Introduction to Arithmetic Algebraic Geometry · Algebraic Geometry The set X(C) or X(K) is called...

Introduction toArithmetic Algebraic Geometry

Sungkon Chang

The Anne and Sigmund HudsonMathematics and ComputingLuncheon Colloquium Series

Diophantine Equations

Let Z denote the set of integers.

Diophantine Equations:

2x2 +3y2 = 4z3 +5w3

where x,y,z,w ∈ Z.

Diophantine Equations

Let Z denote the set of integers.

Diophantine Equations:

x21 + y2

1 = x32 + y3

2 = x43 + y4

3

where xk’s and yk’s ∈ Z.

Rational Solutions

Let Q denote the set of rational numbers.

2x2 +3y2 = 4z3 +5w3

where x,y,z,w ∈Q.

Rational Solutions

Let Q denote the set of rational numbers.

2x2 +3y2 = 4z3 +5w3

where x,y,z,w ∈Q.

Q is a field.

a±b, a ·b, a/b ∈Q where a,b ∈Q.

K-Rational Solutions

Let K be a field.

2x2 +3y2 = 4z3 +5w3

where x,y,z,w ∈ K.


Let K be a field.

X : 2x2 +3y2 = 4z3 +5w3

X(K) = (x,y,z,w) ∈ K4 : 2x2 +3y2 = 4z3 +5w3.


Let K be a field.

X : 2x2 +3y2 = 4z3 +5w3

X(K) = (x,y,z,w) ∈ K4 : 2x2 +3y2 = 4z3 +5w3.

Question: Determine whether X(K) is empty.


Let K be a field.

X : 2x2 +3y2 = 4z3 +5w3

X(K) = (x,y,z,w) ∈ K4 : 2x2 +3y2 = 4z3 +5w3.


Is X(Q) empty?


X : f (x,y) = 0where f (x,y) is a polynomial

with Q-coeff.

X(Q) = (x,y) ∈Q2 : f (x,y) = 0.

Question: Determine whether X(Q) is empty.


X : f (x,y) = 0where f (x,y) is a polynomial

with K-coeff.

X(K) = (x,y) ∈ K2 : f (x,y) = 0.


Geometry

X : x2 + y2 = 1

Geometry

X : x2 + y2 = 1

t 7→ x = cos(t), y = sin(t).

[0,2π) ∼= X(R)

Geometry

X : x2 + y2 = 1

t 7→ x = cos(t), y = sin(t).

[0,2π) ∼= X(R)

What about X(Q)?

Geometry

X : x2 + y2 = 1

t 7→ x =1− t2

1+ t2 , y =2t

1+ t2 .

Q∼= X(Q)\(−1,0)

X is called a rational curve.

Geometry

X : x2 + y2 = 1

Q ∼= X(Q)\(−1,0)

Geometry

X : 64yx4 +56x3y+16x4 +16x3−4y2 +72y2x3 +16y2x2

+8y2x+40y3x3 +12y3x+32y3x2 +64y3x4

+96y2x4 +4y4x+16y4x4 +8y4x3 +16y4x2−4y3 +3y4 = 0.

Geometry

X : 64yx4 +56x3y+16x4 +16x3−4y2 +72y2x3 +16y2x2

+8y2x+40y3x3 +12y3x+32y3x2 +64y3x4

+96y2x4 +4y4x+16y4x4 +8y4x3 +16y4x2−4y3 +3y4 = 0.

x =t2−1

t4 + t +1, y =

2tt4− t +1

.

Geometry

X : x3 + y3 = 1

Is this a rational curve?x = f (t) and y = g(t)

for rational functions f (t) and g(t)?

Geometry

X : x3 + y3 = 1



(a/c)3 +(b/c)3 = 1⇒ a3 +b3 = c3.

By Fermat’s Last Theorem,there are no parametric solutions.

Geometry

X : x3 + y3 = 1



Geometry

X : x3 + y3 = 1



Geometry of X(C)

Geometry

X : x3 + y3 = 1



Geometry

Topology: continuous maps

Differential Geometry: differentiable maps

Algebraic Geometry: rational maps

Geometry

Topology: continuous maps

Differential Geometry: differentiable maps

Algebraic Geometry: rational maps

X(K)−→ Y (K)

Geometry

Algebraic Geometry

X : x3 + y3 = 1Y : x6 +6x4y2 +12x2y4 +8y6

+27x3−27x2y+9xy2− y3 = 1.

