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The Average Value of Bi-Unitary Divisor Function in Function Fields
DOI: 10.12677/PM.2021.1110192, PDF, HTML, XML, 下载: 58  浏览: 127

Abstract: Let Fq be the finite field with q elements. In the function field Fq(T), a monic divisor d of a polynomial f is called unitary, if f=dδ and (d,δ)=1. For polynomials f,g, if d is an unitary divisor of both f and g, it is called the common unitary divisor of them. Let (f,g)∗∗ be the common unitary monic divisor of f and g, whose degree is the largest. We say a divisor g of f is bi-unitary, if f=gh and (g,h)∗∗=1. Let τ∗∗(f) denote the number of bi-unitary divisor of f. We consider the average value of τ∗∗(f) and give a corresponding asymptotic formula.

1. 引言

$\underset{n\le x}{\sum }\tau \left(n\right)=x\left(\mathrm{log}x+2\gamma -1\right)+O\left(\sqrt{x}\right)$,

$\underset{n\le x}{\sum }{\tau }^{\ast }\left(n\right)=\frac{x}{\zeta \left(2\right)}\left(\mathrm{log}x+2\gamma -1-\frac{2{\zeta }^{\prime }\left(2\right)}{\zeta \left(2\right)}\right)+O\left(\sqrt{x}\mathrm{log}x\right)$,

$\underset{n\le x}{\sum }{\tau }^{\ast \ast }\left(n\right)=Ax\left(\mathrm{log}x+2\gamma -1+2\underset{p}{\sum }\frac{{\left(p-1\right)}^{2}{p}^{2}\mathrm{log}p}{\left({p}^{2}+1\right)\left({p}^{4}+2p-1\right)}+2B\right)+O\left({x}^{1/2}\mathrm{log}x\right)$,

${F}_{q}$ 为q元有限域，在函数域 ${F}_{q}\left(T\right)$ 中，类似的我们也可以考虑各类除数函数的均值。对于 $f\in {F}_{q}\left[T\right]$，令 $\tau \left(f\right)$ 表示f的全部首一因式的个数，我们有(见文献 [6] )

$\underset{f}{\sum }\tau \left(f\right)=\left(n+1\right){q}^{n}$,

$\underset{f}{\sum }{\tau }^{\ast }\left(f\right)=\left(n+1\right){q}^{n}+{O}_{\epsilon }\left(n{q}^{n-1/2+\epsilon }\right)$,

1) $d||f,\text{\hspace{0.17em}}d||g$

2) 若有任意的多项式h满足 $h||f,h||g$，则一定有 $h|d$

$\underset{f\in {Μ}_{n}}{\sum }{\tau }^{\ast \ast }\left(f\right)=\left(n+1\right){q}^{n}+{O}_{\epsilon }\left(n{q}^{n-1/2+\epsilon }\right)$.

2. 预备知识与引理

2.1. 函数域上的Zeta函数

${\zeta }_{Α}\left(s\right)=\underset{f\in Μ}{\sum }\frac{1}{{‖f‖}^{s}}$, $\Re \left(s\right)>1$, (2.1)

${\zeta }_{Α}\left(s\right)=\underset{\text{irr}\text{.}\text{\hspace{0.17em}}P\in Μ}{\prod }{\left(1-\frac{1}{{‖P‖}^{s}}\right)}^{-1}$, $\Re \left(s\right)>1$, (2.2)

${\zeta }_{Α}\left(s\right)=\frac{1}{1-{q}^{1-s}}$, $\Re \left(s\right)>1$. (2.3)

${\stackrel{^}{\zeta }}_{Α}\left(u\right):=\frac{1}{1-qu}$, $|u|<{q}^{-1}$. (2.4)

2.2. 双重酉除数函数的性质

${\tau }^{\ast \ast }\left(fg\right)={\tau }^{\ast \ast }\left(f\right){\tau }^{\ast \ast }\left(g\right)$.

${\tau }^{\ast \ast }\left(fg\right)=\underset{\begin{array}{c}d|fg\\ {\left(d,fg/d\right)}^{\ast \ast }=1\end{array}}{\sum }1$.

${\tau }^{\ast \ast }\left(fg\right)=\underset{\begin{array}{c}{d}_{1}{d}_{2}|fg\\ {\left({d}_{1}{d}_{2},fg/\left({d}_{1}{d}_{2}\right)\right)}^{\ast \ast }=1\end{array}}{\sum }1$.

