# Improving the minimum distance bound of Trace Goppa codes

In this article we prove that a class of Goppa codes whose Goppa polynomial is of the form g(x) = x + x^q + ⋯ + x^q^m-1 where m ≥ 3 (i.e. g(x) is a trace polynomial from a field extension of degree m ≥ 3) has a better minimum distance than what the Goppa bound d ≥ 2deg(g(x))+1 implies. Our improvement is based on finding another Goppa polynomial h such that C(L,g) = C(M, h) but deg(h) > deg(g). This is a significant improvement over Trace Goppa codes over quadratic field extensions (i.e. the case m = 2), as the Goppa bound for the quadratic case is sharp.

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## 1 Introduction

Binary Goppa codes are one of the fundamental linear code constructions in Coding Theory. Binary Goppa codes have been extensively studied since their introduction by V.D. Goppa in [1]. Their rich algebraic structure, and good decoding capabilities make binary Goppa codes suitable candidates for cryptography applications. There are also Best Known Linear Codes constructions realized by binary Goppa codes.

Throughout this article we shall assume is a prime, for some natural number , and We focus on binary Goppa codes where the defining polynomial is of the form , or that is

 g(x)=x+xq+xq2+⋯+xqm−1.
###### Definition 1.

[1] Let be a prime power. Let be a power of . Suppose . Let be a univariate polynomial of degree such that , for . The –ary Goppa code is defined as

 C(L,g):={(c1,c2,…,cn)∈Fnp:n∑i=1cix−αi≡0modg(x)}.

Our dimension bound looks slightly different than the classical dimension bound, . This is because in the classical definition, the set is defined over where whereas in our case, the set is defined in where . The reason for the different bound is that our polynomial is also defined taking into consideration the subfield of . Thus there are two subfields to consider: the subfield in which takes values and the subfield over is defined. We shall take , will take values in the subfield and will be defined in . Goppa codes may also be defined as subcodes over any subfield of . However, Goppa codes over the prime subfield (and in particular the binary subfield) remain the most interesting. Although our results hold for any subfield for the sake of simplicity in this article Goppa codes are defined over the prime field . The following bound on the dimension of binary Goppa codes is well known:

###### Proposition 1.

[1] Let be a prime power. Let be a power of . Let . Let be a polynomial of degree such that , for . Then the dimension of is at least and the minimum distance of is at least .

One of the first improvements on the bounds of Binary Goppa codes was given by Goppa in [1]. This improvement is based on establishing that two different Goppa polynomials give the same binary Goppa codes. This allows to use one polynomial to bound the dimension of the code and another polynomial to bound the minimum distance of the code.

###### Proposition 2.

[1] Let . Let be a subset of . Let be a squarefree polynomial of degree such that , for . Then the binary Goppa codes satisfy:

 C(L,g)=C(L,g2).

This proposition improves the distance bound from to . The distance bound on the Goppa code comes from the fact that the codewords of satisfy certain special parity check equations. Sugiyama et. al generalize this equivalence between Goppa codes over arbitrary fields .

###### Proposition 3.

[9] Let be a prime power. Let be a subfield of . Let be a subset of . Let be a squarefree polynomial of degree such that , for . Then the Goppa codes defiend over satisfy:

 C(L,gq0−1)=C(L,gq0).
###### Definition 2.

Let where . Let be a polynomial. We define the evaluation map as

 evL:Fqm[X]→Fnqm,evL(f)=(f(α1),f(α2)…,f(αn)).

The map is a linear map from the polynomial ring

to the vector space

. The kernel is . We find it more illustrating to work with to understand the parity check equations. From the definition of Goppa codes it follows that the parity check equations for may also be written as evaluation maps . We describe those parity check equations as follows.

###### Proposition 4.

[1] Let be a prime power. Let . Let be a polynomial of degree such that , for . Then any codeword satisfies

 n∑i=1ciαjig(αi)=c⋅evL(Xjg(X))=0\makeboxwhere0≤j≤t−1

Goppa codes belong to a class of codes known as Alternant Codes. Alternant codes are subfield subcodes of Generalized Reed–Solomon codes. One of the good things about Alternant codes is that one can get a bound on its minimum distance at follows:

###### Proposition 5.

Let . Let be distinct elements in . Let be nonzero elements in Let be a positive integer. Let be a code of length over .

