riemann hypothesis for function fields bombieri's proof

Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof

Riemann Hypothesis for function fieldsBombieri’s proof

Nivedita Bhaskhar

Chennai Mathematical Institute

June 25,2009

Riemann Hypothesis

Analogously

Credits

Bounds

The language of curves

The correspondence

The proof

A natural construction

ζ(s) :=∑n∈N

• Primes : 2,3,5,7,11,. . .

• Q∗+ : Free abelian group generated by the primes under

multiplication.

• N: Set of non-negative combination of primes.

ζ(s) :=∑n∈N

• Primes : 2,3,5,7,11,. . .

multiplication.

ζ(s) :=∑n∈N

• Primes : 2,3,5,7,11,. . .

multiplication.

ζ(s) :=∑n∈N

• Primes : 2,3,5,7,11,. . .

multiplication.

Properties

• ζ converges absolutely for any s with Re(s) > 1 to an analyticfunction and diverges elsewhere.

• ζ can be analytically extended to all points of C except fors = 1 where it has a simple pole.

• ζ(s) =∏

p, a prime1

1−p−s .

• It satisfies a functional equation ξ(s) = ξ(1− s) where

• ξ(s) = π−s2 Γ

)ζ(s)

• Γ(z) =∫∞0

tz−1e−tdt

Properties

• ζ(s) =∏

p, a prime1

1−p−s .

• ξ(s) = π−s2 Γ

)ζ(s)

• Γ(z) =∫∞0

tz−1e−tdt

Properties

• ζ(s) =∏

p, a prime1

1−p−s .

• ξ(s) = π−s2 Γ

)ζ(s)

• Γ(z) =∫∞0

tz−1e−tdt

Properties

• ζ(s) =∏

p, a prime1

1−p−s .

• ξ(s) = π−s2 Γ

)ζ(s)

• Γ(z) =∫∞0

tz−1e−tdt

Properties

• ζ(s) =∏

p, a prime1

1−p−s .

• ξ(s) = π−s2 Γ

)ζ(s)

• Γ(z) =∫∞0

tz−1e−tdt

Properties

• ζ(s) =∏

p, a prime1

1−p−s .

• ξ(s) = π−s2 Γ

)ζ(s)

• Γ(z) =∫∞0

tz−1e−tdt

Properties

• ζ(s) =∏

p, a prime1

1−p−s .

• ξ(s) = π−s2 Γ

)ζ(s)

• Γ(z) =∫∞0

tz−1e−tdt

The eighth problem

Hypothesis (Riemann, 1859)

All the non-trivial zeroes of the ζ function lie on the lineRe(s) = 1

Riemann Hypothesis

Analogously

Credits

Bounds

The correspondence

The proof

Function fields

K is said to be a function field in one variable over F if ∃x ∈ Ksuch that [K : F (x)] is finite.

K has a non-negative integer g called its genus associated with it.

Assumptions :

• F is a finite field of characteristic p > 0.

• |F | = q.

• F is algebraically closed in K .

Function fields

Assumptions :

• |F | = q.

Function fields

Assumptions :

• |F | = q.

P , Q, . . . ?

• A discrete valuation ring (DVR) is a principal ideal domainwith unique maximal ideal.

• DVR (R,P) is said to be a prime of K if its fraction field is K .

• The DVRs with fraction field Q are precisely

RP :={a

b: GCD(a, b) = 1, p 6 |b

for primes p.

• SK := {P|(R,P) is a prime of K}.

P , Q, . . . ?

RP :={a

b: GCD(a, b) = 1, p 6 |b

for primes p.

P , Q, . . . ?

RP :={a

b: GCD(a, b) = 1, p 6 |b

for primes p.

P , Q, . . . ?

RP :={a

b: GCD(a, b) = 1, p 6 |b

for primes p.

P , Q, . . . ?

RP :={a

b: GCD(a, b) = 1, p 6 |b

for primes p.

An effective construction

ζK (s) :=∑

{A∈DK |A≥0}

(NA)s.

• DK called the divisor group of K is the free abelian group onprimes of K .

• A non-negative combination of primes is called an effectivedivisor.

