Dense families of modular curves, prime numbers and uniform symmetric tensor rank of multiplication in certain finite fields

TL;DR

This paper establishes new uniform bounds for the symmetric tensor rank of multiplication in finite field extensions using dense families of modular curves and advanced prime number theorems.

Contribution

It introduces a novel approach combining modular curves and prime density theorems to improve bounds on tensor rank in finite fields.

Findings

New bounds for symmetric tensor rank in finite fields.

Application of modular curves attaining the Drinfeld-Vladuts bound.

Use of prime number density theorems like Hoheisel and Dudek.

Abstract

We obtain new uniform bounds for the symmetric tensor rank of multiplication in finite extensions of any finite field Fp or Fp2 where p denotes a prime number greater or equal than 5. In this aim, we use the symmetric Chudnovsky-type generalized algorithm applied on sufficiently dense families of modular curves defined over Fp2 attaining the Drinfeld-Vladuts bound and on the descent of these families to the definition field Fp. These families are obtained thanks to prime number density theorems of type Hoheisel, in particular a result due to Dudek (2016).

Equations

$$T=\sum _{i=1}^r{x}_i^{\star }\otimes y_i^{\star }\otimes c_i,$$

$$x\cdot y=\sum _{i=1}^r{x}_i^{\star }(x)y_i^{\star }(y)c_i.$$

$$\mu_{q}(n)=\min\left\{r\;\Big{|}\;T=\sum_{i=1}^{r}x_{i}^{\star}\otimes y_{i}^{\star}\otimes c_{i}\right\}.$$

$$T=\sum _{i=1}^r{x}_i^{\star }\otimes x_i^{\star }\otimes c_i.$$

$$\mu^{\mathrm{sym}}_{q}(n)=\min\left\{r\;\Big{|}\;T=\sum_{i=1}^{r}x_{i}^{\star}\otimes x_{i}^{\star}\otimes c_{i}\right\}.$$

$$M_q^{\mathrm{sym}}=\underset{k\to \infty }{\mathrm{limsup}}\frac{{\mu }_q^{\mathrm{sym}}(k)}{k},$$

$$m_q^{\mathrm{sym}}=\underset{k\to \infty }{\mathrm{liminf}}\frac{{\mu }_q^{\mathrm{sym}}(k)}{k}.$$

$$N_{q}(g)=\max\big{\{}N_{1}(F)\,|\,F\mbox{ is a function field over }\mathbb{F}_{q}\mbox{ of genus }g\big{\}}$$

$$A(q)=\underset{g\to +\infty }{\mathrm{limsup}}\frac{N_q(g)}{g}.$$

$$A(q)\le q^{\frac{1}{2}}-1,$$

$${\mu }_q^{\mathrm{sym}}(n)\le 2n+g-1,$$

$${\mu }_q^{\mathrm{sym}}(n)\le 3n+2g.$$

$${\mu }_q^{\mathrm{sym}}(n)\le C_{q}n.$$

$$l_{k+1}-l_k\le l_k^{\alpha }$$

$${\mu }_{p^2}^{\mathrm{sym}}(n)\le 2\left(1+\frac{1+{ϵ}_p(n)}{p-3}\right)n-\frac{(1+{ϵ}_p(n))(p+1)}{p-3}-1,$$

$${\mu }_{p^2}^{\mathrm{sym}}(n)\le 2\left(1+\frac{1+{ϵ}_p(n)}{p-3}\right)n-\frac{2(1+{ϵ}_p(n))(p-1)}{p-3},$$