${\left({d}_{1}{d}_{2},fg/\left({d}_{1}{d}_{2}\right)\right)}^{\ast \ast }={\left({d}_{1},f/{d}_{1}\right)}^{\ast \ast }{\left({d}_{2},g/{d}_{2}\right)}^{\ast \ast }$ (2.5)

${\tau }^{\ast \ast }\left(fg\right)=\underset{\begin{array}{c}{d}_{1}{d}_{2}|fg\\ {\left({d}_{1},f/{d}_{1}\right)}^{\ast \ast }{\left({d}_{2},g/{d}_{2}\right)}^{\ast \ast }=1\end{array}}{\sum }1=\underset{\begin{array}{c}{d}_{1}|f\\ {\left({d}_{1},f/{d}_{1}\right)}^{\ast \ast }=1\end{array}}{\sum }1\underset{\begin{array}{c}{d}_{2}|g\\ {\left({d}_{2},g/{d}_{2}\right)}^{\ast \ast }=1\end{array}}{\sum }1={\tau }^{\ast \ast }\left(f\right){\tau }^{\ast \ast }\left(g\right)$,

$t={\left({d}_{1}{d}_{2},fg/\left({d}_{1}{d}_{2}\right)\right)}^{\ast \ast }$, $r={\left({d}_{1},f/{d}_{1}\right)}^{\ast \ast }$, $s={\left({d}_{2},g/{d}_{2}\right)}^{\ast \ast }$.

${\tau }^{\ast \ast }\left({P}^{{\alpha }_{i}}\right)=\left\{\begin{array}{cc}{\alpha }_{i}& {\alpha }_{i}\text{ }\text{ }是偶数,\\ {\alpha }_{i}+1& {\alpha }_{i}\text{ }\text{ }是奇数.\end{array}$

2.3. 其他所需引理

$s\in C$，我们定义

$G\left(s\right)=\underset{\text{irr}\text{.}\text{\hspace{0.17em}}P\in Μ}{\prod }\left(1-\frac{1}{{‖P‖}^{2s}}+\frac{1}{{‖P‖}^{3s-1}\left(‖P‖-1\right)}\right)$,(2.6)

$G\left(s\right)=\underset{n=1}{\overset{\infty }{\prod }}\underset{\text{irr}\text{.}\text{\hspace{0.17em}}P\in {Μ}_{n}}{\prod }\left(1-\frac{1}{{‖P‖}^{2s}}+\frac{1}{{‖P‖}^{3s-1}\left(‖P‖-1\right)}\right)$.

$G\left(s\right)=\underset{n=1}{\overset{\infty }{\prod }}\underset{\text{irr}.\text{\hspace{0.17em}}P\in {Μ}_{n}}{\prod }\left(1-\frac{1}{{q}^{2ns}}+\frac{1}{{q}^{n\left(3s-1\right)}\left({q}^{n}-1\right)}\right)=\underset{n=1}{\overset{\infty }{\prod }}{\left(1-\frac{1}{{q}^{2ns}}+\frac{1}{{q}^{n\left(3s-1\right)}\left({q}^{n}-1\right)}\right)}^{{a}_{n}}$,

$\frac{1}{{q}^{n\left(3\Re \left(s\right)-1\right)}\left({q}^{n}-1\right)}<\frac{2}{{q}^{3n\Re \left(s\right)}}<\frac{2}{{q}^{2n\Re \left(s\right)}}$,

$|G\left(s\right)|\le \underset{n=1}{\overset{\infty }{\prod }}{\left(1+\frac{1}{{q}^{2n\Re \left(s\right)}}\right)}^{\left({C}_{1}{q}^{n}\right)/n}$.(2.7)

$|G\left(s\right)|\le \mathrm{exp}\left\{O\left(\underset{n=1}{\overset{\infty }{\sum }}\frac{1}{n{q}^{n\left(2\Re \left(s\right)-1\right)}}\right)\right\}$.

$|G\left(s\right)|\le C\left(\epsilon \right)$, (2.8)

$\underset{f\in Μ}{\sum }\frac{{\tau }^{\ast \ast }\left(f\right)}{{‖f‖}^{s}}={\zeta }_{Α}^{2}\left(s\right)G\left(s\right)$,(2.9)

$F\left(s\right)=\underset{f\in Μ}{\sum }\frac{{\tau }^{\ast \ast }\left(f\right)}{{‖f‖}^{s}}$,(2.10)

$F\left(s\right)=\underset{\text{irr}.\text{\hspace{0.17em}}P\in Μ}{\prod }\left(1+\frac{{\tau }^{\ast \ast }\left(P\right)}{{‖P‖}^{s}}+\frac{{\tau }^{\ast \ast }\left({P}^{2}\right)}{{‖P‖}^{2s}}+\cdots \right)$,

$F\left(s\right)=\underset{\text{irr}.\text{\hspace{0.17em}}P\in Μ}{\prod }\left(1+\frac{2}{{‖P‖}^{s}}+\frac{2}{{‖P‖}^{2s}}+\cdots +\frac{2k}{{‖P‖}^{\left(2k-1\right)s}}+\frac{2k}{{‖P‖}^{2ks}}+\cdots \right)$.