• Degree of a prime (R,P) =[

RP : F

• Degree of a divisor∑

n(P)P =∑

n(P) deg(P).

• Norm of a divisor D is defined as ND = qdeg(D).

ζK (s) :=∑

{A∈DK |A≥0}

(NA)s.

RP : F

n(P)P =∑

n(P) deg(P).

ζK (s) :=∑

{A∈DK |A≥0}

(NA)s.

RP : F

n(P)P =∑

n(P) deg(P).

ζK (s) :=∑

{A∈DK |A≥0}

(NA)s.

RP : F

n(P)P =∑

n(P) deg(P).

ζK (s) :=∑

{A∈DK |A≥0}

(NA)s.

RP : F

n(P)P =∑

n(P) deg(P).

ζK (s) :=∑

{A∈DK |A≥0}

(NA)s.

RP : F

n(P)P =∑

n(P) deg(P).

Rational nature

TheoremThere exists a polynomial LK (u) ∈ Z[u] of degree 2g where g isthe genus of K with LK (0) = 1, such that

ζK (s) =LK (q−s)

(1− q−s)(1− q1−s),

which holds for all s ∈ Re(s) > 1. The right hand side provides ananalytic continuation for ζK to all of C with the only poles ats = 0, s = 1 (which are simple).

and etc.

• ζK (s) =∏

P∈SK

11−(NP)−s .

• ξ(s) = ξ(1− s) where ξ(s) = q(g−1)sζK (s).

and etc.

• ζK (s) =∏

P∈SK

11−(NP)−s .

and etc.

• ζK (s) =∏

P∈SK

11−(NP)−s .

The Riemann hypothesis for function fields

Hypothesis (E.Artin, 1924)

All the zeroes of ζK function lie on the line Re(s) = 12 .

Riemann Hypothesis

Analogously

Credits

Bounds

The correspondence

The proof

Credits

• The case g = 1 was proved by Hasse in 1934.

• Two proofs were given by Weil in the early 1940s which usealgebraic geometry techniques.

• S.A.Stepanov in the 1960s provided a simple proof albeit forspecial cases

• Enrico Bombieri’s proof (1970s) for the general case expandson the above proof.

• The Riemann Hypothesis for a general algebraic variety wasproved by Deligne in 1974.

Credits

Riemann Hypothesis

Analogously

Credits

Bounds

The correspondence

The proof

LK (u) =

2g∏i=1

(1− αiu).

Hypothesis (R.H restated)

Let u = q−s . Then,|αi | =

Equating euler product and rational expression of ζK ,

a1 = q + 1− (α1 + α2 + . . . α2g ),

where a1 is the number of primes of K of degree 1.

LK (u) =

2g∏i=1

(1− αiu).

a1 = q + 1− (α1 + α2 + . . . α2g ),

LK (u) =

2g∏i=1

(1− αiu).

a1 = q + 1− (α1 + α2 + . . . α2g ),

Weil bound

Theorem (Weil bound)

−2g√

q + q + 1 ≤ a1 ≤ 2g√

q + q + 1.

A weaker bound

TheoremLet g be the genus of K and suppose that (g + 1)4 < q and that qis an even power of the characteristic p. Then,

a1 ≤ (2g + 1)√

q + q + 1.

Riemann Hypothesis

Analogously

Credits

Bounds

The correspondence

The proof

A picture is worth ..

.. a thousand words ?

• F denotes the algebraic closure of F .

• (a0, a1, . . . aN) ∼ (b0, b1, . . . , bN) if ∃y ∈ F∗

such thatai = ybi for each i .

• Projective space PN(F ) = FN+1\{0}∼ .

• Projective variety V (defined over F ) is a subset of PN(F )such that the set of polynomials of F [x0, x1, . . . , xN ] for whichit vanishes (I (V )) is a prime ideal generated by homogeneouspolynomials of F [x0, x1, . . . xN ].

• (a0, a1, . . . aN) ∼ (b0, b1, . . . , bN) if ∃y ∈ F∗

K is defined to be the set of all rational functions fg such that:

• f and g are homogeneous polynomials of the same degree inF [x0, x1, . . . , xN ].

