Numerical semigroups and codes

Numerical semigroups and codes

Lecturer: Maria Bras-Amorós

Mon Numerical Semigroups. The Paradigmatic Example of Weierstrass Semigroups

Tue Classification, Characterization and Counting of Semigroups Wed Semigroup and Alegebraic Geometry Codes

Thu Semigroup Ideals and Generalized Hamming Weights Fri R-molds of Numerical Semigroups with Musical Motivation References can be found in

http://crises-deim.urv.cat/∼mbras/cimpa2017

Numerical Semigroups. The Paradigmatic Example of Weierstrass Semigroups

Maria Bras-Amorós

CIMPA Research School Algebraic Methods in Coding Theory

Ubatuba, July 3-7, 2017

Contents

1 Algebraic curves

2 Weierstrass semigroup

3 Examples

4 Bounding the number of points of a curve

Algebraic curves

William Fulton.

Algebraic curves.

Advanced Book Classics. Addison-Wesley Publishing Company Advanced Book Program, Redwood City, CA, 1989.

Massimo Giulietti.

Notes on algebraic-geometric codes.

www.math.kth.se/math/forskningsrapporter/Giulietti.pdf.

Tom Høholdt, Jacobus H. van Lint, and Ruud Pellikaan.

Algebraic Geometry codes, pages 871–961.

North-Holland, Amsterdam, 1998.

Oliver Pretzel.

Codes and algebraic curves, volume 8 ofOxford Lecture Series in Mathematics and its Applications.

The Clarendon Press Oxford University Press, New York, 1998.

Henning Stichtenoth.

Algebraic function fields and codes.

Universitext. Springer-Verlag, Berlin, 1993.

Fernando Torres.

Notes on Goppa codes.

www.ime.unicamp.br/˜ftorres/RESEARCH/ARTS_PDF/codes.pdf.

Plane curves

LetKbe a field with algebraic closureK.¯ LetP²(¯K)be the projective plane overK:¯ P²(¯K) ={[a:b:c] : (a,b,c)∈K¯³\{(0,0,0)}}/

([a:b:c]∼[a^′:b^′:c^′]⇐⇒ ^{(a,b,c) =λ(a′,b′,c′)}

for someλ6=0 )

Plane curves

LetKbe a field with algebraic closureK.¯ LetP²(¯K)be the projective plane overK:¯ P²(¯K) ={[a:b:c] : (a,b,c)∈K¯³\{(0,0,0)}}/

([a:b:c]∼[a^′:b^′:c^′]⇐⇒ ^{(a,b,c) =λ(a′,b′,c′)}

for someλ6=0 )

Affine curve

Letf(x,y)∈K[x,y].

Theaffine curveassociated tof is the set of points {(a,b)∈K¯² :f(a,b) =0}

Plane curves

Projective curve

LetF(X,Y,Z)∈K[X,Y,Z]be ahomogeneouspolynomial.

Theprojective curveassociated toFis the set of points X_F={(a:b:c)∈P²(¯K) :F(a:b:c) =0}

Homogenization and dehomogenization

Affine to projective

Thehomogenizationoff ∈K[x,y]is f^∗(X,Y,Z) =Z^deg(f)f

X Z,Y

Z

.

The points(a,b)∈K¯²of the affine curve defined byf(x,y)correspond to the points(a:b:1)∈P²(¯K)ofX_f^∗.

Homogenization and dehomogenization

Projective to affine

A projective curve defined by a homogeneous polynomialF(X,Y,Z) defines three affine curves withdehomogenizedpolynomials

F(x,y,1),F(1,u,v),F(w,1,z).

The points(X:Y:Z)withZ6=0 (resp.X6=0,Y6=0) ofX_Fcorrespond to the points of the affine curve defined byF(x,y,1)(resp.F(1,u,v), F(w,1,z)). The points withZ=0 are said to beat infinity.

Hermitian example

Letqbe a prime power.

TheHermitian curveH_qoverF_q2is defined by

x^q+1=y^q+yandX^q+1−Y^qZ−YZ^q=0.

Hermitian example

Letqbe a prime power.

TheHermitian curveH_qoverF_q2is defined by

x^q+1=y^q+yandX^q+1−Y^qZ−YZ^q=0.

Exercise

Letq=2,F_q2 =Z₂/(x²+x+1),αthe class ofx. Then, F₄={0,1, α, α²=1+α}.