If for and then the minimum distance of is at least .

Goppa codes are Alternant codes where . The classical Goppa distance bound comes from the consecutive powers from to . Our distance bound improvements come from finding more consecutive powers which are parity check equations for .

Goppa codes are linear codes defined over a small field, . However the parity check equations describing the Goppa codes are defined over the larger field . For denote by

 x(pi)=(xpi1,xpi2,…,xoin)∈Fnqm.

Note that if and then . As it follows that . Thus for each –power, we get the additional parity check equations . As and there are different –powers, this is how the dimension bound is derived. Recall that the trace function takes values in the subfield for any . This implies that and also that . This fact will be important later when we prove that certain –powers of evaluation vectors and are in the dual codes or .

P. Véron has improved bounds on the dimension of Trace Goppa codes. In fact his bounds are sharp for . S. Bezzatev and N. Shekhunova in [8] proved that the classical distance bound is sharp for . We improve the minimum distance for trace Goppa codes when instead.

## 2 Improving the Minimum Distance of Trace Goppa Codes

We improve the minimum distance bound of Trace Goppa codes with by establishing that the Goppa code is equivalent to the Trace Goppa code where

 h(y)=TrFqm∖Fq(ya)modyqm−y

and . If then

 h(y)=yb−1+yb−q+yb−q2+⋯+yb−qm−1.

The degree of is . Denote by

 L={α∈Fqm:TrFqm∖Fq(α)≠0}

and denote by

 M={β∈Fqm : h(β)≠0}.

Now we prove that

 α∈L\makeboxifandonlyifα−1∈M.
###### Lemma 1.

Let . Then if and only if

###### Proof.

Let . This implies that . As , this implies Therefore we divide by and obtain .

This implies . We rewrite the sum as where . Collecting the different powers we obtain which implies and thus .

Now suppose that . Then . From the definiton of , it follows

 h(α−1)=α−a+α−aq+⋯+α−aqm−1≠0.

We rewrite the sum as As it follows that which implies .∎

We’ve established a relation between elements which are not roots of and the elements which are not roots of . Now we describe the relations amongst the parity check equations for and . As stated in Proposition 4 the parity check equations for are vectors of the form for and the corresponding –powers for each of the vectors. Likewise the parity check equations for are generated from parity check equations of the form for and all their –powers. Luckily, the trace polynomial takes values in the subfield . This implies there are relations among the different –powers of the parity check equations. For example and . P. Véron ([4]) used these relations to improve the dimension bound from to . We find relations between the different parity check equations to improve the distance bounds. We begin with the following lemma.

###### Lemma 2.

Suppose . Assume the –ary expansion of where . Then .

The parity check equations for are for and their –powers. We use the fact that for to prove that for may be obtained from a –power of some where .

Let . Then .

###### Proof.

Suppose . The –ary expansion of is of the form where at least one of the entries . Otherwise if each then . If , then . Therefore . ∎

A similar technique proves the following lemma:

.

###### Proof.

Recall that . Thus . As is the –power of and it follows that the evaluation vector . Since is a –power of and contains all of its –powers, follows. ∎

Note that both and have parity check equations of consecutive powers from to . We prove now that there is a change of variables which maps one set of parity check equations to the other.

###### Lemma 5.

The codes and are equal.

###### Proof.

We’ll prove that the map maps the parity check equations for to the parity check equations for . Let for . Let , then

 evL(Xig(X))=evM(Y−ig(Y−1))

We multiply both sides of the fraction by and obtain:

 evM(Y−ig(Y−1))=evM(YaqY−iYaqg(Y−1))=evM(Yaq−ih(Y)).

Setting we obtain that . Therefore

 evL(Xig(X))=evM(Yaq−ih(Y))=evM(Yjh(Y))∈C(M,h)⊥.

