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Algebraic curvesover

finite fieldsRuud Pellikaan

Technical University of Eindhoven

Mastermath Coding TheoryOctober 28, 2008

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Content:§0 Introduction§1 Algebraic curves§2 Local parameters and discrete valuations§3 Bézout’s theorem§4 Divisors

Sections 1, 2.1, 2.2 and 2.3 from:Algebraic geometry codesby T. Høholdt, J.H. van Lint and R. Pellikaanin Handbook of Coding Theory, vol 1, pp. 871-961V.S. Pless and W.C. Huffman eds., Elsevier, Amsterdam 1998.

http://www.win.tue.nl/~ruudp/paper/31.pdfhttp://www.win.tue.nl/~ruudp/lectures/masterm-AGC1.pdf

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§0 Introduction

Consider a geometric object X

with a subset P consisting of n points enumerated by P1, . . . , Pn.

Suppose there is a vector space L over Fq offunctions on X with values in Fq.So f (Pi) ∈ Fq for all i and f ∈ L.

Consider the evaluation map

evP : L −→ Fnq

defined by evP(f ) = (f (P1), . . . , f (Pn)).

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This evaluation map is linear,so its image is a linear code.

The image and its dual are the objects of study.The dimension and the minimum distance ofthese codes and their duals will be considered in these lectures.

Decoding algorithms for these codes will be treated inthe other lectures by Tom Høholdt and Mike O’Sullivan.

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In this generality not much can be said about the parameters of these codes.

In the following, X is a subset of the affine or projective spacewhich is the common set of zeros of some given set of polynomials,called a variety.

P1, . . . , Pn will be rational points of X ,that is points that have coordinates in Fq.

The functions will be polynomials or rational functions,that is to say quotients of polynomials.

We call the above codes algebraic geometry (AG) codesif some theory of the variety X gives bounds onthe dimension of the vector space L and the minimum distance of the code.

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The classical example is given by Reed-Solomon (RS) codes.

Here the geometric object X is the affine line over Fq,the points are n distinct elements of Fq and

L is the vector space of polynomials of degree at most k − 1,with coefficients in Fq.

This vector space has dimension k.Such polynomials have at most k − 1 zeros,so nonzero codewords have at least n− k + 1 nonzeros.

Hence this code has parameters [n, k, n− k + 1] if k ≤ n.

The length of a RS code is at most q.

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Take as geometric object Am the affine space of dimension m over Fq,

for the set P all the qm points of this affine space,

and as vector space all polynomials of degree at most r.

Then we get the Reed-Muller (RM) codesof order r in m variables over Fq.

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Every variety has a dimension.A variety of dimension one is called an algebraic curve.

Let X be an algebraic curve over Fq andP a set of n distinct points of X that are defined over Fq.

Let L be a vector space of rational functions withprescribed behavior of their poles and zeros.

Then we get the geometric Goppa codes.

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The parameters of these codes are determinedby the theorem of Riemann-Roch.

They satisfy the following bound

k + d ≥ n + 1− g,

where g is an invariant of the curve called its genus

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The best codes are obtained for curves of genus zero.

They are in fact extended generalized RS codes.These codes have length at most q + 1.

Therefore they cannot give asymptotically good sequences of codes.

The length n of RM codes is not bounded,but k/n or d/n tends to zero if n→∞.

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The information rate R = k/n and

the relative minimum distance δ = d/n of

geometric Goppa codes satisfy the following inequality

R + δ ≥ 1− g − 1

n.

For good geometric Goppa codes we needcurves of low genus with many rational points.

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By studying the number of rational points onmodular curves over finite fields

It was shown that there exist asymptotically good sequencesof geometric Goppa codes satisfying the

Tsfasman-Vladut-Zink (TVZ) bound

R + δ ≥ 1− 1√q − 1

when q is a square.

This bound is better than the Gilbert-Varshamov (GV) bound when q ≥ 49.

It was the first time that the GV bound could be improved.

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§1 Algebraic curves

A field is denoted by F and its algebraic closure by F.

The finite field with q = pm elements, with p a prime is denoted by Fq.

Let F be a field. The prime field of F is the smallest field contained in F.This is Q the field of rational numbers, or Fp with p prime.Then the characteristic of F is 0 or p, respectively.