DoesH2have points at infinity? Find all the points ofH2.

Hermitian example

Letqbe a prime power.

TheHermitian curveH_qoverF_q2is defined by

x^q+1=y^q+yandX^q+1−Y^qZ−YZ^q=0.

Exercise

Letq=2,F_q2 =Z₂/(x²+x+1),αthe class ofx. Then, F₄={0,1, α, α²=1+α}.

DoesH2have points at infinity? Find all the points ofH2.

The unique point at infinity isP∞= (0:1:0).

Hermitian example

Letqbe a prime power.

TheHermitian curveH_qoverF_q2is defined by

x^q+1=y^q+yandX^q+1−Y^qZ−YZ^q=0.

Exercise

Letq=2,F_q2 =Z₂/(x²+x+1),αthe class ofx. Then, F₄={0,1, α, α²=1+α}.

DoesH2have points at infinity? Find all the points ofH2.

The unique point at infinity isP∞= (0:1:0). For the remaining points, notice that

0q+1=0 1q+1=1 αq+1=1 (α2)q+1=1 0q+0=0 1q+1=0 αq+α=1 (α2)q+α2=1

Hermitian example

Letqbe a prime power.

TheHermitian curveH_qoverF_q2is defined by

x^q+1=y^q+yandX^q+1−Y^qZ−YZ^q=0.

Exercise

Letq=2,F_q2 =Z₂/(x²+x+1),αthe class ofx. Then, F₄={0,1, α, α²=1+α}.

DoesH2have points at infinity? Find all the points ofH2.

The unique point at infinity isP∞= (0:1:0). For the remaining points, notice that

0q+1=0 1q+1=1 αq+1=1 (α2)q+1=1 0q+0=0 1q+1=0 αq+α=1 (α2)q+α2=1

Then the points are:

P1= (0:0:1)≡(0,0),P2= (0:1:1)≡(0,1),P3= (1:α:1)≡(1, α),P4= (1:α2:1)≡(1, α2), P5= (α:α:1)≡(α, α),P6= (α:α2:1)≡(α, α2),P7= (α2:α:1)≡(α2, α),P8= (α2:α2:1)≡ (α2, α2)

Irreducibility

IfFcan factor in a field extension ofKthen the curve is a proper union of at least two curves.

Irreducibility

IfFcan factor in a field extension ofKthen the curve is a proper union of at least two curves.

Hence, we imposeFto be irreducible in any field extension ofK.

Irreducibility

IfFcan factor in a field extension ofKthen the curve is a proper union of at least two curves.

Hence, we imposeFto be irreducible in any field extension ofK.

In this case we say thatFisabsolutely irreducible.

Function field

G(X,Y,Z)−H(X,Y,Z) =mF(X,Y,Z) =⇒G(a,b,c) =H(a,b,c)for all(a:b:c)∈ X_F.

Function field

G(X,Y,Z)−H(X,Y,Z) =mF(X,Y,Z) =⇒G(a,b,c) =H(a,b,c)for all(a:b:c)∈ X_F. So, we consider

K(X,Y,Z)/(F) ={G(X,Y,Z)∈K(X,Y,Z)}/_(G∼H⇐⇒G−H=mF)

Function field

G(X,Y,Z)−H(X,Y,Z) =mF(X,Y,Z) =⇒G(a,b,c) =H(a,b,c)for all(a:b:c)∈ X_F. So, we consider

K(X,Y,Z)/(F) ={G(X,Y,Z)∈K(X,Y,Z)}/_(G∼H⇐⇒G−H=mF)

SinceFis irreducible,K(X,Y,Z)/(F)is an integral domain and we can construct its field of fractionsQF.

Function field

G(X,Y,Z)−H(X,Y,Z) =mF(X,Y,Z) =⇒G(a,b,c) =H(a,b,c)for all(a:b:c)∈ X_F. So, we consider

K(X,Y,Z)/(F) ={G(X,Y,Z)∈K(X,Y,Z)}/_(G∼H⇐⇒G−H=mF)

SinceFis irreducible,K(X,Y,Z)/(F)is an integral domain and we can construct its field of fractionsQF.

For evaluating one such fraction at a projective point we want the result not to depend on the representative of the projective point.

Hence, we require the numerator and the denominator to have one representative each, which is a homogeneous polynomial and both having the same degree.