Algebraic Geometry

X : x3 + y3 = 1Y : x6 +6x4y2 +12x2y4 +8y6

+27x3−27x2y+9xy2− y3 = 1.

Y : (x2 +2y2)3 +(3x− y)3 = 1

Y (R)−→ X(R) given by(x,y) 7→ (x2 +2y2,3x− y).

Algebraic Geometry

X : x3 + y3 = 1Y : x6 +6x4y2 +12x2y4 +8y6

+27x3−27x2y+9xy2− y3 = 1.

Y : (x2 +2y2)3 +(3x− y)3 = 1


Y (Q)−→ X(Q).

Algebraic Geometry

X : x3 + y3 = 1Y : x6 +6x4y2 +12x2y4 +8y6

+27x3−27x2y+9xy2− y3 = 1.

Y : (x2 +2y2)3 +(3x− y)3 = 1


Y (Q)−→ X(Q).

Y (Q) is empty.

Algebraic Geometry

Algebraic Geometry

X : x3 + y3 = 1Y : 2x3−3x2y+15xy2−7y3 +3x4y+3x5y2

+x6y3−18x3y2−18x4y3−6x5y4 +36x2y3 +36x3y4 +12x4y5

−24y4x−24y5x2−8y6x3−3xy−3x2y2− x3y3 = 1.

(x+y1+xy

)3+(x−2y)3 = 1

Y (C)\(±1,∓1) −→ X(C) given by

(x,y) 7→(

x+y1+xy, x−2y

).

Algebraic Geometry

X : x3 + y3 = 1Y : 2x3−3x2y+15xy2−7y3 +3x4y+3x5y2

+x6y3−18x3y2−18x4y3−6x5y4 +36x2y3 +36x3y4 +12x4y5

−24y4x−24y5x2−8y6x3−3xy−3x2y2− x3y3 = 1.

(x+y1+xy

)3+(x−2y)3 = 1

Y (C)\(±1,∓1) −→ X(C) given by

(x,y) 7→(

x+y1+xy, x−2y

).

Y (Q) = (±1,∓1)

Algebraic Geometry

The set X(C) or X(K) is called an algebraic setwhere K = K.

Fundamental Theorem of Algebra:All polynomials with C-coeff

have all their zeros in C, i.e.,C is algebraically closed.

Algebraic Geometry


For each field K, there isa (large) smallest field extension K

which is algebraically closed.

Algebraic Geometry


For each field K, there isa (large) smallest field extension K

which is algebraically closed.

Examples: finite fields, fields of functions,number fields, p-adic fields, ...

Algebraic Geometry treats all fields of all characteristics.

Algebraic Geometry


Algebraic Geometry


Example: The set of equivalent classes ofcurves of genus 1 with C-coeff.

Algebraic Geometry


Moduli Problem (Representability):

The category of equivalent classes ofcertain algebraic sets

↔ an algebraic set X (C).

X (C) is called a moduli space.

Algebraic Geometry


Arithmetic Geometry Problem:

• The moduli space for the category ofequivalent classes of curves of genus 1 with Q-coeff.

• The moduli space for the category ofequivalent classes of curves of genus 1 with Q-coeff.,

which contain a rational point.

X (Q) ∼= [C] : C(Q) 6= /0, genus 1

Algebraic Geometry


Arithmetic Geometry Problem:

Given a curve of genus 1 with Q-coeff.,determine whether it has a rational point.

Algebraic Geometry

The ring of functions on an algebraic set

Examples:The set of polynomial functions K[x] forms a ring R:

f ,g ∈ R⇒ f ±g, f ·g ∈ R.

f (x) = x3 + x2 + x+1 ∈Q[x].f : Q−→Qf : R−→ Rf : C−→ Cf : K −→ K

Algebraic Geometry


Example: f (x) = x3 + x2 + x+1.

D = 0,1. Then, f |D ≡−(x−1)+4x |D.

Algebraic Geometry


Example: f (x) = x3 + x2 + x+1.

D = 0,1. Then, f |D ≡−(x−1)+4x |D.

For any polynomial f (x) ∈ C[x],

f |D ≡ ax+b |D.

Let R be the ring of functions on D.

C[x]−→ R, but C[x] 6∼= R.

Algebraic Geometry


Example: D = 0,1, C[x], R = f |D : f ∈ C[x]

Note: f (x) = x(x−1) ·Q(x)+ r(x). So,f (x)≡ r(x) as functions on D.

We use the division algorithm.

