
(R)
such
that
corresponds
to
(A
,
.
.
.
,
A
)
→
7
(A
2
1
e0
e0
1
i:K,→R
op
0
2
∼
(In these types, K = 0 implies D = D via . So [D] = 1 in the Brauer group.)
Type IV: K is an imaginary quadratic extension of K0 ; for every finite place v of K, [D ⊗K Kv ] ∈
Br(Kv ) ∼
= Q/Z. Let invv D ∈ Q/Z be [D ⊗K Kv ]. Then invv D + invσ(v) D = 0, where σ : K → K
is the automorphism induced
by 0 . If σ(v) = v, then in fact invv D = 0. In this case, there is an
Q
isomorphism D ⊗Q R ∼
= K,→C/∼ Md (C) such that 0 goes to (A1 , . . . , Ae0 ) 7→ (A∗1 , . . . , A∗e0 ).
Let S = {x ∈ D : x = x0 } and η =
dimQ S
dimQ D .
Then if D = End0 A for A simple over k:
40
Type
I
II
III
IV
e
e0
e0
e0
2e0
d
1
2
2
d
η
1
1
4
3
4
1
2
Restrictions if char(k) = 0
e|g
2e|g
2e|g
e0 d2 |g
Restrictions if char(k) = p > 0
e|g
e|g
2e|g
e0 d|g
Proposition 12.4. Let A be simple, D = End0 (A). If L ⊆ S is a subfield, then [L : Q]|g.
ϕ
∼
b −
b Pick λ = ϕL a polarization. Then Hom0 (A, A)
→ End0 (A)
Proof. N S(A) ⊗ Q −
→ Hom0,sym (A, A).
0
∼
r
b is isomorphically mapped to S. Therefore N S(A)⊗Q −
by φ 7→ λ−1 φ, and Hom0,sym (A, A)
→ S. Now
∼
χ : N S(A) ⊗ Q → Q by N 7→ χ(N ) is a homogeneous polynomial function of degree g. We obtain
1
a homogeneous polynomial function of degree d, f = χ(L
r∗ χ : S → Q. Explicitly, if φ = λ−1 ϕN ,
0)
then f (φ) =
χ(N )
χ(L0 ) .
Claim. f |L is a norm.
The claim implies the proposition. To show the claim, it’s enough to show that f is multiplicative
(and in fact, that f 2 is multiplicative). But
f (φ)2 =
deg ϕN
χ(N )2
=
= deg φ.
2
χ(L0 )
deg ϕL0
(12.10)
b symmetric implies over k, φ = φM for some M (we did not finish the proof).
Remark. φ : A → A
We know that 2φ = φL . Let N = L2 . Then A[4] ⊆ K(N ) = ker φN . A[4] acts on A, and we have a
commutative pairing A[4] × A[4] → µ4 ⊆ Gm . A[2] is totally isotropic, so the action of A[2] on A
lifts to an action of A[2] on N . By faithfully flat descent, N = 2∗A M for some M on A.
13
Abelian Varieties Over Finite Fields
Consider k = Fq and X a variety over k. We have the Frobenius:
X
F (1)
X (1)
X
(13.1)
k
F
k
For q = pm , we have F m = 1. So X (m) = X and F (m) is an automorphism of X. If X = A, then
denote F (m) by πA ∈ End(A). Let pA = charπA , so
pA (t) = t2g − tr(πA )t2g−1 + · · · + q g .
Theorem 13.1.
(13.2)
• Q[πA ] ⊆ End0 (A) is semisimple.
• (Riemann Hypothesis) Let ω be a root of pA . Then under any embedding Q(ω) ,→ C, we have
1
|ω| = q 2 .
41
b be a polarization. Then π 0 πA = qA . To show this, we need to show that
Proof. Let λ : A → A
A
λ−1 π
bA λπA = qA ⇐⇒ π
bA λπA = qAbλ.
(13.3)
L
b for L a
But for any f : X → Y , πY f = f πX , so it’s enough to show π
bA πAb = qAb. Given S −
→A
(m)
(m)
line bundle on A × S, π
bA ◦ L = (1 × FS )∗ L, then applying π
bA , we get (πA × 1)∗ (1 × FS )∗ L. But
(m)
(m)
(m)
0 π =q .
