riemann hypothesis for function fields bombieri's proof
Post on 12-Sep-2021
4 Views
Preview:
TRANSCRIPT
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
Riemann Hypothesis
Analogously
Credits
Bounds
The language of curves
The correspondence
The proof
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
A natural construction
ζ(s) :=∑n∈N
1
ns
• Primes : 2,3,5,7,11,. . .
• Q∗+ : Free abelian group generated by the primes under
multiplication.
• N: Set of non-negative combination of primes.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
A natural construction
ζ(s) :=∑n∈N
1
ns
• Primes : 2,3,5,7,11,. . .
• Q∗+ : Free abelian group generated by the primes under
multiplication.
• N: Set of non-negative combination of primes.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
A natural construction
ζ(s) :=∑n∈N
1
ns
• Primes : 2,3,5,7,11,. . .
• Q∗+ : Free abelian group generated by the primes under
multiplication.
• N: Set of non-negative combination of primes.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
A natural construction
ζ(s) :=∑n∈N
1
ns
• Primes : 2,3,5,7,11,. . .
• Q∗+ : Free abelian group generated by the primes under
multiplication.
• N: Set of non-negative combination of primes.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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
(s2
)ζ(s)
• Γ(z) =∫∞0
tz−1e−tdt
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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
(s2
)ζ(s)
• Γ(z) =∫∞0
tz−1e−tdt
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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
(s2
)ζ(s)
• Γ(z) =∫∞0
tz−1e−tdt
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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
(s2
)ζ(s)
• Γ(z) =∫∞0
tz−1e−tdt
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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
(s2
)ζ(s)
• Γ(z) =∫∞0
tz−1e−tdt
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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
(s2
)ζ(s)
• Γ(z) =∫∞0
tz−1e−tdt
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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
(s2
)ζ(s)
• Γ(z) =∫∞0
tz−1e−tdt
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
The eighth problem
Hypothesis (Riemann, 1859)
All the non-trivial zeroes of the ζ function lie on the lineRe(s) = 1
2 .
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
Riemann Hypothesis
Analogously
Credits
Bounds
The language of curves
The correspondence
The proof
Riemann Hypothesis Analogously Credits Bounds The language of curves 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 .
Riemann Hypothesis Analogously Credits Bounds The language of curves 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 .
Riemann Hypothesis Analogously Credits Bounds The language of curves 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 .
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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}.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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}.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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}.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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}.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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}.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
An effective construction
ζK (s) :=∑
{A∈DK |A≥0}
1
(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).
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
An effective construction
ζK (s) :=∑
{A∈DK |A≥0}
1
(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).
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
An effective construction
ζK (s) :=∑
{A∈DK |A≥0}
1
(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).
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
An effective construction
ζK (s) :=∑
{A∈DK |A≥0}
1
(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).
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
An effective construction
ζK (s) :=∑
{A∈DK |A≥0}
1
(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).
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
An effective construction
ζK (s) :=∑
{A∈DK |A≥0}
1
(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).
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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).
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
and etc.
• ζK (s) =∏
P∈SK
11−(NP)−s .
• ξ(s) = ξ(1− s) where ξ(s) = q(g−1)sζK (s).
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
and etc.
• ζK (s) =∏
P∈SK
11−(NP)−s .
• ξ(s) = ξ(1− s) where ξ(s) = q(g−1)sζK (s).
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
and etc.
• ζK (s) =∏
P∈SK
11−(NP)−s .
• ξ(s) = ξ(1− s) where ξ(s) = q(g−1)sζK (s).
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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 language of curves The correspondence The proof
Riemann Hypothesis
Analogously
Credits
Bounds
The language of curves
The correspondence
The proof
Riemann Hypothesis Analogously Credits Bounds The language of curves 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.
Riemann Hypothesis Analogously Credits Bounds The language of curves 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.
Riemann Hypothesis Analogously Credits Bounds The language of curves 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.
Riemann Hypothesis Analogously Credits Bounds The language of curves 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.
Riemann Hypothesis Analogously Credits Bounds The language of curves 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.
Riemann Hypothesis Analogously Credits Bounds The language of curves 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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
Riemann Hypothesis
Analogously
Credits
Bounds
The language of curves
The correspondence
The proof
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
LK (u) =
2g∏i=1
(1− αiu).
Hypothesis (R.H restated)
Let u = q−s . Then,|αi | =
√q.
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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
LK (u) =
2g∏i=1
(1− αiu).
Hypothesis (R.H restated)
Let u = q−s . Then,|αi | =
√q.
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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
LK (u) =
2g∏i=1
(1− αiu).
Hypothesis (R.H restated)
Let u = q−s . Then,|αi | =
√q.
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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
Weil bound
Theorem (Weil bound)
−2g√
q + q + 1 ≤ a1 ≤ 2g√
q + q + 1.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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 language of curves The correspondence The proof
Riemann Hypothesis
Analogously
Credits
Bounds
The language of curves
The correspondence
The proof
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
A picture is worth ..
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
.. 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 ].
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
.. 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 ].
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
.. 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 ].
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
.. 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 ].
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
.. 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 ].
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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 .
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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 .
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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 .
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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 .
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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 .
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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
qN ]
• F -rational points are also the fixed points of φ.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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
qN ]
• F -rational points are also the fixed points of φ.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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
qN ]
• F -rational points are also the fixed points of φ.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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
qN ]
• F -rational points are also the fixed points of φ.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
Riemann Hypothesis
Analogously
Credits
Bounds
The language of curves
The correspondence
The proof
Riemann Hypothesis Analogously Credits Bounds The language of curves 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) =
{f
g∈ K |g(α) 6= 0
}.
is the discrete valuation ring associated with it.