Algebraic Geometry


Example: D = 0,1, C[x], R = f |D : f ∈ C[x]

Note: f (x) = x(x−1) ·Q(x)+ r(x). So,f (x)≡ r(x) as functions on D.

We use the division algorithm.

R∼= C×C

Ring of Functions

Example: D = (x,y) ∈ R2 : x2 + y2 = 1 and R[x,y].

R = f |D : f ∈ R[x,y].

f (x,y) = 3x3 +3x2y+ x2 +2y2x+4y3

+y2−2x−3y−1 ∈ R[x,y]f (x,y)≡ x3 + y3 on D

Ring of Functions


R = f |D : f ∈ R[x,y].

f (x,y) = 3x3 +3x2y+ x2 +2y2x+4y3

+y2−2x−3y−1 ∈ R[x,y]f (x,y)≡ x3 + y3 on D

f (x,y) = (x2 + y2−1)(2x+3y+1)+ x3 + y3.f (x,y)≡ (x+ y)(1− xy) on D

Ring of Functions


R = f |D : f ∈ R[x,y].

The ring R has no zero divisors:f ·g≡ 0 on D ⇒ f or g≡ 0 on D.

Ring of Functions


R = f |D : f ∈ R[x,y].


R → R(t)where R(t) = f /g : f ,g ∈ R[t],

x 7→ 2t1+t2 , y 7→ 1−t2

1+t2 .

R∼= R[

2t1+t2 ,

1−t2

1+t2

].

Ring of Functions


R = f |D : f ∈ R[x,y].


There are no injective ring homomorphisms: R→Q(t).

Ring of Functions

Ring of Functions and Topology

Example: D = (x,y) ∈Q2 : x2 + y2 = 1.

R = f |D : f ∈Q[x,y].

Topology: local geometry,defined by declaring open neighberhoods.


Example: D = (x,y) ∈Q2 : x2 + y2 = 1.

R = f |D : f ∈Q[x,y].

Topology: ∀ P = (a,b) ∈ D,

RP := f ∈ R : f (P) = 0.


Example: D = (x,y) ∈Q2 : x2 + y2 = 1.

R = f |D : f ∈Q[x,y].


RP := f ∈ R : f (P) = 0.

RP = (x−a) · f +(y−b) ·g : f ,g ∈ R.• This set of linear combinations is called an ideal,

denoted by 〈x−a,y−b〉.• The ideals RP’s are all maximal, and

D↔Max(R)?


Example: D = (x,y) ∈Q2 : x2 + y2 = 1.

R = f |D : f ∈Q[x,y].


RP := f ∈ R : f (P) = 0.

RP = (x−a) · f +(y−b) ·g : f ,g ∈ R.• This set of linear combinations are called an ideal,

denoted by 〈x−a,y−b〉.• The ideals RP’s are all maximal, and

D→Max(R).〈x2−1/2,y2−1/2〉 & 〈x2 +1,y2−2〉 are maximal.


Example: D = (x,y) ∈Q2 : x2 + y2 = 1.

R = f |D : f ∈Q[x,y].

Topology: ∀ V ⊂ D,

RV := f ∈ R : f (P) = 0, ∀P ∈V.


Example: D = (x,y) ∈Q2 : x2 + y2 = 1.

R = f |D : f ∈Q[x,y].


RV := f ∈ R : f (P) = 0, ∀P ∈V.

RV = 〈 f1, . . . , fn〉.

For two subsets V1 and V2 of D with V1 ⊂V2,RV2 ⊂ RV1.


Example: D = (x,y) ∈Q2 : x2 + y2 = 1.

R = f |D : f ∈Q[x,y].


RV := f ∈ R : f (P) = 0, ∀P ∈V.

RV = 〈 f1, . . . , fn〉.

For two subsets V1 and V2 of D with V1 ⊂V2,RV2 ⊂ RV1.

The chains of ideals ⇒ topology.


Example: D = (x,y) ∈Q2 : x2 + y2 = 1.

R = f |D : f ∈Q[x,y].

Zariski Topology:• Each ideal in Max(R) forms a (singleton) closed subset.• Let I be an ideal.

J ∈Max(R) : I ⊂ J forms a closed subset.Note that VJ ⊂VI.

Example: I = 〈(x−1)(x−4/5)〉 is contained in〈x−1,y〉, 〈x−4/5,y−3/5〉, and 〈x−4/5,y+3/5〉.


Example: D = (x,y) ∈Q2 : x2 + y2 = 1.

R = f |D : f ∈Q[x,y].