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Let X/F be a variety X over a field F

As a setX consists of points defined over F with a topology of closed subsets

X (F) are the F-rational points or points defined over F

F[X ] coordinate ring or regular functions on X

F(X ) function field or rational functions on X

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An/F is the n-dimensional affine space over F

coordinates (x1, x2, . . . , xn)

An(F) = Fn is the n-dimensional affine space over F

The closed sets are the algebraic sets.

F[An] = F[X1, X2, . . . , Xn] is the coordinate ring of polynomials

F(An) = F(X1, X2, . . . , Xn) is the function field of rational functions

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Variables: X1, X2, . . . , Xn

Monomial: X i = X i11 · · ·X in

n

Polynomial: F (X1, X2, . . . , Xn) =∑

i fiXi

Coefficients: fi in F

F[X1, X2, . . . , Xn] is the ring of all polynomials

in the variables X1, X2, . . . , Xn with coefficients in F

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Let F1, . . . , Fm ∈ F[X1, X2, . . . , Xn].Define the set of zeros of F1, . . . , Fm by

V (F1, . . . , Fm) = { x ∈ Fn | Fi(x) = 0 for all i }.

Let I be an ideal in F[X1, X2, . . . , Xn]. Define the zero set of I by

V (I) := { x ∈ Fn | F (x) = 0 for all F ∈ I }.

Let I = 〈F1, . . . , Fm〉 be the ideal generated by F1, . . . , Fm

Then V (F1, . . . , Fm) = V (I).

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Let V be a subset of Fn.

Define the vanishing ideal of V by

I(V ) := { F ∈ F[X1, X2, . . . , Xn] | F (x) = 0 for all x ∈ V }

Then I(V ) is a radical ideal,

this means that:

F n ∈ I(V ), n ∈ N0 ⇒ F ∈ I(V ).

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Let I be an ideal. Define the radical of I by

rad(I) = {F | F n ∈ I for some n ∈ N0 }.

Hilbert’s Nullstellensatz.

I(V (I)) = rad(I)

If I is a radical ideal, then it consists of all the polynomials that vanish on V (I).

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An algebraic set in An is a subset of the form V (I).

The algebraic sets form the closed sets of the Zarisky topology,since

V ({0}) = Fn, V (〈1〉) = ∅,V (I) ∪ V (J) = V (I ∩ J),

∩a∈AV (Ia) = V (∑

a∈A Ia).

The complement of a closed set is called open.

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An algebraic set is called irreducible if it cannot be writtenas the union of two proper algebraic subsets.

Every algebraic set is the finite union of irreducible components

An ideal I is called prime if

FG ∈ I ⇒ F ∈ I or G ∈ I.

If I is a radical ideal, thenV (I) is irreducible if and only if I is a prime ideal.

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Example Let

F (X, Y ) = X2 − Y 2 and I = 〈F 〉.

The corresponding algebraic set in A2

is the union of two lines with equations:

Y = X and Y = −X.and each of these lines is an irreducible algebraic set in the plane .

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Example

Suppose that−1 is not a square in F.

Let

F (X, Y ) = X2 + Y 2 and I = 〈F 〉.

Then I is a prime ideal and V (I) is irreducible.

But I is not prime in F[X, Y ] and V (I) is reducible over F.

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An algebraic set is absolutely irreducible if it is irreducible over F.

An affine variety is an absolutely irreducible algebraic set.

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Example Let

F (X, Y ) = X2 + Y 2 + Z2 − 1 and I = 〈F 〉.

The corresponding algebraic set in A3

is the is the affine variety consistingof all (x, y, z) ∈ F3 such that

x2 + y2 + z2 = 1.

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Let X = V (I) be an affine variety in An.

Then the coordinate ring of X is defined by

F[X ] := F[X1, X2, . . . , Xn]/I

Two polynomials that differ by an element of I will havethe same value in each point of X .

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We adopt the following convention:

capital letters X1, . . . , Xn, Y and Z denote variables.

Polynomials are denoted by F , G and H

and their cosets modulo the ideal I

are denoted by small letters f , g and h.

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Let I be a prime ideal of an affine variety X = V (I).

Then the coordinate ring F[X ] is an integral domain,

that means:

if fg = 0 , then f = 0 or g = 0.

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The quotient field or the field of fractions of the ring F[X ]

is defined by

F(X ) :=

{f

g| f, g ∈ F[X ], g 6= 0

}

and it is called the function field of X .