Function field

G(X,Y,Z)−H(X,Y,Z) =mF(X,Y,Z) =⇒G(a,b,c) =H(a,b,c)for all(a:b:c)∈ X_F. So, we consider

K(X,Y,Z)/(F) ={G(X,Y,Z)∈K(X,Y,Z)}/_(G∼H⇐⇒G−H=mF)

SinceFis irreducible,K(X,Y,Z)/(F)is an integral domain and we can construct its field of fractionsQF.

For evaluating one such fraction at a projective point we want the result not to depend on the representative of the projective point.

Hence, we require the numerator and the denominator to have one representative each, which is a homogeneous polynomial and both having the same degree.

Thefunction fieldofX_F, denotedK(X_F), is the set of elements ofQ_F admitting one such representation.

Function field

G(X,Y,Z)−H(X,Y,Z) =mF(X,Y,Z) =⇒G(a,b,c) =H(a,b,c)for all(a:b:c)∈ X_F. So, we consider

K(X,Y,Z)/(F) ={G(X,Y,Z)∈K(X,Y,Z)}/_(G∼H⇐⇒G−H=mF)

SinceFis irreducible,K(X,Y,Z)/(F)is an integral domain and we can construct its field of fractionsQF.

For evaluating one such fraction at a projective point we want the result not to depend on the representative of the projective point.

Hence, we require the numerator and the denominator to have one representative each, which is a homogeneous polynomial and both having the same degree.

Thefunction fieldofX_F, denotedK(X_F), is the set of elements ofQ_F admitting one such representation.

Its elements are therational functionsofX_F.

Regular functions

We say that a rational functionf ∈K(X_F)isregular in a pointPif there exists a representation of it as a fraction ^G(X,Y,Z)_H(X,Y,Z)withH(P)6=0.

Regular functions

We say that a rational functionf ∈K(X_F)isregular in a pointPif there exists a representation of it as a fraction ^G(X,Y,Z)_H(X,Y,Z)withH(P)6=0.

In this case we define

f(P) = G(P) H(P).

Regular functions

We say that a rational functionf ∈K(X_F)isregular in a pointPif there exists a representation of it as a fraction ^G(X,Y,Z)_H(X,Y,Z)withH(P)6=0.

In this case we define

f(P) = G(P) H(P).

The ring of all rational functions regular inPis denotedO_P.

Regular functions

We say that a rational functionf ∈K(X_F)isregular in a pointPif there exists a representation of it as a fraction ^G(X,Y,Z)_H(X,Y,Z)withH(P)6=0.

In this case we define

f(P) = G(P) H(P).

The ring of all rational functions regular inPis denotedO_P. Again it is an integral domain and this time its field of fractions is K(XF).

Singularities

LetP∈ X_Fbe a point. If all the partial derivativesFX,FY,FZvanish at PthenPis said to be asingular point. Otherwise it is said to be a simple point.

Singularities

LetP∈ X_Fbe a point. If all the partial derivativesFX,FY,FZvanish at PthenPis said to be asingular point. Otherwise it is said to be a simple point.

Curves without singular points are callednon-singular,regularor smooth curves.

Singularities

LetP∈ X_Fbe a point. If all the partial derivativesFX,FY,FZvanish at PthenPis said to be asingular point. Otherwise it is said to be a simple point.

Curves without singular points are callednon-singular,regularor smooth curves.

Thetangent lineat a singular pointPofX_Fis defined by the equation F_X(P)X+F_Y(P)Y+F_Z(P)Z=0.

Singularities

LetP∈ X_Fbe a point. If all the partial derivativesFX,FY,FZvanish at PthenPis said to be asingular point. Otherwise it is said to be a simple point.

Curves without singular points are callednon-singular,regularor smooth curves.

Thetangent lineat a singular pointPofX_Fis defined by the equation F_X(P)X+F_Y(P)Y+F_Z(P)Z=0.

From now on we will assume thatFis absolutely irreducible and that X_Fis smooth.

Hermitian example

Letqbe a prime power.

TheHermitian curveH_qoverF_q2is defined by

x^q+1=y^q+yandX^q+1−Y^qZ−YZ^q=0.

Hermitian example

Letqbe a prime power.

TheHermitian curveH_qoverF_q2is defined by

x^q+1=y^q+yandX^q+1−Y^qZ−YZ^q=0.