• g 6∈ I (C )

• Two functions fg and f ′

g ′ are identified if fg ′ − f ′g ∈ I (C ).

A projective curve (defined over F ) is a projective variety Cdefined over F such that the corresponding K has transcendencedegree 1 over F .

• g 6∈ I (C )

Rational points

• F -rational points of C form the set

{[f0, f1, . . . fN ] ∈ PN(F )|fi ∈ F∀i ≤ N}.

• φ : C → C sends [a0, a1, . . . aN ] [aq0 , a

q1 , . . . , a

• F -rational points are also the fixed points of φ.

Rational points

{[f0, f1, . . . fN ] ∈ PN(F )|fi ∈ F∀i ≤ N}.

q1 , . . . , a

Rational points

{[f0, f1, . . . fN ] ∈ PN(F )|fi ∈ F∀i ≤ N}.

q1 , . . . , a

Rational points

{[f0, f1, . . . fN ] ∈ PN(F )|fi ∈ F∀i ≤ N}.

q1 , . . . , a

Riemann Hypothesis

Analogously

Credits

Bounds

The correspondence

The proof

Points and DVRs

Take a ‘smooth’ projective curve C and its corresponding K .

Given a point α of C ,

(Oα,Pα) =

g∈ K |g(α) 6= 0

is the discrete valuation ring associated with it.

Points and DVRs

Take a ‘smooth’ projective curve C and its corresponding K .Given a point α of C ,

(Oα,Pα) =

g∈ K |g(α) 6= 0

is the discrete valuation ring associated with it.

T : C → Primes of K

deg(Pα) = |T−1(Pα)| = | Galois orbit of α| = |{α1, α2, . . . , αr}|.

Rational points correspond exactly to the primes of degree 1 of K

Riemann Hypothesis

Analogously

Credits

Bounds

The correspondence

The proof

Zeroes and poles

• ordPα(f ) = max{n|f ∈ Pnα}.

• f is said to have a zero at α if ordPα(f ) > 0 and a pole ifordPα(f ) < 0.

• Divisor associated with f ∈ K ∗ = (f ) =∑

P∈SKordP(f )P.

(f )0 =∑

{P| ordP(f )>0}

ordP(f )P

(f )∞ =∑

{P| ordP(f )<0}

− ordP(f )P.

deg((f )0) = deg((f )∞)

Zeroes and poles

P∈SKordP(f )P.

(f )0 =∑

{P| ordP(f )>0}

ordP(f )P

(f )∞ =∑

{P| ordP(f )<0}

− ordP(f )P.

deg((f )0) = deg((f )∞)

Zeroes and poles

P∈SKordP(f )P.

(f )0 =∑

{P| ordP(f )>0}

ordP(f )P

(f )∞ =∑

{P| ordP(f )<0}

− ordP(f )P.

deg((f )0) = deg((f )∞)

Zeroes and poles

P∈SKordP(f )P.

(f )0 =∑

{P| ordP(f )>0}

ordP(f )P

(f )∞ =∑

{P| ordP(f )<0}

− ordP(f )P.

deg((f )0) = deg((f )∞)

Zeroes and poles

P∈SKordP(f )P.

(f )0 =∑

{P| ordP(f )>0}

ordP(f )P

(f )∞ =∑

{P| ordP(f )<0}

− ordP(f )P.

deg((f )0) = deg((f )∞)

Zeroes and poles

P∈SKordP(f )P.

(f )0 =∑

{P| ordP(f )>0}

ordP(f )P

(f )∞ =∑

{P| ordP(f )<0}

− ordP(f )P.

deg((f )0) = deg((f )∞)

Idea of the proof

Find a function f with a zero at almost all rational points but veryfew poles (that too, of small order). For some small s,

(a1 + O(1)) ≤ deg((f )0) = deg((f )∞) ≤ s

Idea of the proof

Find a function f with a zero at almost all rational points but veryfew poles (that too, of small order). For some small s,

(a1 + O(1)) ≤ deg((f )0) = deg((f )∞) ≤ s

Search the kernel

If we can construct a map ψ which sends h ◦ φ h for h ◦ φ in thedomain of ψ and if h ◦ φ is in the kernel of ψ,

h ◦ φ(β) = h(β) = 0

for any rational point β where h is defined.