(1 × FS ) ◦ (πA × 1) = FA×S , so (πA × 1)∗ (1 × FS )∗ = Lq . We get π
bA πAb = qAb, so πA
A
A
0 π = q implies π is invertible in End0 (A). As the constant term is
For Q[πA ] ⊆ End0 (A), πA
A
A
A
−1
nonzero, πA
∈ Q[πA ], so Q[πA ]0 = Q[πA ]. Let a ⊆ Q[πA ] be an ideal; then a0 is also an ideal. Let
b = (a0 )⊥ with respect to B(x, y) = tr(xy 0 ). Then a ∩ b = 0, but b is also an ideal. By dimension
reasons, a + b = Q[πA ]. So Q[πA ] is semisimple.
Now as Q[πA ] is also commutative, it has the form K1 × · · · × Kr for Ki finite field extensions of
Q. Note that 0 permutes the Ki . Positivity of trace implies 0 fixes Ki . So 0 is an automorphism of Ki ,
and then positivity implies Ki is either totally real or a CM field. For any j : Ki ,→ C, j(x0 ) = j(x).
Now suppose ω is a root of pA . Then Q(ω) is one of the Ki . So now
q = j(ωω 0 ) = j(ω)j(ω 0 ) = j(ω)j(ω) = |j(ω)|2 .
(13.4)
A Weil q-number is an algebraic integer π such that for every embedding j : Q(π) ,→ C, we have
1
∼
|j(π)| = q 2 . Two Weil q-numbers π and π 0 are conjugate if there exists Q −
→ Q(π 0 ) sending π to π 0 .
Let W (q) be the set of Weil q-numbers and W (q)/ ∼ the set of Weil q-numbers up to conjugacy.
Corollary 13.2. If A is simple, then πA is a Weil q-number.
Let Σ(AVF0q ) be the isomorphism classes of simple objects in AVF0q . Then Σ(AVF0q ) → W (q)/ ∼.
Theorem 13.3 (Honda-Tate). This map is a bijection.
Lemma 13.4. Let π be a Weil q-number. Then one of the following holds:
1. q = p2m and π = ±pm .
√
2. q = p2m+1 and π = ±pm p, so Q(π) totally real with [Q(π) : Q] = 2.
3. For any embedding j : Q(π) ,→ C, j(π) 6∈ R, and Q(π) is a CM field.
The only real numbers with absolute value
is totally real.
√
√
q are ± q. In case (3), take α = π + πq , so Q(α)
Recall Hom(A, B) ⊗ Z` ,→ HomG (T` (A), T` (B)) for G = Gal(k/k) and ` 6= p.
Theorem 13.5 (Tate). This map is an isomorphism.
Corollary 13.6. Let A, B/Fq . Then the following are equivalent:
1. B is isogenous to a sub-abelian variety of A defined over k.
42
2. There is a G-equivariant map V` (B) ,→ V` (A).
3. pB |pA .
This shows injectivity in Honda-Tate.
Proof of corollary. (1) =⇒ (2) is clear. (2) =⇒ (1) follows from Tate. (2) =⇒ (3) is clear. And
for (3) =⇒ (2), we have Q` [x]/pA [x] ∼
= V` (A) and semisimplicity.
Theorem 13.7.
1. The center of End0 (A) is Q[πA ].
2. Every abelian variety over k is CM.
3. Assume A is simple. Let D = End0 (A) and K = Q[πA ], For a place v of K,

1


2


invv D = ordv πA [K : Q ]
v
p


ordv q



0
Proof.
v real
(13.5)
v|p
otherwise
1. We have Q[πA ] ⊆ Z ⊆ End0 (A). After tensoring with Q` ,
Q` [πA ] ⊆ Z ⊗ Q` ⊆ End0 (A) ⊗ Q` = EndG (V` (A))
(13.6)
which by the double commutant theorem has center Q` [πA ].
2. We can assume that A is simple. Then K = Q[πA ] is the center. Let e = [K : Q] and
d2 = [D : K] so de|2g. Let K ⊗ Q` = Kv1 × · · P
· × Kvr and ei = [Kvi : Q` ]. We have
V` (A) = V1 × · · · × Vr with dimKi Vi = di . We have i di ei = g. Also
D ⊗ Q` = EndK⊗Q` (V` (A)) =
Y
EndKi (Vi )
(13.7)
i
implying ed2 =
2
i ei di .
Finally,
4g 2 =
X
P
P
i ei
= e. Now
!2
i
di ei
!
≤
X
i
ei
!