Exercise

Find the partial derivatives ofH_q Are there singular points?

What is the tangent line atP∞?

Hermitian example

Letqbe a prime power.

TheHermitian curveH_qoverF_q2is defined by

x^q+1=y^q+yandX^q+1−Y^qZ−YZ^q=0.

Exercise

Find the partial derivatives ofH_qF_X=X^q,F_Y=−Z^q,F_Z=−Y^q Are there singular points?No

What is the tangent line atP∞?

F_X(P∞)X+F_Y(P∞)Y+F_Z(P∞)Z=−Z=0.

Genus

Thegenusof a smooth plane curveX_Fmay be defined as g= (deg(F)−1)(deg(F)−2)

2 .

Genus

Thegenusof a smooth plane curveX_Fmay be defined as g= (deg(F)−1)(deg(F)−2)

2 .

For general curves the genus is defined using differentials on a curve.

Genus

Thegenusof a smooth plane curveX_Fmay be defined as g= (deg(F)−1)(deg(F)−2)

2 .

For general curves the genus is defined using differentials on a curve.

Exercise

What is in general the genus ofHq?

Genus

Thegenusof a smooth plane curveX_Fmay be defined as g= (deg(F)−1)(deg(F)−2)

2 .

For general curves the genus is defined using differentials on a curve.

Exercise

What is in general the genus ofHq? ^q(q−₂¹⁾

Weierstrass semigroup

Valuation at a point

Theorem

Consider a point P in the projective curveX_F. There exists t∈ O_Psuch that for any non-zero f ∈K(X_F)there exists a unique integer v_P(f)with

f =t^vP(f)u for some u∈ O_Pwith u(P)6=0.

Valuation at a point

Theorem

Consider a point P in the projective curveX_F. There exists t∈ O_Psuch that for any non-zero f ∈K(X_F)there exists a unique integer v_P(f)with

f =t^vP(f)u for some u∈ O_Pwith u(P)6=0.

The valuev_P(f)depends only onX_F,P.

Valuation at a point

Theorem

Consider a point P in the projective curveX_F. There exists t∈ O_Psuch that for any non-zero f ∈K(X_F)there exists a unique integer v_P(f)with

f =t^vP(f)u for some u∈ O_Pwith u(P)6=0.

The valuev_P(f)depends only onX_F,P.

IfG(X,Y,Z)andH(X,Y,Z)are two homogeneous polynomials of degree 1 such thatG(P) =0,H(P)6=0, andGis not a constant multiple ofFX(P)X+FY(P)Y+FZ(P)Z, then we can taketto be the class inO_Pof ^G(X,Y,Z)_H(X,Y,Z).

Valuation at a point

An element such astis called alocal parameter.

Valuation at a point

An element such astis called alocal parameter.

The valuevP(f)is called thevaluationoff atP.

Valuation at a point

An element such astis called alocal parameter.

The valuevP(f)is called thevaluationoff atP.

The pointPis said to be azeroof multiplicitymifv_P(f) =m>0 and apoleof multiplicity−mifv_P(f) =m<0.

Valuation at a point

An element such astis called alocal parameter.

The valuevP(f)is called thevaluationoff atP.

The pointPis said to be azeroof multiplicitymifv_P(f) =m>0 and apoleof multiplicity−mifv_P(f) =m<0.

The valuation satisfies thatvP(f)>0 if and only iff ∈ O_Pand that in this casevP(f)>0 if and only iff(P) =0.

Valuation at a point

Lemma

1 vP(f) =∞if and only if f =0

2 vP(λf) =vP(f)for all non-zeroλ∈K

3 v_P(fg) =v_P(f) +v_P(g)

4 vP(f+g)>min{vP(f),vP(g)}and equality holds if vP(f)6=vP(g)

5 If v_P(f) =v_P(g)>0then there existsλ∈K such that v_P(f−λg)>v_P(f).

Valuation at a point

LetL(mP)be the set of rational functions having only poles atPand with pole order at mostm.

Valuation at a point

LetL(mP)be the set of rational functions having only poles atPand with pole order at mostm.

LetA=S

m>0L(mP), that is,Ais the ring of rational functions having poles only atP.

Valuation at a point

LetL(mP)be the set of rational functions having only poles atPand with pole order at mostm.