$$M_{p^2}^{\mathrm{sym}}\le 2\left(1+\frac{1}{p-3}\right).$$

$$(p-1)(l_{k+1}+1)>2n+2l_{k+1}-2$$

$$(p-1)(l_k+1)\le 2n+2l_k-2$$

$$N_1(F_{k+1}/{\mathbb{F}}_{p^2})\ge (p-1)(l_{k+1}+1)>2n+2l_{k+1}-2.$$

$$q^{l_k\frac{p-3}{4}+\frac{p-1}{4}}(q^{\frac{1}{2}}-1)\ge 2(1+ϵ(l_k))l_k+1,$$

$$q^{x\frac{p-3}{4}+\frac{p-1}{4}}(q^{\frac{1}{2}}-1)-4x-1\ge 0$$

$${\mu }_{p^2}^{\mathrm{sym}}(n)\le 2n+l_{k+1}-1\le 2n+(1+ϵ(l_k))l_k-1,$$

$$N_1(X_k({\mathbb{F}}_{p^2}))\ge \frac{{\mu }_N(p-1)/12}{\mathrm{deg}{\lambda }_N}\ge 2(p-1)(l_k+1)$$

$$2(p-1)(l_{k+1}+1)>2n+2(2l_{k+1}+1)-2$$

$$2(p-1)(l_k+1)\le 2n+2(2l_k+1)-2,$$

$$(p-1)(l_{k+1}+1)>n+2l_{k+1}$$

$$(p-1)(l_k+1)\le n+2l_k.$$

$$N_1(F_{k+1}/{\mathbb{F}}_{p^2})\ge 2(p-1)(l_{k+1}+1)>2n+4l_{k+1}.$$

$$11^{4l_k+\frac{9}{2}}(11^{\frac{1}{2}}-1)\ge 8l_k+3.$$

Full text

The second author was partially supported by ANR Globes ANR-12-JS01-0007-01 and by the Russian Academic Excellence Project ’5-100’.

1. Introduction

1.1. Notation

Let $q=p^{s}$ be a prime power, $\mathbb{F}_{q}$ be the finite field with $q$ elements and $\mathbb{F}_{q^{n}}$ be the degree $n$ extension of $\mathbb{F}_{q}$. The multiplication of two elements of $\mathbb{F}_{q^{n}}$ is an $\mathbb{F}_{q}$-bilinear application from $\mathbb{F}_{q^{n}}\times\mathbb{F}_{q^{n}}$ onto $\mathbb{F}_{q^{n}}$. It can be considered as an $\mathbb{F}_{q}$-linear application from the tensor product ${\mathbb{F}_{q^{n}}\otimes_{\mathbb{F}_{q}}\mathbb{F}_{q^{n}}}$ onto $\mathbb{F}_{q^{n}}$. Consequently it can be also viewed as an element $T$ of ${(\mathbb{F}_{q^{n}}\otimes_{\mathbb{F}_{q}}\mathbb{F}_{q^{n}})^{\star}\otimes_{\mathbb{F}_{q}}\mathbb{F}_{q^{n}}}$, namely an element of ${\mathbb{F}_{q^{n}}^{\,\star}\otimes_{\mathbb{F}_{q}}\mathbb{F}_{q^{n}}^{\,\star}\otimes_{\mathbb{F}_{q}}\mathbb{F}_{q^{n}}}$. More precisely, when $T$ is written

[TABLE]

where the $r$ elements $x_{i}^{\star}$ and the $r$ elements $y_{i}^{\star}$ are in the dual $\mathbb{F}_{q^{n}}^{\,\star}$ of $\mathbb{F}_{q}$ and the $r$ elements $c_{i}$ are in $\mathbb{F}_{q^{n}}$, the following holds for any ${x,y\in\mathbb{F}_{q^{n}}}$:

[TABLE]

Definition 1**.**

The minimal number of summands in a decomposition of the multiplication tensor $T$ is called the rank of the tensor of the multiplication in the extension field $\mathbb{F}_{q^{n}}$ (or bilinear complexity of the multiplication) and is denoted by $\mu_{q}(n)$:

[TABLE]

It is known that the tensor $T$ can have a symmetric decomposition:

[TABLE]

Definition 2**.**

The minimal number of summands in a symmetric decomposition of the multiplication tensor $T$ is called the symmetric tensor rank of the multiplication (or the symmetric bilinear complexity of the multiplication) and is denoted by $\mu^{\mathrm{sym}}_{q}(n)$:

[TABLE]

From an asymptotical point of view, let us define the following

[TABLE]

[TABLE]

Let $F/\mathbb{F}_{q}$ be a function field of genus $g$ over the finite field $\mathbb{F}_{q}$ and $N_{k}(F)$ be the number of places of degree $k$ of $F/\mathbb{F}_{q}$.