Fix a rational point α.

Any function of

Rm = {k ∈ K ∗|(k) + mPα ≥ 0}

can have a pole only at α (and of order atmost m).

Search the kernel

h ◦ φ(β) = h(β) = 0

Any function of

Rm = {k ∈ K ∗|(k) + mPα ≥ 0}

Search the kernel

h ◦ φ(β) = h(β) = 0

Any function of

Rm = {k ∈ K ∗|(k) + mPα ≥ 0}

Search the kernel

h ◦ φ(β) = h(β) = 0

Any function of

Rm = {k ∈ K ∗|(k) + mPα ≥ 0}

Search the kernel

h ◦ φ(β) = h(β) = 0

Any function of

Rm = {k ∈ K ∗|(k) + mPα ≥ 0}

Search the kernel

h ◦ φ(β) = h(β) = 0

Any function of

Rm = {k ∈ K ∗|(k) + mPα ≥ 0}

Search the kernel

h ◦ φ(β) = h(β) = 0

Any function of

Rm = {k ∈ K ∗|(k) + mPα ≥ 0}

Done!?

• Rm ◦ φ = {f ◦ φ|f ∈ Rm}• Rm ◦ φ ⊆ Rmq

What about ψ : Rm ◦ φ→ Rm sending f ◦ φ f ?

Alas ! It is an isomorphism.

Luther King : We must accept finite disappointment, but never lose infinite hope.

Done!?

• Rm ◦ φ = {f ◦ φ|f ∈ Rm}

• Rm ◦ φ ⊆ Rmq

Done!?

• Rm ◦ φ = {f ◦ φ|f ∈ Rm}• Rm ◦ φ ⊆ Rmq

Done!?

• Rm ◦ φ = {f ◦ φ|f ∈ Rm}• Rm ◦ φ ⊆ Rmq

Done!?

• Rm ◦ φ = {f ◦ φ|f ∈ Rm}• Rm ◦ φ ⊆ Rmq

Done!?

• Rm ◦ φ = {f ◦ φ|f ∈ Rm}• Rm ◦ φ ⊆ Rmq

Done!?

• Rm ◦ φ = {f ◦ φ|f ∈ Rm}• Rm ◦ φ ⊆ Rmq

When in doubt, multiply

For A, B, subspaces of C ,

aibi |ai ∈ A, bi ∈ B

ψ : Rl(Rm ◦ φ) → RlRm.

Any f ∈ kernel(ψ) has a zero at all rational points except maybe α.

(a1 − 1) ≤ deg((f )0) = deg((f )∞) ≤ l + mq.

Can we do better ?

ψ : Rpe

l (Rm ◦ φ) → Rpe

Choose e such that pe < q so that any element in the domain is ape th power.

Any f ∈ kernel(ψ) has a zero of order atleast pe at all rationalpoints except maybe α.

Can we do better ?

ψ : Rpe

Can we do better ?

ψ : Rpe

Can we do better ?

ψ : Rpe

It works!

• ∃ a ‘nice’ basis {f1, f2, . . . ft} of Rm whereordPα(fi ) < ordPα(fi+1) for all i .

• Rpe

l ⊗ (Rm ◦ φ) ∼=F Rpe

l (Rm ◦ φ) if lpe < q.

• ψ : Rpe

l Rm which sends∑gpe

i (fi ◦ φ) gpe

i fi .

• Domain is a subspace of Rlpe+mq and hence any element ofthe domain can have a pole only at α of order atmostlpe + mq.

• Any element in the domain is defined for all rational pointsexcept maybe at α. So a zero of order atleast pe is assured atall rational points except α.

pe(a1 − 1) ≤ deg((f )0) = deg((f )∞) ≤ lpe + mq.

It works!

• Rpe

• ψ : Rpe

i (fi ◦ φ) gpe

i fi .

It works!

• Rpe

• ψ : Rpe

i (fi ◦ φ) gpe

i fi .

It works!

• Rpe

• ψ : Rpe

i (fi ◦ φ) gpe

i fi .