The elements of F(X ) are called rational functions.

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The dimension of the variety X is defined equivalently by

1) transcendence degree of F(X ) over F.

2) the maximal d such that there is a chain of proper inclusions ofsubvarieties

X0 ⊂ X1 ⊂ · · · ⊂ Xd

X is called an algebraic curve if this dimension is 1.

So points and the curve itself are the only subvarieties of a curve.

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Example Consider the parabola X in A2/F with equation

Y 2 = X.

The coordinate ring F[X ] consists of all expressions of the form

A + By,

where A and B are in F[x] and y satisfies y2 = x.

The function field F(X ) is an algebraic extension of F(x)of degree 2, by the element y.

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In projective space Pn we have homogeneous coordinates.

A point(x0 : x1 : · · · : xn)

in Pn is the line in An+1 through the origin and (x0, x1, . . . , xn) 6= 0.

So(x0 : x1 : · · · : xn) = (y0 : y1 : · · · : yn)

if and only if(x0, x1, . . . , xn) = λ(y0, y1, . . . , yn)

for some λ ∈ F∗.

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A polynomial F of the form

F (X0, X1 . . . , Xn) =∑

i0+i1+···+in=d

fiXi00 X

i11 · · ·X in

n

is called homogeneous of degree d. Then

F (λx0, λx1, . . . , λxn) = λdF (x0, x1, . . . , xn)

Hence it makes sense to consider the zero set in Pn

of homogeneous polynomials.

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Let I be an ideal in F[X0, X1, . . . , Xn] generated byhomogeneous polynomials.

Define the zero set of I in Pn by

V (I) :=

{ (x0 : x1 : · · · : xn) | x ∈ Fn+1,x 6= 0, F (x) = 0, F homogeneous in I }.

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For rational functions to have a meaning one takes only

those quotients F/G for which the numerator F and

the denominator G are homogeneous polynomials of the same degree.

If G(x) 6= 0, thenF (λx)

G(λx)=λdF (x)

λdG(x)=F (x)

G(x).

So the value of F/G is well defined at all point x of Pn such that G(x) 6= 0.

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A projective variety X is the zero set in Pn

of a homogeneous prime ideal I in F[X0, X1, . . . , Xn].

Consider the subring R(X ) of F(X0, X1, . . . , Xn)consisting of the fractions F/G, where F and G arehomogeneous polynomials of the same degree and G /∈ I .

Then R(X ) has a unique maximal ideal M(X )consisting of all those F/G with F ∈ I .

Define the function field

F(X ) := R(X )/M(X ).

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Let X be a projective variety. Let P be a point on X .

Then a rational function φ on X is calledregular at the point P if one can find F and G,homogeneous polynomials of the same degree,such that G(P ) 6= 0 and φ is the coset of F/G.

The functions that are regular at every point of the set Uform a ring, denoted by F[U ].

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Example Consider the variety X in P2 defined by

XZ − Y 2 = 0.

The function x/y is equal to y/z on the curve,hence it is regular in the point P = (0 : 0 : 1).

The function2xz + z2

y2 + z2

is regular in P . By replacing y2 by xz, we see that

2xz + z2

y2 + z2=

2x + z

x + z.

and therefore it is also regular at Q = (1 : 0 : 0).

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Let X be a projective variety.Then the only regular functions on X are constant.

Let U be an affine open subset of X .Then the coordinate ring F[U ] coincideswith the ring of regular functions on U .

So there is no ambiguity in the notation F[U ].

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The local ringOP orOP (X )

of P ∈ X isthe set of all rational functions that are regular at P .

This is indeed a “local ring”, sinceit has the unique maximal ideal,

MP = { φ ∈ OP | φ(P ) = 0 }.

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Embedding of an affine variety in a projective variety.

Associate with F ∈ F[X1, . . . , Xn] the homogeneous polynomial F ∗

defined by

F ∗(X0, X1, . . . , Xn) := Xd0F (X1/X0, . . . , Xn/X0),

where d is the degree of F .

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Let X be an affine variety in An

defined by the prime ideal I .

Let I∗ be the ideal generated by {F ∗|F ∈ I}.

Then I∗ is a homogeneous prime idealdefining the projective variety X ∗ in Pn.

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LetX ∗0 = { (x0 : x1 : · · · : xn) ∈ X ∗ | x0 6= 0}.