X
ei d2i
= e2 d2 ≤ 4g 2 .
(13.8)
i
So we require ed = 2g and all di equal (to d).
Q
3. First D ⊗ Q` = EndKi (Vi ) implies invvi D = 0 for vi |`. If v is real, then K = Q(πA ) has a
| {z }
√D⊗Ki
real place, so π = ± q. K is totally real, so D is of Type I, II, or III. Type I implies d = 1,
so e = 2g, not possible since e|g. So d = 2 and e = g. But Type III requires 2e|g, which isn’t
possible. So D is of Type II implying invv D = 12 .
We omit the case of vi |p, which uses a result on p-divisible groups similar to the Tate conjecture.
43
Here are some examples:
1. Q(π) = Q, q = p2m , π = ±pm . Then D/QP
is a quaternion algebra and d = 2. We have
invv D = 0 for v 6= ∞, p, and inv∞ D = 12 , so v invv D = 0 forces invp D = 12 . This uniquely
determines D = Dp∞ . Also 2g = ed = 2 implies A is an elliptic curve.
Claim. The p-rank of A is zero.
Otherwise, A would have p-rank 1, so D ⊗ Qp Vp,ét (A) of rank 1. But D ⊗ Qp is a nonsplit
quaternion algebra, so we cannot have a 1-dimensional representation, a contradiction.
So A is supersingular. Also, if πA = pm , then A ∼ A0 /Fp2 ⊗ Fp2m , since A0 can be found with
πA0 = p. Then πA0 /Fp2 ⊗Fp2m = pm = πA implying A ∼ A0 ⊗ Fp2m .
√
2. [Q(π) : Q] = 2, Q(π) real, q = p2m+1 , π = ±pm p, d = 2 and e = 2, so g = 2. It turns out that
A⊗Fq2 is isogenous to a product of two supersingular elliptic curves. (For pA⊗F2q (t) = (t−q)ϕ .)
Fix Me What is this? (3)
3. [Q(π) : Q] = 2 and K = Q(π) a quadratic imaginary.
• If p does not split in K, then D = K. So g = 1. We claim A is supersingular (using
N ∈ Q.
the action of D ⊗ Qp on Tp,ét ). We also claim that there exists N such that πA
Equivalently,
2
πA
q
is a root of unity.
π2
For every place v|`, qA has absolute value 1. (For α = π + πq ∈ Z, π 2 − απ + q = 0 so
q = π(π − α). We get the result for ` 6= q. By the product formula, we’re done.) Now
A ⊗ FqN is an elliptic curve as in (1).
• If p splits in K, let v1 , v2 be the two places over p. From π(π − α) = q, either |π|vi = |q|vi
or |π − α|vi = |q|vi . This implies D = K. So A is an elliptic curve, but this time its
p-rank is 1, so A is ordinary. (Otherwise, A[p∞ ] would be a simple p-divisible group, and
D ⊗ Qp ,→ End(A[p∞ ]). But End(A[p∞ ]) is a division algebra, while D ⊗ Qp ∼
= Qp × Qp .
Theorem 13.8. Let E be an elliptic curve over Fq . Then there are three possibilities:
• Q(π) = Q, E supersingular, and D = Qp∞ .
• Q(π) is quadratic imaginary over which p does not split, E is supersingular, and D = Q(π).
• Q(π) is quadratic imaginary over which p splits, E is ordinary, and D = Q(π).
Theorem 13.9. All supersingular elliptic curves over Fq are isogenous, and D = Qp∞ .
Theorem 13.10. Let A/k be an abelian variety with k = k and char(k) > 0. Then if A is CM,
then A is isogenous to an abelian variety defined over a finite field.
Theorem 13.11. Let E be an elliptic curve over k = k with char(k) = 0. Then D = End0 E = Q
if and only if E cannot be defined over a finite field.
Proof. We already know the “only if” implication. Now assume that E cannot be defined over Fp .
For simplicity, assume p > 2, Then E has the form Eλ : y 2 = x(x − 1)(x − λ) for some λ 6= 0, 1.
We must have that λ is transcendental over Fp . Let K0 = Fp ⊆ k. Regard this as a function field of
44
S0 = A1F \ {0, 1}, R0 = Fp [S0 ]. We have the Legendre family E : y 2 = x(x − 1)(x − λ) → S0 with
p
Eλ the generic fiber. So D = End0 (Eλ ), K ⊆ k finite over K0 such that all endomorphisms of Eλ
are defined over K. Let S be the normalization of S0 in K. Then:
Lemma 13.12. Let (A, m, κ) be a DVR and K the fraction field of A. Let A/ Spec A be an abelian
scheme (a smooth proper group scheme with connected fibers). Then End(AK ) → End(A) →
End(Aκ ).