It works!

• Rpe

• ψ : Rpe

i (fi ◦ φ) gpe

i fi .

It works!

• Rpe

• ψ : Rpe

i (fi ◦ φ) gpe

i fi .

It works!

• Rpe

• ψ : Rpe

i (fi ◦ φ) gpe

i fi .

How do we ensure that the kernel is nonzero ?

Riemann-Roch to the rescue

L(D) = {k ∈ K ∗|(k) + D ≥ 0} and dimF (L(D)) = l(D).

Theorem (Riemann inequality)

For any divisor D

l(D) ≥ deg(D)− g + 1

where g is the genus of K.

Corollary

For a divisor D whose degree is strictly greater than 2g − 2,

l(D) = deg(D)− g + 1.

For any divisor D

l(D) ≥ deg(D)− g + 1

Corollary

l(D) = deg(D)− g + 1.

For any divisor D

l(D) ≥ deg(D)− g + 1

Corollary

l(D) = deg(D)− g + 1.

Various dimensions

ψ : Rpe

If l ,m ≥ g ,

l (Rm ◦ φ))

= dimF

l ⊗ (Rm ◦ φ))

= dimF

)× dimF (Rm ◦ φ)

= dimF (Rl)× dimF (Rm)

≥ (l − g + 1)(m − g + 1)

dimF (image(ψ)) ≤ dimF (Rpe

≤ dimF (Rlpe+m)

= lpe + m − g + 1

Various dimensions

ψ : Rpe

If l ,m ≥ g ,

l (Rm ◦ φ))

= dimF

l ⊗ (Rm ◦ φ))

= dimF

≥ (l − g + 1)(m − g + 1)

≤ dimF (Rlpe+m)

= lpe + m − g + 1

Various dimensions

ψ : Rpe

If l ,m ≥ g ,

l (Rm ◦ φ))

= dimF

l ⊗ (Rm ◦ φ))

= dimF

≥ (l − g + 1)(m − g + 1)

≤ dimF (Rlpe+m)

= lpe + m − g + 1

Various dimensions

ψ : Rpe

If l ,m ≥ g ,

l (Rm ◦ φ))

= dimF

l ⊗ (Rm ◦ φ))

= dimF

≥ (l − g + 1)(m − g + 1)

≤ dimF (Rlpe+m)

= lpe + m − g + 1

Various dimensions

ψ : Rpe

If l ,m ≥ g ,

l (Rm ◦ φ))

= dimF

l ⊗ (Rm ◦ φ))

= dimF

≥ (l − g + 1)(m − g + 1)

≤ dimF (Rlpe+m)

= lpe + m − g + 1

All is well that ends well

dimF kernel(ψ) ≥ (l − g + 1)(m − g + 1)− (lpe + m − g + 1).

Choose e, l ,m such that

• lpe < q.

• l ,m ≥ g

• (l − g + 1)(m − g + 1) > (lpe + m − g + 1).

pe =√

q , m =√

q + 2g and l =[

]+ g + 1 does the job.

• lpe < q.

• l ,m ≥ g

• (l − g + 1)(m − g + 1) > (lpe + m − g + 1).

pe =√

q , m =√

q + 2g and l =[

• lpe < q.

• l ,m ≥ g

• (l − g + 1)(m − g + 1) > (lpe + m − g + 1).

pe =√

q , m =√

q + 2g and l =[

• lpe < q.

• l ,m ≥ g

• (l − g + 1)(m − g + 1) > (lpe + m − g + 1).

pe =√

q , m =√

q + 2g and l =[

• lpe < q.

• l ,m ≥ g

• (l − g + 1)(m − g + 1) > (lpe + m − g + 1).

pe =√

q , m =√

q + 2g and l =[

• lpe < q.

• l ,m ≥ g

• (l − g + 1)(m − g + 1) > (lpe + m − g + 1).

pe =√

q , m =√

q + 2g and l =[

• lpe < q.

• l ,m ≥ g

• (l − g + 1)(m − g + 1) > (lpe + m − g + 1).

pe =√

q , m =√

q + 2g and l =[

riemann hypothesis for function fields bombieri's proof

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