Then X is isomorphic with X ∗0under the map

(x1, . . . , xn) 7→ (1 : x1 : · · · : xn).

The points (x0 : x1 : · · · : xn) ∈ X ∗ such that x0 = 0are called the points at infinity of X .

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The isomorphism of function fields

F(X ) ∼= F(X ∗)

is given by the mapf

g7→ xm0 f

∗

g∗,

where m = deg(g)− deg(f ).

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For any point P of a projective variety X and

any hyperplaneH not containing P

the complementX \ H

is an affine variety containing P .

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LetF =

∑aijX

iY j.

The partial derivative FX of F with respect to Xis defined by

FX =∑

iaijXi−1Y j

and FY is defined similarly.

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Let X be the affine plane curve defined by F (X, Y ) = 0.

Let P ∈ X . IfFX(P ) = FY (P ) = 0,

then P is called singular, otherwise simple or nonsingular

A curve is called nonsingular or smoothif all its points are nonsingular.

Let P = (a, b) be a nonsingular point on X .The tangent line TP at P is defined by

FX(a, b)(X − a) + FY (a, b)(Y − b) = 0.

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The definitions for a projective plane curve are similar.

Let a projective plane curve be defined bythe homogeneous equation F = 0.

Let P ∈ X . If at least one of the derivatives

FX, FY , or FZ

is not zero in P , then P is called a simple or nonsingular.

Let P = (a : b : c) be a nonsingular point.Then the tangent line at P has equation

FX(P )X + FY (P )Y + FZ(P )Z = 0.

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The Fermat curve Fmis the projective plane curve with defining equation

F := Xm + Y m + Zm = 0.

The partial derivatives of F are

mXm−1, mY m−1, and mZm−1.

This curve is nonsingular if and only ifm is relatively prime to the characteristic.

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Let q = r2. The Hermitian curveHr over Fq

is defined by the affine equation

U r+1 + V r+1 + 1 = 0.

The corresponding homogeneous equation is

U r+1 + V r+1 + W r+1 = 0.

Hence it has r + 1 points at infinity.It is the Fermat curve Fm over Fq with m = r + 1.

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The conjugate of a ∈ Fq over Fr with q = r2

is obtained bya = ar.

So the equationU r+1 + V r+1 + W r+1 = 0.

can be rewritten asUU + V V + WW = 0.

Like a Hermitian form over the complex numbers.

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Choose an element b ∈ Fq such that

br+1 = −1.

There are exactly r + 1 of these, since q = r2.

Let P = (1 : b : 0). Then P is a point of the Hermitian curve.The tangent line at P has equation

U + brV = 0.

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Multiplying the equation U + brV = 0 with b gives

V = bU.

Substituting V = bU in the defining equation gives that

W r+1 = 0.

So P is the only intersection point ofthe Hermitian curve with the tangent line at P .

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New homogeneous coordinates are chosen such that this tangent linebecomes the line at infinity.

Let X1 = W , Y1 = U and Z1 = bU − V .Then the curve has homogeneous equation

Xr+11 = brY r

1 Z1 + bY1Zr1 − Zr+1

1

in the coordinates X1, Y1 and Z1.Choose an element a ∈ Fq such that

ar + a = −1.

There are r of these.Let X = X1, Y = bY1 + aZ1 and Z = Z1.

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Then the curve has homogeneous equation

Xr+1 = Y rZ + Y Zr

with respect to X , Y and Z .Hence the Hermitian curve has affine equation

Xr+1 = Y r + Y

with respect to X and Y .

This last equation has (0 : 1 : 0) as the only point at infinity.

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The Klein curve has homogeneous equation

X3Y + Y 3Z + Z3X = 0.

More generally define the curve Kmby the equation

XmY + Y mZ + ZmX = 0.

The partial derivatives are

mXm−1Y + Zm, mY m−1Z + Xm and mZm−1X + Y m.

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Let m2 −m + 1 be relatively prime to the characteristic.

Let (x : y : z) ∈ Km be a singular point.

If m is divisible by the characteristic, then

xm = ym = zm = 0.

So x = y = z = 0, a contradiction.

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If m is relatively prime to the characteristic, then

xmy = −mymz = m2zmx.

So(m2 −m + 1)zmx = xmy + ymz + zmx = 0.