LetA=S

m>0L(mP), that is,Ais the ring of rational functions having poles only atP.

L(mP)is aK-vector space and so we can definel(mP)=dimK(L(mP)).

Valuation at a point

LetL(mP)be the set of rational functions having only poles atPand with pole order at mostm.

LetA=S

m>0L(mP), that is,Ais the ring of rational functions having poles only atP.

L(mP)is aK-vector space and so we can definel(mP)=dimK(L(mP)).

One can prove thatl(mP)is eitherl((m−1)P)orl((m−1)P) +1 and l(mP) =l((m−1)P) +1⇐⇒ ∃f ∈Awithv_P(f) =−m

Valuation at a point

LetL(mP)be the set of rational functions having only poles atPand with pole order at mostm.

LetA=S

m>0L(mP), that is,Ais the ring of rational functions having poles only atP.

L(mP)is aK-vector space and so we can definel(mP)=dimK(L(mP)).

One can prove thatl(mP)is eitherl((m−1)P)orl((m−1)P) +1 and l(mP) =l((m−1)P) +1⇐⇒ ∃f ∈Awithv_P(f) =−m DefineΛ ={−vP(f) :f ∈A\ {0}}.

Valuation at a point

LetL(mP)be the set of rational functions having only poles atPand with pole order at mostm.

LetA=S

m>0L(mP), that is,Ais the ring of rational functions having poles only atP.

L(mP)is aK-vector space and so we can definel(mP)=dimK(L(mP)).

One can prove thatl(mP)is eitherl((m−1)P)orl((m−1)P) +1 and l(mP) =l((m−1)P) +1⇐⇒ ∃f ∈Awithv_P(f) =−m DefineΛ ={−vP(f) :f ∈A\ {0}}.

Obviously,Λ⊆N₀.

Weierstrass semigroup

Lemma

The setΛ⊆N₀satisfies

1 0∈Λ

2 m+m^′∈Λwhenever m,m^′∈Λ

3 N₀\Λhas a finite number of elements Proof:

1 Constant functionsf =ahave no poles and satisfyv_P(a) =0 for allP∈ XF. Hence, 0∈Λ.

Weierstrass semigroup

Lemma

The setΛ⊆N₀satisfies

1 0∈Λ

2 m+m^′∈Λwhenever m,m^′∈Λ

3 N₀\Λhas a finite number of elements Proof:

2 Ifm,m^′∈Λthen there existf,g∈Awithv_P(f) =−m, vP(g) =−m^′.

vP(fg) =−(m+m^′) =⇒m+m^′∈Λ.

Weierstrass semigroup

Lemma

The setΛ⊆N₀satisfies

1 0∈Λ

2 m+m^′∈Λwhenever m,m^′∈Λ

3 N₀\Λhas a finite number of elements Proof:

3 The well-known Riemann-Roch theorem implies that l(mP) =m+1−g

ifm>2g−1.

On one hand this means thatm∈Λfor allm>2g, and on the other hand, this means thatl(mP) =l((m−1)P)only forgvalues ofm.

=⇒#(N₀\Λ) =g.

Weierstrass semigroup

Lemma

The setΛ⊆N₀satisfies

1 0∈Λ

2 m+m^′∈Λwhenever m,m^′∈Λ

3 N₀\Λhas a finite number of elements

The three properties of a subset ofN₀in the lemma constitute the definition of anumerical semigroup.

Weierstrass semigroup

Lemma

The setΛ⊆N₀satisfies

1 0∈Λ

2 m+m^′∈Λwhenever m,m^′∈Λ

3 N₀\Λhas a finite number of elements

The three properties of a subset ofN₀in the lemma constitute the definition of anumerical semigroup.

The particular numerical semigroup of the lemma is called the Weierstrass semigroupatPand the elements inN₀\Λare called the Weierstrass gaps.

Numerical semigroups

Example: What amounts can be withdrawn?

Numerical semigroups

Example: What amounts can be withdrawn?

0e, 20e, 40e, 50e, 60e, 70e, 80e, 90e, 100e, . . .

Numerical semigroups

Example: What amounts can be withdrawn?

0e, 20e, 40e, 50e, 60e, 70e, 80e, 90e, 100e, . . .

0 in the set

Numerical semigroups

Example: What amounts can be withdrawn?

0e, 20e, 40e, 50e, 60e, 70e, 80e, 90e, 100e, . . .