Let us define:

[TABLE]

and

[TABLE]

We know that (Drinfeld–Vladuts bound):

[TABLE]

the bound being attained if $q$ is a square.

1.2. Known results

The original algorithm of D.V. and G.V. Chudnovsky introduced in [10] is symmetric by definition and leads to the two following results from [3], [8] and [7]:

Theorem 3**.**

Let $q$ be a prime power and let $n>1$ be an integer. Let $F/\mathbb{F}_{q}$ be an algebraic function field of genus $g$ and $N_{k}$ be the number of places of degree $k$ in $F/\mathbb{F}_{q}$. If $F/\mathbb{F}_{q}$ is such that $2g+1\leq q^{\frac{n-1}{2}}(q^{\frac{1}{2}}-1)$ then:

if $N_{1}>2n+2g-2$, then

[TABLE] 2. 2)

if ${N_{1}+2N_{2}>2n+2g-2}$ and there exists a non-special divisor of degree $g-1,$ then

[TABLE]

Theorem 4**.**

Let $q$ be a power of a prime $p$ and let $n$ be an integer. Then the symmetric tensor rank $\mu^{\mathrm{sym}}_{q}(n)$ of multiplication in any finite field $\mathbb{F}_{q^{n}}$ is linear with respect to the extension degree; more precisely, there exists a constant $C_{q}$ such that for any integer $n>1$,

[TABLE]

From different versions of symmetric algorithms of Chudnovsky type applied to good towers of algebraic function fields of type Garcia–Stichtenoth attaining the Drinfeld–Vladuts bounds of order one, two or four, different authors have obtained uniform bounds for the tensor rank of multiplication, namely general expressions for $C_{q}$, such as the following best currently published estimates:

Theorem 5**.**

Let ${q=p^{r}}$ be a power of a prime $p$ and let $n$ be an integer $>1$. Then:

1. (i)

If ${q=2}$, then $\displaystyle{\mu^{\mathrm{sym}}_{q}(n)\leq 15.46n}$ (cf. **[6, Corollary 29]** and **[9]**) 2. (ii)

If ${q=3}$, then $\displaystyle{\mu^{\mathrm{sym}}_{q}(n)\leq 7.732n}$ (cf. **[6, Corollary 29]** and **[9]**) 3. (iii)

If ${q\geq 4}$, then $\displaystyle{\mu^{\mathrm{sym}}_{q}(n)\leq 3\left(1+\frac{\frac{4}{3}p}{q-3+2(p-1)\frac{q}{q+1}}\right)n}$ (cf. **[7]**) 4. (iv)

If $p\geq 5$, then $\displaystyle{\mu^{\mathrm{sym}}_{p}(n)\leq 3\left(1+\frac{8}{3p-5}\right)n}$ (cf. **[7]**) 5. (v)

If ${q\geq 4}$, then $\displaystyle{\mu^{\mathrm{sym}}_{q^{2}}(n)\leq 2\left(1+\frac{p}{q-3+(p-1)\frac{q}{q+1}}\right)n}$ (cf. **[1]** and **[7]**) 6. (vi)

If $p\geq 5$, then $\displaystyle{\mu^{\mathrm{sym}}_{p^{2}}(n)\leq 2\left(1+\frac{2}{p-\frac{33}{16}}\right)n}$ (cf. **[7]**)