Therefore z = 0 or x = 0, sincem2 −m + 1 is relatively prime to the characteristic.

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But z = 0 impliesxm = −mym−1z = 0.

Furthermoreym = −mzm−1x.

So x = y = z = 0, which is a contradiction.Similarly x = 0 leads to a contradiction.

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Hence Km with equation

XmY + Y mZ + ZmX = 0.

is nonsingular if

gcd(m2 −m + 1, p) = 1.

where p is the characteristic.

Show that the converse is also true.

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§2 Local parameters and discrete valuations

Let X be a smooth curve in A2 defined by the equation F = 0.

Let P = (a, b) ∈ X . The maximal idealMP

is generated by x− a and y − b.

dPF = FX(P )(x− a) + FY (P )(y − b) = 0.

Hence the F-vector spaceMP/M2

P

has dimension 1 and thereforeMP has one generator.

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HenceMP ofOP is a principal ideal.

Let g ∈ F[X ] be the coset of a polynomial G.

Then g is a generator ofMP if and only if

dPG is not a constant multiple of dPF .

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Let t be a generating element ofMP .Then t is called a local parameter or uniformizing parameter.

Every element z ∈ OP has a unique expression

z = utm,

where u is a unit and m ∈ N0.

If m > 0, then P is a zero of multiplicity m.

We writem = ordP (z) = vP (z).

Convention vP (0) =∞ > n for all n ∈ N0.

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Theorem The mapvP : OP → N0 ∪ {∞}

is a discrete valuation, that isthe map is surjective and satisfies:

(i) vp(f ) =∞ if and only if f = 0,(ii) vP (λf ) = vP (f ) for all nonzero λ ∈ F,(iii) vP (f + g) ≥ min{vP (f ), vP (g)}

and equality holds when vP (f ) 6= vP (g),(iv) vP (fg) = vP (f ) + vP (g).(v) If vP (f ) = vP (g),

then there exists a nonzero λ ∈ Fsuch that vP (f − λg) > vP (g).

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Extend the function vP to F(X ) by defining

vP (f/g) = vP (f )− vP (g).

If vP (z) = m > 0, then z has a zero of order m in P .

If vP (z) = −m < 0, then z has a pole of order m in P .

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If z ∈ F(X ) with vP (z) = m, then we can write

z = atm + z′,

where a ∈ F, a 6= 0 and vP (z′) > m.

In this way, one can show that z has a (formal) Laurent series

z =∑i≥m

aiti,

where ai ∈ F for all i and am 6= 0.

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Example Let P1 be the projective line.

A local parameter in the point P = (1 : 0) is

x1/x0.

The rational functionx2

0 − x21

x21

has a pole of order 2 at P .

If F does not have characteristic 2, then (1 : 1) and (−1 : 1)are zeros with multiplicity 1.

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Example Let the characteristic of F be unequal to 2.

Let X be in A2 with equation

X2 + Y 2 = 1.

Let P = (1, 0). Let z = 1− x.

This function is 0 in P , so it is inMP .

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We claim that z has order 2.

Observe that y is a local parameter in P ,because the line

Y = 0

is not equal to the tangent line

X = 1

in P . Furthermore, on X we have

1− x = y2/(1 + x)

and the function1/(1 + x)

is a unit inOP .

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Example Let X be the plane curve with equation

X3 + Y 3 + Z3 = 0

over the field F2. Then Q = (0 : 1 : 1) ∈ X .

Take t = x/z as local parameter at Q.

The expression x/(y + z) does not make sense at Q.

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On X we have

x

y + z=x(y2 + yz + z2)

y3 + z3= t−2 · y

2 + yz + z2

z2,

where the second factor on the right is regular and not 0 in Q.

Hence f has a pole of order 2 in Q.

Exercisey/(y + z) has a pole of order 3 in Q.

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Let X be a curve defined over Fq,the defining equations have coefficients in Fq.

Then points on X with all their coordinates in Fq

are called Fq-rational points.

The set of all Fq-rational points of X is denoted by:

X (Fq).

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Example Consider the Klein quartic with equation

X3Y + Y 3Z + Z3X = 0

over the algebraic closure of F2.

Over F2 the rational points are

(1 : 0 : 0), (0 : 1 : 0), and (0 : 0 : 1).

LetF4 = {0, 1, α, α2},

whereα2 = 1 + α.