0 in the set

s,s^′in the set=⇒s+s^′in the set

Numerical semigroups

Example: What amounts can be withdrawn?

0e, 20e, 40e, 50e, 60e, 70e, 80e, 90e, 100e, . . .

0 in the set

s,s^′in the set=⇒s+s^′in the set If we just consider multiples of 10 then

Numerical semigroups

Example: What amounts can be withdrawn?

0e, 20e, 40e, 50e, 60e, 70e, 80e, 90e, 100e, . . .

0 in the set

s,s^′in the set=⇒s+s^′in the set If we just consider multiples of 10 then

only 10e, 30earenotin the set (#(N₀\(S/10))<∞)

Examples

Hermitian example

Letqbe a prime power.

TheHermitian curveH_qoverF_q2is defined by

x^q+1=y^q+yandX^q+1−Y^qZ−YZ^q=0.

Hermitian example

Letqbe a prime power.

TheHermitian curveH_qoverF_q2is defined by

x^q+1=y^q+yandX^q+1−Y^qZ−YZ^q=0.

FX=X^q,FY=−Z^q,FZ=−Y^q=⇒no singular points.

Hermitian example

Letqbe a prime power.

TheHermitian curveH_qoverF_q2is defined by

x^q+1=y^q+yandX^q+1−Y^qZ−YZ^q=0.

FX=X^q,FY=−Z^q,FZ=−Y^q=⇒no singular points.

P∞= (0:1:0)is the unique point ofH_qat infinity (withZ=0).

Hermitian example

Letqbe a prime power.

TheHermitian curveH_qoverF_q2is defined by

x^q+1=y^q+yandX^q+1−Y^qZ−YZ^q=0.

FX=X^q,FY=−Z^q,FZ=−Y^q=⇒no singular points.

P∞= (0:1:0)is the unique point ofH_qat infinity (withZ=0).

t=^X_Yis a local parameter atP∞sinceFX(P∞)X+FY(P∞)Y+FZ(P∞)Z=−Z.

Hermitian example

Letqbe a prime power.

TheHermitian curveH_qoverF_q2is defined by

x^q+1=y^q+yandX^q+1−Y^qZ−YZ^q=0.

FX=X^q,FY=−Z^q,FZ=−Y^q=⇒no singular points.

P∞= (0:1:0)is the unique point ofH_qat infinity (withZ=0).

t=^X_Yis a local parameter atP∞sinceFX(P∞)X+FY(P∞)Y+FZ(P∞)Z=−Z.

X

Z,^Y_Z are regular everywhere except atP∞(⇒ ^X_Z,^Y_Z ∈ ∪_m>0L(mP∞)).

Hermitian example

Letqbe a prime power.

TheHermitian curveH_qoverF_q2is defined by

x^q+1=y^q+yandX^q+1−Y^qZ−YZ^q=0.

FX=X^q,FY=−Z^q,FZ=−Y^q=⇒no singular points.

P∞= (0:1:0)is the unique point ofH_qat infinity (withZ=0).

t=^X_Yis a local parameter atP∞sinceFX(P∞)X+FY(P∞)Y+FZ(P∞)Z=−Z.

X

Z,^Y_Z are regular everywhere except atP∞(⇒ ^X_Z,^Y_Z ∈ ∪_m>0L(mP∞)).

To find their valuation...

Hermitian example

Letqbe a prime power.

TheHermitian curveH_qoverF_q2is defined by

x^q+1=y^q+yandX^q+1−Y^qZ−YZ^q=0.

FX=X^q,FY=−Z^q,FZ=−Y^q=⇒no singular points.

P∞= (0:1:0)is the unique point ofH_qat infinity (withZ=0).

t=^X_Yis a local parameter atP∞sinceFX(P∞)X+FY(P∞)Y+FZ(P∞)Z=−Z.

X

Z,^Y_Z are regular everywhere except atP∞(⇒ ^X_Z,^Y_Z ∈ ∪_m>0L(mP∞)).

To find their valuation...

t^q+1= ^Z_Yq

+_Y^Z⇒v_P∞( ^Z_Yq

+^Z_Y) =q+1⇒v_P∞(^Z_Y) =q+1⇒ vP∞(^Y_Z) =−(q+1).

Hermitian example

Letqbe a prime power.