1.3. New results

The main goal of the paper is to improve the upper bounds for $\mu^{\mathrm{sym}}_{q}(n)$ from the previous theorem for the assertions concerning the extensions of finite fields $\mathbb{F}_{p^{2}}$ and $\mathbb{F}_{p}$ where $p$ is a prime number. Note that one of main ideas used in this paper was introduced in [4] by the first author thanks to the use of the Chebyshev Theorem (or also called the Bertrand Postulat) to bound the gaps between prime numbers in order to construct families of modular curves as dense as possible. Later, motivated by [4], the approach of using such bounds on gaps between prime numbers (e.g. Baker-Harman-Pintz) was also used in the preprint [12] in order to improve the upper bounds of $\mu^{\mathrm{sym}}_{p^{2}}(n)$ where $p$ is a prime number. In our paper, we improve all the known uniform upper bounds for $\mu^{\mathrm{sym}}_{p^{2}}(n)$ and $\mu^{\mathrm{sym}}_{p}(n)$ for $p\geq 5$.

2. New upper bounds

In this section, we give new better upper bounds for the symmetric tensor rank of multiplication in certain extensions of finite fields $\mathbb{F}_{p^{2}}$ and $\mathbb{F}_{p}$. In order to do that, we construct suitable families of modular curves defined over $\mathbb{F}_{p^{2}}$ and $\mathbb{F}_{p}$.

Theorem 6**.**

Let $l_{k}$ be the $k$-th prime number. Then there exists a real number $\alpha<1$ such that the difference between two consecutive prime numbers $l_{k}$ and $l_{k+1}$ satisfies

[TABLE]

for any prime $l_{k}\geq x_{\alpha}.$

In particular, one can take $\alpha=\frac{21}{40}$ with the value of $x_{\alpha}$ that can in principle be determined effectively, or $\alpha=\frac{2}{3}$ with $x_{\alpha}=\exp(\exp(33.3)).$

Proof.

It is known that for all $x>x_{\alpha}$, the interval $[x-x^{\alpha},x]$ with $\alpha=\frac{21}{40}$ contains prime numbers by a result of Baker, Harman and Pintz [2, Theorem 1]. Moreover, the value of $x_{\alpha}$ can in principle be determined, according to the authors. However, to our knowledge, this computation has not been realized yet.

For a bigger $\alpha=\frac{2}{3}$, Dudek obtained recently in [11] an explicit bound $x_{\alpha}\geq\exp(\exp(33.3)).$

∎∎

2.1. The case of the quadratic extensions of prime fields

Proposition 7**.**

Let $p\geq 5$ be a prime number, and let $x_{\alpha}$ be the constant from Theorem 6.

1. (1)

If $p\neq 11,$ then for any integer $\displaystyle n\geq\frac{p-3}{2}x_{\alpha}+\frac{p+1}{2}$ we have

[TABLE]

where $\displaystyle\epsilon_{p}(n)=\left(\frac{2n}{p-3}\right)^{\alpha-1}.$ 2. (2)

For $p=11$ and $\displaystyle n\geq(p-3)x_{\alpha}+p-1=8x_{\alpha}+10$ we have

[TABLE]

where $\displaystyle\epsilon_{p}(n)=\left(\frac{n}{p-3}\right)^{\alpha-1}.$ 3. (3)

Asymptotically the following inequality holds for any $p\geq 5$:

[TABLE]

Proof.

First, let us consider the characteristic $p$ such that $p\neq 11$. Then it is known ([15, Corollary 4.1.21] and [14, proof of Theorem 3.9]) that the modular curve $X_{k}=X_{0}(11l_{k})$, where $l_{k}$ is the $k$-th prime number, is of genus $g_{k}=l_{k}$ and satisfies ${N_{1}(X_{k}(\mathbb{F}_{p^{2}}))\geq(p-1)(g_{k}+1)},$ where $N_{1}(X_{k}(\mathbb{F}_{p^{2}}))$ denotes the number of rational points over $\mathbb{F}_{p^{2}}$ of the curve $X_{k}$. Let us consider an integer $n>1$. Then there exist two consecutive prime numbers $l_{k}$ and $l_{k+1}$ such that