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Over F4, there are two more points:

(1 : α : 1 + α) and (1 : 1 + α : α).

ExerciseCompute the number of rational pointsof the Klein quartic over F8.

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Define F8 as F2(ξ), whereξ3 = ξ + 1.

If a rational point has a coordinate 0,it must be one of the points over F2.

Ifxyz 6= 0,

we can take z = 1 by symmetry.

Ify = ξi, 0 ≤ i ≤ 6,

then write x = ξ3iη.

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Substitution in the equation gives

η3 + η + 1 = 0,

that is, η is one of the elements

ξ, ξ2 or ξ4.

So we find a total of3 + 7 ∗ 3 = 24

rational points over F8.

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Example Consider the Hermitian curveH2.

X3 + Y 3 + Z3 = 0

over F4.

Since a third power of an element of F4 is 0 or 1,all the rational points have one coordinate 0.

Take one of the others to be 1, and the third one any nonzero element of F4.

So we find nine (projective) points.

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Exercise Let q = r2.

Consider the Hermitian curve with affine equation

Xr+1 = Y r + Y

Show that it has r3 + 1 points over Fq.

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§3 Bézout’s theorem

A polynomial of degree m in one variable,with coefficients in a field has at most m zeros.

If the field is algebraically closedand if the zeros are counted with multiplicities,then the number of zeros is equal to m.

Bézout’s theorem is a generalization of these factsto polynomials in several variables.

The degree of a projective plane curve isthe degree of the defining polynomial.It is the maximal number of points in the intersection with a line,not containing a component of the curve.

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Consider the intersection of irreduciblenonsingular projective curves X and Y of degrees l and m.

Let G = 0 be the defining equation of Y .

Let P be a point of X .Let H be a homogeneous linear form such that H(P ) 6= 0.Let h be the class of H modulo the ideal defining X .

Then the intersection multiplicity

I(P ;X ,Y)

of X and Y at P is defined by vP (g/hm).

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This definition of the intersection multiplicitydoes not depend on the choice of H ,since h/h′ is a unit inOP for any other choice of alinear form H ′ that is not zero in P .

If P is also a point of Y , then

I(P ;X ,Y) = I(P ;Y ,X ).

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Bézout’s Theorem

Let X and Y be two plane curves that have no component in common.Then X and Y intersect in exactly lm points(counted with multiplicities).

Corollary

The intersection of two projective plane curves is not empty.

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Corollary

A nonsingular projective plane curve is irreducible.

Exercisea) Proof this corollary.b) Give an example of a nonsingular affine plane curve that is reducible.

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Proof

If F = GH is a factorization of F with factors of positive degree, we get

FX = GXH + GHX

by the product or Leibniz rule for the partial derivative.

SoFX ∈ (G,H),

and similarly for the other two partial derivatives. Hence

(FX, FY , FZ, F ) ⊆ (G,H).

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The zero sets satisfy

V (G,H) ⊆ V (FX, FY , FZ, F ).

Now V (G,H) is the intersection of the curves with equationsG = 0 and H = 0, and this not empty by the above Corollary.

Therefore the curve has a singular point.

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Notice that the assumption that the curveis a projective plane curve is essential.

The equation X2Y −X = 0 defines a nonsingular affine plane curve,but is clearly reducible.

However if F = 0 is an affine plane curve and the homogenization F ∗

defines a nonsingular projective curve, then F is irreducible.

The affine curve with equation X2Y −X = 0 has the points(1 : 0 : 0) and (0 : 1 : 0)at infinity,and (0 : 1 : 0) is a singular point.

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LetVl = {F ∈ Fq[X, Y ] | deg(F ) ≤ l },

the vector space of bivariate polynomials of degree at most land coefficients in Fq.

Consider an element G of degree m in Fq[X, Y ] such thatthe homogeneous form G∗ defines a nonsingular curve.

Then G is irreducible in F[X, Y ], where F is the algebraic closure of Fq.

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Let P1, P2, . . . , Pn be rational points on the plane curvedefined by the equation G = 0, So

Pi = (ai, bi) ∈ F2q and G(Pi) = 0 for 1 ≤ i ≤ n.

Define the code C by

C = {(F (P1), F (P2), . . . , F (Pn)) |F ∈ Fq[X, Y ], deg(F ) ≤ l}.

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Theorem Let n > lm.

Denote the minimum distance of this code by d and its dimension by k.