TheHermitian curveH_qoverF_q2is defined by

x^q+1=y^q+yandX^q+1−Y^qZ−YZ^q=0.

FX=X^q,FY=−Z^q,FZ=−Y^q=⇒no singular points.

P∞= (0:1:0)is the unique point ofH_qat infinity (withZ=0).

t=^X_Yis a local parameter atP∞sinceFX(P∞)X+FY(P∞)Y+FZ(P∞)Z=−Z.

X

Z,^Y_Z are regular everywhere except atP∞(⇒ ^X_Z,^Y_Z ∈ ∪_m>0L(mP∞)).

To find their valuation...

t^q+1= ^Z_Yq

+_Y^Z⇒v_P∞( ^Z_Yq

+^Z_Y) =q+1⇒v_P∞(^Z_Y) =q+1⇒ vP∞(^Y_Z) =−(q+1).

X Z

q+1

= ^Y_Zq

+^Y_Z ⇒(q+1)vP∞(^X_Z) =−q(q+1)⇒vP∞(^X_Z) =−q.

Hermitian example

⇒q,q+1∈Λ.

Hermitian example

⇒q,q+1∈Λ.

⇒Λcontains what we will call later the semigroup generated by q,q+1.

Hermitian example

⇒q,q+1∈Λ.

⇒Λcontains what we will call later the semigroup generated by q,q+1.

The complement inN₀of the semigroup generated byq,q+1 has

q(q−1)

2 =gelements.

Exercise

Can you prove that?

Hermitian example

⇒q,q+1∈Λ.

⇒Λcontains what we will call later the semigroup generated by q,q+1.

The complement inN₀of the semigroup generated byq,q+1 has

q(q−1)

2 =gelements.

Exercise

Can you prove that?The number of gaps is(q−1) + (q−2) +· · ·+1=q(q−1) 2

Hermitian example

⇒q,q+1∈Λ.

⇒Λcontains what we will call later the semigroup generated by q,q+1.

The complement inN₀of the semigroup generated byq,q+1 has

q(q−1)

2 =gelements.

Exercise

Can you prove that?The number of gaps is(q−1) + (q−2) +· · ·+1=q(q−1) 2

Since we know that the complement ofΛinN₀also hasgelements, this means that both semigroups are the same.

Klein example

TheKlein quarticoverF_qis defined by

x³y+y³+x=0andX³Y+Y³Z+Z³X=0

Klein example

TheKlein quarticoverF_qis defined by

x³y+y³+x=0andX³Y+Y³Z+Z³X=0 FX=3X²Y+Z³,FY=3Y²Z+X³,FZ=3Z²X+Y³.

Klein example

TheKlein quarticoverF_qis defined by

x³y+y³+x=0andX³Y+Y³Z+Z³X=0 FX=3X²Y+Z³,FY=3Y²Z+X³,FZ=3Z²X+Y³.

If the characteristic ofF_q2is 3 thenF_X=F_Y=F_Z=0 implies X³ =Y³ =Z³ =0⇒X=Y=Z=0⇒ Khas no singularities.

Klein example

TheKlein quarticoverF_qis defined by

x³y+y³+x=0andX³Y+Y³Z+Z³X=0 FX=3X²Y+Z³,FY=3Y²Z+X³,FZ=3Z²X+Y³.

If the characteristic ofF_q2is 3 thenF_X=F_Y=F_Z=0 implies X³ =Y³ =Z³ =0⇒X=Y=Z=0⇒ Khas no singularities.

If the characteristic ofF_q2is different than 3 thenFX=FY=FZ=0 impliesX³Y=−3Y³ZandZ³X=−3X³Y=9Y³Z.

Klein example

TheKlein quarticoverF_qis defined by

x³y+y³+x=0andX³Y+Y³Z+Z³X=0 FX=3X²Y+Z³,FY=3Y²Z+X³,FZ=3Z²X+Y³.

If the characteristic ofF_q2is 3 thenF_X=F_Y=F_Z=0 implies X³ =Y³ =Z³ =0⇒X=Y=Z=0⇒ Khas no singularities.

If the characteristic ofF_q2is different than 3 thenFX=FY=FZ=0 impliesX³Y=−3Y³ZandZ³X=−3X³Y=9Y³Z.

From the equation of the curve−3Y³Z+Y³Z+9Y³Z=7Y³Z=0.