Now Eλ ⊗K0 K ,→ ES , D = End(Eλ ⊗K0 K) ,→ End(Es ) for s a closed point of S. Choose two
s’s, one with Es1 supersingular and End(Es1 ) = Qp∞ , and the other with Es2 ordinary such that p
splits. We get D = Q.
Proposition 13.13. Let Eλ be y 2 = x(x − 1)(x − λ). Then Eλ is supersingular if and only if
p−1
h(λ) =
2 p−1 2
X
2
i
i=1
λi = 0.
(13.9)
(Here p > 2.)
Corollary 13.14. Over k, there are finitely many supersingular elliptic curves up to isomorphism.
All are defined over Fp .
As an example, for p = 3, h(λ) = λ + 1, so y 2 = x3 − x is√the only supersingular elliptic curve in
characteristic 3 (up to F3 -isomorphism). We have πE−1 = ± −3 by counting F3 -points. So E−1 ⊗F9
2
0
has π = −3. We claim that E 0 : dy 2 = x3√
− x for d ∈ (F9 )× \ (F×
9 ) has π = 3. For E and E−1 are
isomorphic over F81 via ϕ : (x, y) 7→ (x, y d), and ϕ−1 πE−1 ϕ(x, y) = πE 0 (x, −y) = −πE 0 (x, y).
In general, given E : y 2 = x3 + ax + b over k, we have E 0 : dy 2 = x3 + ax + b for d ∈
0
× 2
0
(k × )\(k
√ ) . If ρ, ρ : Gal(k/k) → GL2 (Q` ), then ρ = ρ⊗η for η the quadratic character Gal(k/k) →
∼
Gal(k( d)/k) = Z/2Z.
14
Good Reduction of Abelian Varieties
∼
Lemma 14.1. For A/S, S = Spec R with R a DVR having fraction field L, then End(A) −
→
End(AL ).
Proof. The map is clearly injective. For surjectivity, given ϕ : AL → AL , get Γϕ ⊆ AL × AL . Let Γ
be the closure of Γϕ in A ×S A. Γ is a proper flat group subscheme of A ×S A.
Claim. Γ0 is smooth. Γ0 is an abelian scheme, such that
Γ0
(14.1)
∼
A
A
extends ϕ.
45
To show the claim, let m be the maximal ideal of R and κ the residue field. Let η be the generic
point of Aκ , Oη the local ring of A at η, and ξ the generic point of Oη . By the valuative criterion,
ϕ
ξ → Aκ −
→ A × A extends uniquely to Spec Oη → A × A. So ϕ extends to U → Aκ for U ⊆ Aκ
open. Therefore Γ0κ |U ×Aκ is the graph of the extension. Γ0κ is generically reduced, so by flatness, we
have the claim.
Let A be an abelian variety over L and v a discrete valuation of L. Consider A[ : Sm/S → Grp/S
by T 7→ A(T ×S Spec L) for T a smooth S-scheme. If A[ is representable, then we say A[ is the
Néron model of A.
Theorem 14.2. Néron models exist.
Proposition 14.3. If A → S = Spec R is an abelian scheme, then A is the Néron model of AL for
L the fraction field of R. In this case, we say that AL has good reduction.
14.1
Complex Multiplication
Let K be a CM field of degree 2g over Q. Complex conjugation acts on the set {K ,→ C}. We say
(K, Φ) is a CM type if Φ ⊆ {K ,→ C} is such that {K ,→ C} is the disjoint union of Φ and Φ0 . Let
AΦ = CΦ /ι(OK ) for ι : K → CΦ via the embeddings in Φ. Then AΦ is an abelian variety with CM
by K.
Now if A/C is an abelian variety of dimension g and CM by K, then K acts on H1 (A; Q) = Λ⊗Q,
if A = V /Λ. H1 (A; Q) is a 1-dimensional K-space. So
Y
C = K ⊗ C H1 (A; C) = H1 (A; Q) ⊗ C.