Thend ≥ n− lm,

and

k =

{ (l+22

)if l < m,

lm + 1−(m−1

2

)if l ≥ m.

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Proof

The monomials of the form XαY β with α + β ≤ l form a basis of Vl.

Hence Vl has dimension(l+22

).

Let F ∈ Vl.If G is a factor of F , then the codeword in C corresponding to F is zero.

Conversely, if this codeword is zero, then the curves with equationsF = 0 and G = 0 have degree l′ ≤ l and m respectively,and they have the n points P1, P2, . . . , Pn in their intersection.

Bézout’s theorem and the assumption lm < n imply thatF and G have a common factor.

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Since G is irreducible, F must be divisible by G.Hence the functions F ∈ Vl that yield the zero codewordform the subspace GVl−m.

This implies that if l < m, then k =(l+22

),

and if l ≥ m, then

k =

(l + 2

2

)−(l −m + 2

2

)= lm + 1−

(m− 1

2

).

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The same argument with Bézout’s theorem shows that

a nonzero codeword has at most lm coordinates equal to 0,

that is to say, it has weight at least n− lm.

Hence

d ≥ n− lm.

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Remark

Let F1, . . . , Fk is a basis for Vl modulo GVl−m. Then

(Fi(Pj) | 1 ≤ i ≤ k, 1 ≤ j ≤ n)

is a generator matrix of C .

So it is a parity check matrix for the dual of C .The minimum distance d⊥ of C⊥ is equal tothe minimal number of dependent columns of this matrix.

Hence for all t < d⊥ and every subsetQ of P = {P1, . . . , Pn}consisting of t distinct points,the corresponding k × t submatrix has maximal rank t.

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Let Ll = Vl/GVl−m.

Then the map that evaluates polynomialsat the points ofQ induces a surjective map

Ll −→ Ftq.

Denote the kernel by Ll(Q),it is the space of all functions F ∈ Vlthat are zero at the points ofQmodulo GVl−m.So

dim(Ll(Q)) = k − t if t < d⊥.

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Conversely, the dimension of Ll(Q) is at least k − tfor all t-subsetsQ of P .

But in order to get a bound for d⊥, we have to know thatdim(Ll(Q)) = k − t for all t < d⊥.

The theory developed so far is not sufficient to get such a bound.

The theorem of Riemann-Roch gives an answer to this question.

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Notice that the following inequality holds

k + d ≥ n + 1− g,

where g = (m− 1)(m− 2)/2.

We will see that g is the genus.

Later g is also the number of gaps ofthe Weierstrass semigroup of a point.

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§4 Divisors

In the following, X is an irreducible,nonsingular, projective plane curve.

A divisor is a formal sumD =

∑P∈X

nPP,

with nP ∈ Z and nP = 0 for all but a finite number of points P ∈ X .

The support of this divisor is the set

{ P | nP 6= 0 }.

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Let E =∑

P∈XmPP be another divisor. Define

D ≤ E ⇔ np ≤ mP for all P.

A divisor D is called effective if

0 ≤ D,

that is if all coefficients nP are nonnegative.

Define the degree of the divisor D by

deg(D) =∑

nP .

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Let X and Y be projective plane curves defined by the equations F = 0 andG = 0 of degrees l and m.

Define the intersection divisor by

X · Y =∑

I(P ;X ,Y)P,

where I(P ;X ,Y) is the intersection multiplicity.

Bézout’s theorem tells us that X · Yis indeed a divisor and

deg(X · Y) = lm.

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Let P ∈ X .

Let vP be the discrete valuation defined for functions on X .

Let f be a nonzero rational function on X .Define the divisor of f by

(f ) =∑P∈X

vP (f )P

The divisor of f is a bookkeeping device that tells uswhere the zeros and poles of f are and what their multiplicities are.

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Theorem

deg(f ) = 0

The degree of a divisor of a nonzero rational function is zero.

Proof

Let X be a plane curve of degree l.Let f be a rational function on the curve X .Then f is represented by a quotient A/B oftwo homogeneous polynomials of the same degree, say m.

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Let Y and Z be the curves defined by the equations A = 0 and B = 0.Then

vP (f ) = I(P ;X ,Y)− I(P ;X ,Z),

sincef = a/b = (a/hm)(b/hm)−1,

whereH is a homogeneous linear form representing h such thatH(P ) 6= 0.Hence

(f ) = X · Y − X · Z.So (f ) is a divisor and its degree is zero, since it is the differenceof two intersection divisors of the same degree lm.