Therefore . All steps are reversible, which implies equality. ∎

The equality between the Goppa codes and leads to improved bounds on the minimum distance . Lemma 5 implies that for binary Goppa code even though itself is not a square free polynomial. This leads to a significant improvement of the distance bound.

###### Corollary 1.

The codes and are equal.

###### Proof.

The proof is the same is the one in Lemma 5 using and instead of and and using intead of . All other steps are equal. ∎

###### Corollary 2.

Let . Suppose is a binary Goppa code. The minimum distance of is at least .

###### Proof.

Let be a power of . Note that is a squarefree polynomial. Therefore the binary Goppa code is equal to . As and , the bound follows. ∎

This lemma states a sufficient condition for a funcion of the form to be in .

###### Lemma 6.

Let be a monic polynomial of degree . Let . Suppose that for all . Then

 evL(Xi+tg(X))∈C(L,g)⊥\makeboxifandonlyifevL(Xi)∈C(L,g)⊥.
###### Proof.

Suppose that for all .

Note that

 evL(Xi)−evL(Xi+tg(X))=evL(Xi−Xi+tg(X))=evL(Xig(X)−gtXi+tg(X)).

As is a polynomial of degree , all terms of have degree less than . By the hypothesis of this Lemma; . As the difference

 evL(Xi)−evL(Xi+tg(X))∈C(L,g)⊥

it follows that if and only if . ∎

In the next Lemma we find more consecutive parity check equations of the form where is larger than .

Let . Then .

###### Proof.

The definition of the binary Goppa code establishes

 evM(Yjh(Y)2)∈C(M,h2)⊥,0≤j≤2aq−1.

Now we consider larger . If then the –base expression of , where has at least one of the satisfy . Otherwise implies . Therefore there is a number of the form which is smaller than . In this case

 evM(Yjh(Y)2)=evM((Yj′)qm−rh(Y)2)=evM(Yj′h(Y)2)(qm−r)∈C(M,h2)⊥.

For , note that

 h(Y)2Yh(Y)2=Y\makeboxandh(Y)2Y2h(Y)2=Y2.

However can be written as the sum of plus other smaller powers of . Likewise can be written as the sum of plus other smaller powers of . Therefore . As is a binary Goppa code both and contain all –powers of their codewords. Therefore . Lemma 6 implies . As is the sum of plus other lower powers of it follows that the evaluation vector . ∎

Now we shall prove that contains additional consecutive parity check equations for negative powers of , which further improves the minimum distance.

###### Lemma 8.

Let be a power of , Let . Then

 evM(Y−ih(X)2)∈C(M,h2)⊥.
###### Proof.

We shall prove that

 evM(Y−ih(X)2)(q2)∈C(M,h2)⊥,0

We proceed by induction and start with . Since . We take the –power of

 evM(Y−1h(X)2)q2=evM⎛⎝Y−q2h(X)⎞⎠=evM⎛⎜⎝Y−q2h(X)h(X)2⎞⎟⎠.

As the lowest degree term of is , all terms of are between and . Thus .

Now let us suppose that for all , . Furthermore, suppose that .

Now take the –power of As , we have that which implies

###### Theorem 1.

The minimum distance of the –ary Goppa code is at least . If then the minimum distane of the binary Goppa code is at least .

###### Proof.

Lemma 5 established that . The degree of is . Therefore

 d(C(L,g))≥qm−1+qm−2+⋯+q+1.

In the case of binary Goppa codes, Corollary 1 implies . Lemma 7 implies has consecutive powers of the form as parity check equations. This implies

 d(C(M,h2))≥2(qm−1+qm−2+⋯+q+1)+2.

Lemma 8 implies there are an additional parity check equations for of the form . As we obtain

 d(C(L,g))≥2(qm−1+qm−2+⋯+q+1)+⌊1+q+q2+⋯+qm−2q2−1⌋.

## 3 Further improvements for m=3

So far we have improved the bound on in two ways. First we established where . This leads to an increase of the distance bound from to . Then we found additional consecutive powers in the dual code leading to

 d(C(L,g)≥2(1+q+q2+…+qm−2+qm−1)+⌊1+q+q2+⋯+qm