Example: I = 〈(x2−3)(x2−5)〉 is contained in〈x2−3,y2 +3〉 and 〈x2−5,y2 +4〉.


Example: D = (x,y) ∈Q2 : x2 + y2 = 1.

R = f |D : f ∈Q[x,y].



The Zariski Topology is a topology on Max(R).

Ring of Functions and Geometry

DimensionLet R be the ring of functions on an algebraic set.

• Topology on Max(R).

• The chain of ideals → the underlying geometry.

I

===

====

=

J1

J2

???

????

J3 J4 J5 J6





0→ p1 → p2 → ··· →Q

where Q ∈Max(R).A chain of prime ideals





Example:D = (x,y,z) ∈ R3 : x2 + y2 + z2 = 1, R = f |D : f ∈ R[x,y,z].

0→ p = 〈z〉 →Q = 〈x−1,y,z〉0→ p = 〈z〉 →Q′ = 〈x− 3

5,y−45,z〉






0→ p = 〈z〉 →Q = 〈x−1,y,z〉0→ p = 〈z〉 →Q′ = 〈x− 3

5,y−45,z〉

〈z〉 corresponds to 〈x−a,y−b,z〉 : a2 +b2 = 1.






0→ 〈z〉 → 〈x−1,y,z〉0→ 〈z〉 → 〈x−3/5,y−4/5,z〉

〈z〉 corresponds to a curve (of dimension 1).





Example:





0→ p1 → p2 → ··· →Q

where Q ∈Max(R).

The length of this chain suggeststhe dimension of the algebraic set,

called the Krull dimension of R.





Spec(R) := p⊂ Spec(R) : p is prime idealwith Zariski topology.


Singularity/Regularity:

dimK Q/Q2 6= dimR ⇒ singular at Q.

Genus: dimK ΩX

where ΩX is the canonical differentials on X .

MATH5900U: Special Topics Course

Integrality of a Ring of Functions

Factorization of Ideals/Galois Covers

Integrality of a ring of functions

Ideal Class Group/Picard Group

Arithmetic Geometry

Global fields K:number fields and function fields of a curve over a finite field.

• K is the field of quotient of a Dedekind domain S.(Notherian, dimension 1, integrally closed).

• S/pS forms a finite field where p is prime.

Example: Let k be the field F3 with 3 elements.• T = (x,y) ∈ F3 : x3 + y3 = 1 and S = f |D : f ∈ k[x,y].

Let K be the field of quotients of S.

Arithmetic Geometry

Global fields K:number fields and function fields of a curve over a finite field.

• K is the field of quotient of a Dedekind domain S.(Notherian, dimension 1, integrally closed).

• S/pS forms a finite field where p is prime.

Example: Let k be the field F3 with 3 elements.• T = (x,y) ∈ F3 : x3 + y3 = 1 and S = f |D : f ∈ k[x,y].

Let K be the field of quotients of S.S ∼= k[t] and K ∼= k(t).

• S/(t2 +1)S ∼= F9.

Arithmetic Geometry

Curves X :

• g(X) = 0: X(K) = /0 or parametrized.

• g(X) = 1: X(K) = /0 orX(K) forms a finitely generated abelian group,

i.e., an elliptic curve.

• g(X) > 2:Mordell-Conjecture: X(Q) is finite.

proved by Gerd Faltings, 1986 Fields Medal

Arithmetic Geometry

Curves X :

• g(X) = 0: X(K) = /0 or parametrized.

• g(X) = 1: X(K) = /0 orX(K) forms a finitely generated abelian group,

i.e., an elliptic curve.

• g(X) > 2:Mordell-Conjecture: X(Q) is finite.

proved by Gerd Faltings, 1986 Fields Medal

Conjecture: For g > 1, there is a constant cg such that#X(Q) < cg for all curves X with genus g.

Arithmetic Geometry

Let k be a finite field w/ #k = q, and kn be a finite ext of k.Let X be a curve with k-coeff, and Nn := #X(kn).

Arithmetic Geometry


• Hasse: X an elliptic curve ⇒|N1− (q+1)| ≤ 2

√q.

• Weil: X a curve with genus g ⇒|Nn− (qn +1)| ≤ 2g

√q.

and he proved a lot more...

Arithmetic Geometry


Weil: X a curve with genus g ⇒ |Nn− (qn +1)| ≤ 2g√

q.

Z(T ) := exp(∑∞n=1 Nn T n/n), a generating function.

Arithmetic Geometry



q.

Z(T ) := exp(∑∞n=1 Nn T n/n), a generating function.