[TABLE]

and

[TABLE]

(here we use the fact that $p\geq 5$). Let us consider the algebraic function field $F_{k+1}/\mathbb{F}_{p^{2}}$ associated to the curve $X_{k+1}$ of genus $l_{k+1}$ defined over $\mathbb{F}_{p^{2}}$. Denoting by $N_{i}(F_{k}/\mathbb{F}_{p^{2}})$ the number of places of degree $i$ of $F_{k}/\mathbb{F}_{p^{2}}$, we get

[TABLE]

We also know that $l_{k+1}-l_{k}\leq l_{k}^{\alpha},$ when $l_{k}\geq x_{\alpha}$ by Theorem 6. Thus $l_{k+1}\leq(1+\epsilon(l_{k}))l_{k},$ with $\epsilon(l_{k})=l_{k}^{\alpha-1}$.

It is easy to check that the inequality $2g+1\leq q^{\frac{n-1}{2}}(q^{\frac{1}{2}}-1)$ of Theorem 3 holds for any prime power $q\geq 5.$ Indeed, it is enough to verify that

[TABLE]

which is true since

[TABLE]

for any $x\geq 0.$

Thus, for any integer $n\geq\frac{p-3}{2}x_{\alpha}+\frac{p+1}{2}$ the function field $F_{k+1}/\mathbb{F}_{p^{2}}$ satisfies Theorem 3, so

[TABLE]

with $l_{k}\leq\frac{2n}{p-3}-\frac{p+1}{p-3}$ by (6).

Let us remark that, as $l_{k}\leq\frac{2n}{p-3}$, $\epsilon(l_{k})\leq\epsilon_{p}(n)=(\frac{2n}{p-3})^{\alpha-1},$ which gives the first inequality.

Now, let us consider the characteristic $p=11$. Take the modular curve $X_{k}=X_{0}(23l_{k})$, where $l_{k}$ is the $k$-th prime number. By [15, Proposition 4.1.20], we easily compute that the genus of $X_{k}$ is $g_{k}=2l_{k}+1$. It is also known that the curve $X_{k}$ has good reduction modulo $p$ outside $23$ and $l_{k}$. Moreover, by using [15, Proof of Theorem 4.1.52], we obtain that the number of $\mathbb{F}_{p^{2}}$-rational points over of the reduction $X_{k}/p$ modulo $p$ satisfies

[TABLE]

in the notation of loc. cit.

Let us take an integer $n>1$. There exist two consecutive prime numbers $l_{k}$ and $l_{k+1}$ such that

[TABLE]

and

[TABLE]

i.e.

[TABLE]

and

[TABLE]

Let us consider the algebraic function field $F_{k+1}/\mathbb{F}_{p^{2}}$ associated to the curve $X_{k+1}$ of genus $g_{k+1}=2l_{k+1}+1$ defined over $\mathbb{F}_{p^{2}}$. We have

[TABLE]

As before $l_{k+1}\leq(1+\epsilon(l_{k}))l_{k},$ with $\epsilon(l_{k})=l_{k}^{\alpha-1}$.

It is also easy to check that the inequality $2g+1\leq q^{\frac{n-1}{2}}(q^{\frac{1}{2}}-1)$ of Theorem 3 holds when $q$ is a power of $11,$ which follows from the fact that

[TABLE]

Thus, for any integer $n\geq(p-3)x_{\alpha}+p-1$, the algebraic function field $F_{k+1}/\mathbb{F}_{p^{2}}$ satisfies Theorem 3, so

[TABLE]

with $l_{k}\leq\frac{n}{p-3}-\frac{p-1}{p-3}$ by (8).

We remark that as $l_{k}\leq\frac{n}{p-3}$, $\epsilon(l_{k})\leq\epsilon_{p}(n)=(\frac{n}{p-3})^{\alpha-1},$ which gives the second inequality of the proposition.