Riemann Hypothesis Analogously Credits Bounds The language of curves 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) =
{f
g∈ K |g(α) 6= 0
}.
is the discrete valuation ring associated with it.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
T : C → Primes of K
αr
•
•
α2
α1
Pα
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 language of curves The correspondence The proof
T : C → Primes of K
αr
•
•
α2
α1
Pα
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 language of curves The correspondence The proof
T : C → Primes of K
αr
•
•
α2
α1
Pα
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 language of curves The correspondence The proof
Riemann Hypothesis
Analogously
Credits
Bounds
The language of curves
The correspondence
The proof
Riemann Hypothesis Analogously Credits Bounds The language of curves 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 )∞)
Riemann Hypothesis Analogously Credits Bounds The language of curves 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 )∞)
Riemann Hypothesis Analogously Credits Bounds The language of curves 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 )∞)
Riemann Hypothesis Analogously Credits Bounds The language of curves 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 )∞)
Riemann Hypothesis Analogously Credits Bounds The language of curves 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 )∞)
Riemann Hypothesis Analogously Credits Bounds The language of curves 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 )∞)
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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).
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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).
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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).
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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).
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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).
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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).
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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).
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
When in doubt, multiply
For A, B, subspaces of C ,
AB :=
{r∑
i=1
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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
When in doubt, multiply
For A, B, subspaces of C ,
AB :=
{r∑
i=1
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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
When in doubt, multiply
For A, B, subspaces of C ,
AB :=
{r∑
i=1
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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
Can we do better ?
ψ : Rpe
l (Rm ◦ φ) → Rpe
l Rm.
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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
Can we do better ?
ψ : Rpe
l (Rm ◦ φ) → Rpe
l Rm.
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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
Can we do better ?
ψ : Rpe
l (Rm ◦ φ) → Rpe
l Rm.
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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
Can we do better ?
ψ : Rpe
l (Rm ◦ φ) → Rpe
l Rm.
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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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 ◦ φ) → 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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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 ◦ φ) → 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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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 ◦ φ) → 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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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 ◦ φ) → 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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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 ◦ φ) → 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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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 ◦ φ) → 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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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 ◦ φ) → 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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
How do we ensure that the kernel is nonzero ?
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
Various dimensions
ψ : Rpe
l (Rm ◦ φ) → Rpe
l Rm.
If l ,m ≥ g ,
dimF
(Rpe
l (Rm ◦ φ))
= dimF
(Rpe
l ⊗ (Rm ◦ φ))
= dimF
(Rpe
l
)× dimF (Rm ◦ φ)
= dimF (Rl)× dimF (Rm)
≥ (l − g + 1)(m − g + 1)
dimF (image(ψ)) ≤ dimF (Rpe
l Rm)
≤ dimF (Rlpe+m)
= lpe + m − g + 1
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
Various dimensions
ψ : Rpe
l (Rm ◦ φ) → Rpe
l Rm.
If l ,m ≥ g ,
dimF
(Rpe
l (Rm ◦ φ))
= dimF
(Rpe
l ⊗ (Rm ◦ φ))
= dimF
(Rpe
l
)× dimF (Rm ◦ φ)
= dimF (Rl)× dimF (Rm)
≥ (l − g + 1)(m − g + 1)
dimF (image(ψ)) ≤ dimF (Rpe
l Rm)
≤ dimF (Rlpe+m)
= lpe + m − g + 1
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
Various dimensions
ψ : Rpe
l (Rm ◦ φ) → Rpe
l Rm.
If l ,m ≥ g ,
dimF
(Rpe
l (Rm ◦ φ))
= dimF
(Rpe
l ⊗ (Rm ◦ φ))
= dimF
(Rpe
l
)× dimF (Rm ◦ φ)
= dimF (Rl)× dimF (Rm)
≥ (l − g + 1)(m − g + 1)
dimF (image(ψ)) ≤ dimF (Rpe
l Rm)
≤ dimF (Rlpe+m)
= lpe + m − g + 1
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
Various dimensions
ψ : Rpe
l (Rm ◦ φ) → Rpe
l Rm.
If l ,m ≥ g ,
dimF
(Rpe
l (Rm ◦ φ))
= dimF
(Rpe
l ⊗ (Rm ◦ φ))
= dimF
(Rpe
l
)× dimF (Rm ◦ φ)
= dimF (Rl)× dimF (Rm)
≥ (l − g + 1)(m − g + 1)
dimF (image(ψ)) ≤ dimF (Rpe
l Rm)
≤ dimF (Rlpe+m)
= lpe + m − g + 1
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
Various dimensions
ψ : Rpe
l (Rm ◦ φ) → Rpe
l Rm.
If l ,m ≥ g ,
dimF
(Rpe
l (Rm ◦ φ))
= dimF
(Rpe
l ⊗ (Rm ◦ φ))
= dimF
(Rpe
l
)× dimF (Rm ◦ φ)
= dimF (Rl)× dimF (Rm)
≥ (l − g + 1)(m − g + 1)
dimF (image(ψ)) ≤ dimF (Rpe
l Rm)
≤ dimF (Rlpe+m)
= lpe + m − g + 1
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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√
qg+1
]+ g + 1 does the job.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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√
qg+1
]+ g + 1 does the job.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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√
qg+1
]+ g + 1 does the job.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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√
qg+1
]+ g + 1 does the job.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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√
qg+1
]+ g + 1 does the job.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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√
qg+1
]+ g + 1 does the job.
Riemann Hypothesis Analogously Credits Bounds The language of curves The correspondence The proof
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√
qg+1
]+ g + 1 does the job.
top related