(14.2)
K,→C
But we also have the Hodge filtration
0 → H 0 (A, ΩA ) → H 1 (A; C) → H 1 (A, OA ) → 1.
| {z }
| {z }
(14.3)
b ∨ → H1 (A; C) → Lie(A) → 0.
1 → (Lie(A))
(14.4)
(T` A)∨
b
Lie(A)
Dualizing gives
K ⊗ C acts on Lie(A) via
K ⊗C∼
=
Y
K,→C
C→
Y
C,
|Ψ| = g.
(14.5)
Ψ⊆{K,→C}
But Hodge theory implies H1 (A; C) = Lie(A) ⊕ Lie(A) so Ψ q Ψ = {K ,→ C}. So (K, Ψ) is a
CM type. We have that A is isogenous to AΨ by considering Λ ⊗ Q.
Proposition 14.4. Let A/C be CM by K. Then A is defined over Q. In fact, there is a unique
model of A over Q.
46
Proof. If k ⊆ Ω are two algebraically closed fields of characteristic zero, then we have
∼
Homk (A, B) −
→ HomΩ (AΩ , BΩ )
(14.6)
for any abelian varieties A, B/k. This gives uniqueness.
We have A(k)tor ∼
= A(Ω)tor since A[N ] ∼
= (Z/N Z)2g . If f : AΩ → BΩ , and a ∈ Aut(Ω/k), we
need to show that f a = af . This holds for torsion points, which are Zariski dense.
Now if A/C has CM type (K, Φ), let R ⊆ C finitely generated over Q such that A, K ,→ End0 (A)
are defined over R. Then
K As
A
(14.7)
Spec Q
s
S
Spec R
Claim. As ⊗ C has CM type (K, Φ).
Q
For i:K,→C OSC ∼
= K ⊗ OSC Lie(ASC ).
So As ⊗ C and A are isogenous. We have N ,→ As ⊗ C → A, for N ⊆ As (C)tor ∼
= As (Q)tor , so
N is defined over Q.
Proposition 14.5. Let A be a CM abelian variety over a number field k. Let p be a prime of k
over p. Then after a finite extension of k, A has good reduction at p.
Proof. We’ll use the following:
Theorem 14.6. A/k hsa good reduction at p if and only if the inertia subgroup Ip of Gal(Q/k) acts
trivially on V` A for ` 6= p.
In our case, Gal(Q/k) ,→ EndK⊗Q` (V` (A))× ∼
= (K ⊗ Q` )× . This factors through Gal(Q/k)ab →
×
(K ⊗ Q` ) whose image is a compact subgroup, so contained in (OK ⊗ Z` )× . Up to finite index,
(OK ⊗ Z` )× is a pro-` group while the image of Ip in Gal(Q/k)ab is a pro-p group. So the image of
Ip → EndK⊗Q` (V` A) is finite. After a finite extension, it will be trivial.
Now suppose A/k has good reduction at p, with CM type (K, Φ). Let A = A[ ⊗ κ(p) over
κ(p) = Fq , and associate the Weil q-number πA . If A ∼ A0 , then they are isogenous over some finite
N = π N . So (K, Φ) should determine π up to a root of unity.
extension, so for some N , πA
A
A0
Lemma 14.7. πA ⊆ K ,→ End0 (A).
Proof. C(K) ⊆ End0 (A) is K, and πA ∈ C(K).
Theorem 14.8 (Shimura-Taniyama formula). Let v|p be a place of K. Assume that k contains K
and is Galois, and such that A/k has good reduction at p. Then
ordv (πA )
#(Φ ∩ Hv )
=
,
ordv (q)
#Hv
Hv = {i : K ,→ k : i−1 (p) = v}.
47
(14.8)
Sketch of surjectivity in Honda-Tate. We can assume L = Q(π) is CM. Let D be the division algebra
over L having the expected invariants. Let e = [L : Q], d2 = [D : L], and 2g = de.
Lemma 14.9. There exists L ⊆ K ⊆ D a CM field of degree 2g over Q.
Fix Qp , then
H = Hom(K, Qp ) ∼
=
a
Hv ,
Hv = Hom(Kv , Qp ).
v|p
Claim. There exists a CM type (K, Φ) such that
ordv (π)
ordv (q)
48
=
#(Hv ∩Φ)
#Hv .
(14.9)
To do. . .
2 1 (p. 18): Fix Me Where does this proof end? It must end at or before here.
2 2 (p. 20): Fix Me Is this correct?
2 3 (p. 44): Fix Me What is this?
49