Finally, when $n\rightarrow+\infty,$ the prime numbers $l_{k}\rightarrow+\infty,$ thus both for $p\neq 11$ and $p=11$ the corresponding $\epsilon_{p}(n)\rightarrow 0.$ So in the two cases we obtain

[TABLE]

∎∎

Remark 8**.**

It is easy to see that the bounds obtained in Proposition 7 are generally better than the published best known bounds (v) and (vi) recalled in Theorem 5. Indeed, it is sufficient to consider the asymptotic bounds which are deduced from them and to see that for any prime $p\geq 5$ we have $\frac{1}{p-3}<\frac{p}{p-3+(p-1)\frac{p}{p+1}}$ and $\frac{1}{p-3}<\frac{2}{p-\frac{33}{16}}$ respectively.

Remark 9**.**

Note that the bounds obtained in [12, Corollary 28] also concern the symmetric tensor rank of multiplication in the finite fields even if it is not mentioned. Indeed, the distinction between $\mu^{\mathrm{sym}}_{q}(n)$ and $\mu_{q}(n)$ was exploited only from [13]. So, we can compare our proposition 7 with Corollary 8 there. Firstly, note that the bounds in [12, Corollary 28] are only valid for $p\geq 7$. Moreover, the only bound which is best than our bounds is the asymptotic bound [12, Corollary 28, Bound (vi)] given for an unknown sufficiently large $n$, contrary to our uniform bound with $\alpha=\frac{2}{3}$ for $n\geq exp(\exp(33.3))$.

2.2. The case of prime fields

Proposition 10**.**

Let $p\geq 5$ be a prime number, let $x_{\alpha}$ be defined as in Lemma 6, and $\epsilon_{p}(n)$ as in Proposition 7.

1. (1)

If $p\neq 11,$ then for any integer $\displaystyle n\geq\frac{p-3}{2}x_{\alpha}+\frac{p+1}{2}$ we have

[TABLE] 2. (2)

For $p=11$ and $\displaystyle n\geq(p-3)x_{\alpha}+p-1=8x_{\alpha}+10$ we have

[TABLE] 3. (3)

Asymptotically the following inequality holds for any $p\geq 5$:

[TABLE]

Proof.

It suffices to consider the same families of curves as in the proof of Proposition 7.

When $p\neq 11$ we take $X_{k}=X_{0}(11l_{k})$, where $l_{k}$ is the $k$-th prime number. These curves are defined over $\mathbb{F}_{p}$, hence, we can consider the associated algebraic function fields $F_{k}/\mathbb{F}_{p}$ defined over $\mathbb{F}_{p}$ and we have $N_{1}(F_{k}/\mathbb{F}_{p^{2}})=N_{1}(F_{k}/\mathbb{F}_{p})+2N_{2}(F_{k}/\mathbb{F}_{p})\geq(p-1)(l_{k}+1),$ since $F_{k}/\mathbb{F}_{p^{2}}=F_{k}/\mathbb{F}_{p}\otimes_{\mathbb{F}_{p}}\mathbb{F}_{p^{2}}$ for any $k$. Note that the genus of the algebraic function fields $F_{k}/\mathbb{F}_{p}$ is also $g_{k}=l_{k},$ since the genus is preserved under descent.

Given an integer $n>1$, there exist two consecutive prime numbers $l_{k}$ and $l_{k+1}$ such that

[TABLE]

and

[TABLE]

Let us consider the algebraic function field $F_{k+1}/\mathbb{F}_{p}$ associated to the curve $X_{k+1}$ of genus $l_{k+1}$ defined over $\mathbb{F}_{p}$. We get

[TABLE]