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Example

Consider the plane curve X with equation over F4

X3 + Y 3 + Z3 = 0

Let Q = (0 : 1 : 1). Take t = x/z as local parameter at Q.

We saw thatvQ(x/(y + z)) = −2.

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Let α ∈ F4 with α2 = 1 + α.

The line L with equation X = 0 intersects the curve in three points:

P1 = (0 : α : 1), P2 = (0 : α2 : 1) and Q.

SoX · L = P1 + P2 + Q.

The lineM with equation Y = 0 intersects the curve in three points:

P3 = (1 : 0 : 1), P4 = (α : 0 : 1) and P5 = (α2 : 0 : 1).

SoX ·M = P3 + P4 + P5.

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The lineN with equation Y + Z = 0 intersects the curve in Q with multi-plicity 3. So

X · N = 3Q.

Hence(x/(y + z) = P1 + P2 − 2Q

and(y/(y + z) = P3 + P4 + P5 − 3Q.

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Example

Let X be the Klein quartic with equation

X3Y + Y 3Z + Z3X = 0

LetP1 = (1 : 0 : 0), P2 = (0 : 1 : 0) and P3 = (0 : 0 : 1).

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Let L be the line with equation X = 0.Then L intersects X in the points

P2 and P3.

The line L is not tangent in P2, so

I(P2;X ,L) = 1 and I(P3;X ,L) = 3,

since the multiplicities add up to 4.Hence

X · L = P2 + 3P3.

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X · L = P2 + 3P3.

Similarly we get for the linesM andN with equations Y = 0 and Z = 0

X ·M = 3P1 + P3 and X · N = 3P2 + P1.

Therefore(x/z) = 3P3 − P1 − 2P2

and(y/z) = 2P1 + P3 − 3P2.

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The divisor of a rational function is called a principal divisor.

Two divisors D and E are called linearly equivalentif and only if D − E is a principal divisor, that is

D − E = (f )

for some rational function f .

NotationD ≡ E.

This is indeed an equivalence relation.

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Let D be a divisor on a curve X .

Define the vector space L(D) over F by

L(D) = {f ∈ F(X )∗ | (f ) + D ≥ 0} ∪ {0}.

Definel(D) = dimFL(D).

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If

D =

r∑i=1

niPi −s∑j=1

mjQj

with all ni,mj > 0, then L(D) consists of 0 andthe rational functions that have

zeros of order at least mj at Qj

poles of order at most ni at Pi

and no other poles.

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If D ≡ E and g is a rational function with

(g) = D − E, then the mapf 7→ fg

shows that

L(D) ∼= L(E)

as vector spaces over F.

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Theorem

(i) l(D) = 0 if deg(D) < 0,

(ii) l(D) ≤ 1 + deg(D).

Proof

(i) If deg(D) < 0, then for any function f ,we have deg(f ) + D) < 0, so f /∈ L(D).

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(ii) If f is not 0 and f ∈ L(D), then

E := D + (f )

is an effective divisor for which

D ≡ E so l(D) = l(E).

So we may assume that D is effective, say

D =

r∑i=1

niPi,

where ni ≥ 0 for all i.

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Let f 6= 0 and f ∈ L(D).Let ti be a local parameter at Pi.Map f onto the corresponding element in

(t−nii OPi

)/OPi,

This space has dimension ni.

Thus we obtain a mapping

L(D)→ ⊕ri=1(t

−nii OPi

)/OPi.

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This is a linear mapping.

Suppose that f is in the kernel.This means that f does not have a pole in all the points Pi, that is to say,f is a constant function. It follows that

l(D) ≤ 1 +

r∑i=1

ni = 1 + deg(D).

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Example Consider the plane curve X with equation

X3 + Y 3 + Z3 = 0

as before. We saw that

(x/(y + z) = P1 + P2 − 2Q

and(y/(y + z) = P3 + P4 + P5 − 3Q.

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Let f = x/(y + z) and g = y/(y + z).

So the functions 1, f and g havemutually distinct pole orders at Q andare linearly independent elements of L(3Q).

Hence3 ≤ l(3Q) ≤ 1 + 3.

We will see that l(3Q) = 3.