Then,

Z(T ) =f (T )

(1−qT )(1−T ).

Arithmetic Geometry



q.

Z(T ) := exp(∑∞n=1 Nn T n/n) =

f (T )(1−qT )(1−T )

.

• f (T ) has coefficients in Z with degree 2g.

• f (T ) = ∏2gi=1(1−ωiT ) where ωi ∈ C.

• Riemann Hypothesis for curves over a finite field:

Arithmetic Geometry



q.

Z(T ) := exp(∑∞n=1 Nn T n/n) =

f (T )(1−qT )(1−T )

.



• Riemann Hypothesis for curves over a finite field:|ωi|= q1/2.

Arithmetic Geometry



q.

Z(T ) := exp(∑∞n=1 Nn T n/n) =

f (T )(1−qT )(1−T )

.

• Riemann Hypothesis for curves over a finite field:|ωi|= q1/2.

log Z(T ) = ∑∞n=1

(qn +1−∑

2gi=1 ωi

)T n

n ⇒ Nn = qn +1−∑2gi=1 ωi.

Therefore, |Nn− (qn +1)| ≤ 2g√

q.

Arithmetic Geometry



q.

Z(T ) := exp(∑∞n=1 Nn T n/n) =

f (T )(1−qT )(1−T )

.



• RH for curves over a finite field: |ωi|= q1/2.

• Functional Equations: Z(1/qT ) = (qT )1−gZ(T ).

Arithmetic Geometry



q.

Z(T ) := exp(∑∞n=1 Nn T n/n) =

f (T )(1−qT )(1−T )

.

Example: Let X be a compact smooth cubic curve.Then, X(k) must be nonempty.

f (T ) = 1+(N1−q−1)T +qT 2, so f (1) = N1,while f (T ) = (1−ω1T )(1−ω2T )⇒

f (1) = (1−ω1)(1−ω2) 6= 0 since |ωi|=√

q.

Arithmetic Geometry



q.

Z(T ) := exp(∑∞n=1 Nn T n/n) =

f (T )(1−qT )(1−T )

.

Example: Let X be a compact smooth cubic curve.Then, X(k) must be nonempty.

f (T ) = 1+(N1− p−1)T + pT 2 = 1+apT + pT 2.

F ap : p gave birth to Taniyama-Shimura-Weil Conj.

Arithmetic Geometry


Weil Conjecture: X a smooth projective variety of dimension n.

Z(T ) := exp(∑∞n=1 Nn T n/n).

(1) Rationality: Z(T ) ∈Q(T ).

(2) Functional Equation: Z(1/qn T ) =±qnε/2 T ε Z(T ).

(3) Rieman Hypothesis:

Z(T ) =P1(T ) · · ·P2n−1(T )P0(T ) · · ·P2n(T )

where Pi(T ) ∈ Z[T ], Pi(T ) = ∏ j(1−αi jT ) with∣∣αi j

∣∣ = qi/2

for 1≤ i≤ 2n−1,

and P0(T ) = 1−T and P2n(T ) = 1−qnT .

Arithmetic Geometry


Weil Conjecture: X a smooth projective variety of dimension n.

Z(T ) := exp(∑∞n=1 Nn T n/n).

Rationality, Functional Equation, Rieman Hypothesis

• Weil, 1949: proved for abelian varieties.

• Dwork, 1960: the rationality, using p-adic functional analysis.

• Artin, Grothendieck, et. al.: `-adic cohomology theory;

proved the functional equation.

• Deligne, 1973: proved the Riemann Hypothesis, Fields Medal

Arithmetic Geometry

Higher Dimensional X :

• An abelian variety X(K) is a compact algebraic group.Mordell-Weil Theorem:

X(K) is a finitely generated abelian group.

• Let C be a curve.Abel-Jacobi: C(C) → X(C) where X is an abelian variety.Weil, et. al.: C(K) → X(K) s.t. C(K) → X(K).

Arithmetic Geometry




• Let C be a curve.Abel-Jacobi: C(C) → X(C) where X is an abelian variety.Weil, et. al.: C(K) → X(K) s.t. C(K) → X(K).

X is called a Jacobian variety of C.

Arithmetic Geometry




• Let C be a curve, and J be its Jacobian variety.Manin-Mumford Conjecture:

C(K)∩ J(K)tor is finite.proved by Michel Raynaud

Introduction to Arithmetic Algebraic Geometry · Algebraic Geometry The set X(C) or X(K) is called...

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