As before $l_{k+1}\leq(1+\epsilon(l_{k}))l_{k},$ with $\epsilon(l_{k})=l_{k}^{\alpha-1},$ and from the proof of the previous proposition we know that the inequality $2g+1\leq q^{\frac{n-1}{2}}(q^{\frac{1}{2}}-1)$ of Theorem 3 holds. Consequently, for any integer $n\geq\frac{p-3}{2}x_{\alpha}+\frac{p+1}{2}$, the algebraic function field $F_{k+1}/\mathbb{F}_{p}$ satisfies Theorem 3, 2) since by [5, Theorem 11 (i)] there always exists a non-special divisor of degree $g_{k+1}-1$ for $p\geq 5$. So

[TABLE]

with $l_{k}\leq\frac{2n}{p-3}-\frac{p+1}{p-3}$ by (10). As before, $\epsilon(l_{k})\leq\epsilon_{p}(n)=(\frac{2n}{p-3})^{\alpha-1}.$

When $p=11$ we use once again the family of curves $X_{k}=X_{0}(23l_{k}).$ They are defined over $\mathbb{F}_{p},$ hence we can consider the associated algebraic function fields $F_{k}/\mathbb{F}_{p}$ over $\mathbb{F}_{p}$ and we have $N_{1}(F_{k}/\mathbb{F}_{p^{2}})=N_{1}(F_{k}/\mathbb{F}_{p})+2N_{2}(F_{k}/\mathbb{F}_{p})\geq(p-1)(l_{k}+1).$ The genus of the algebraic function fields $F_{k}/\mathbb{F}_{p}$ defined over $\mathbb{F}_{p}$ is also $g_{k}=2l_{k}+1$ since the genus is preserved under descent.

Given an integer $n>1$, there exist two consecutive prime numbers $l_{k}$ and $l_{k+1}$ such that

[TABLE]

and

[TABLE]

i.e.

[TABLE]

and

[TABLE]

Let us consider the algebraic function field $F_{k+1}/\mathbb{F}_{p}$ associated to the curve $X_{k+1}$ of genus $g_{k+1}=2l_{k+1}+1$ defined over $\mathbb{F}_{p}$. We get

[TABLE]

As above $l_{k+1}\leq(1+\epsilon(l_{k}))l_{k},$ with $\epsilon(l_{k})=l_{k}^{\alpha-1},$ and the inequality $2g+1\leq q^{\frac{n-1}{2}}(q^{\frac{1}{2}}-1)$ of Theorem 3 holds. Consequently, for any integer $n\geq(p-3)x_{\alpha}+p-1$, the algebraic function field $F_{k+1}/\mathbb{F}_{p}$ satisfies Theorem 3, 2) since, as before, there exists a non-special divisor of degree $g_{k+1}-1$ by [5, Theorem 11 (i)]. So,

[TABLE]

with $l_{k}\leq\frac{n}{p-3}-\frac{p-1}{p-3}$ by (12). We can also bound $\epsilon(l_{k})\leq\epsilon_{p}(n)=(\frac{n}{p-3})^{\alpha-1}.$

Finally, when $n\rightarrow+\infty,$ the prime numbers $l_{k}\rightarrow+\infty,$ thus both for $p\neq 11$ and $p=11,$ $\epsilon_{p}(n)\rightarrow 0.$ So we obtain $M^{sym}_{p}\leq 3\left(1+\frac{\frac{4}{3}}{p-3}\right)$. ∎

∎

Remark 11**.**

It is easy to see that the bounds obtained in Proposition 10 are generally better than the best known bounds (iii) and (iv) recalled in Theorem 5. Indeed, it is sufficient to consider the asymptotic bounds which are deduced from them and to see that for any prime $p\geq 5$ we have $\frac{\frac{4}{3}}{p-3}<\frac{\frac{4}{3}p}{p-3+\frac{2(p-1)p}{p+1}}$ and $\frac{\frac{4}{3}}{p-3}<\frac{8}{3p-5}$ respectively.

Acknowledgments

The first author wishes to thank Sary Drappeau, Olivier Ramaré, Hugues Randriambololona, Joël Rivat and Serge Vladuts for valuable discussions.