$F\left(s\right)={\zeta }_{Α}^{2}\left(s\right)G\left(s\right)$.

$\Re \left(s\right)\ge \frac{1+\epsilon }{2}$ 时，我们可以将 $G\left(s\right)$ 写成如下Dirichlet级数的形式

$G\left(s\right)=\underset{f\in Μ}{\sum }\frac{h\left(f\right)}{{‖f‖}^{s}}$, (2.11)

$G\left(s\right)=\underset{l=0}{\overset{\infty }{\sum }}\underset{f\in {Μ}_{l}}{\sum }h\left(f\right){q}^{-ls}$.

$u={q}^{-s}$，上式可以写成

$\stackrel{^}{G}\left(u\right):=\underset{l=0}{\overset{\infty }{\sum }}{h}_{l}{u}^{l}$, $|u|\le {q}^{-1/2-\epsilon }$,(2.12)

${h}_{l}:=\underset{f\in {Μ}_{l}}{\sum }h\left(f\right)$.(2.13)

$|{h}_{l}|\le C\left(\epsilon \right){q}^{\left(l+l\epsilon \right)/2}$.

${h}_{l}=\frac{1}{2\pi i}{\oint }_{\Gamma }\frac{\stackrel{^}{G}\left(u\right)}{{u}^{l+1}}\text{d}u$,(2.14)

$|{h}_{l}|\le \frac{1}{2\pi }{\oint }_{\Gamma }\frac{|\stackrel{^}{G}\left(u\right)|}{|{u}^{l+1}|}|\text{d}u|\le \frac{1}{2\pi }C\left(\epsilon \right){q}^{\left(1+\epsilon \right)\left(l+1\right)/2}{\oint }_{\Gamma }|\text{d}u|$.

$|{h}_{l}|\le \frac{1}{2\pi }C\left(\epsilon \right){q}^{\left(1+\epsilon \right)\left(l+1\right)/2}2\pi {q}^{-\left(1+\epsilon \right)/2}=C\left(\epsilon \right){q}^{\left(l+l\epsilon \right)/2}$.

3. 定理1.1的证明

$F\left(s\right)$ 定义(2.10)式可得

$F\left(s\right)=\underset{n=0}{\overset{\infty }{\sum }}\underset{f\in {Μ}_{n}}{\sum }{\tau }^{\ast \ast }\left(f\right){q}^{-ns}$.

$u={q}^{-s}$，上式可写为

$\stackrel{^}{F}\left(u\right):=\underset{n=0}{\overset{\infty }{\sum }}\underset{f\in {Μ}_{n}}{\sum }{\tau }^{\ast \ast }\left(f\right){u}^{n}$,(3.1)

$\stackrel{^}{F}\left(u\right)={\stackrel{^}{\zeta }}_{Α}^{2}\left(u\right)\stackrel{^}{G}\left(u\right)$.

$\stackrel{^}{F}\left(u\right)={\left(\underset{r=0}{\overset{\infty }{\sum }}{q}^{r}{u}^{r}\right)}^{2}\underset{l=0}{\overset{\infty }{\sum }}{h}_{l}{u}^{l}=\underset{n=0}{\overset{\infty }{\sum }}\underset{r+m+l=n}{\sum }{q}^{r+m}{h}_{l}{u}^{r+m+l}$.(3.2)

$\underset{f\in {Μ}_{n}}{\sum }{\tau }^{\ast \ast }\left(f\right)=\underset{r+m+l=n}{\sum }{q}^{r+m}{h}_{l}$.

$\underset{f\in {Μ}_{n}}{\sum }{\tau }^{\ast \ast }\left(f\right)=\underset{r=0}{\overset{n}{\sum }}{q}^{r}\underset{m=0}{\overset{n-r}{\sum }}{q}^{m}{h}_{n-r-m}=\underset{r=0}{\overset{n}{\sum }}{q}^{r}\underset{m=0}{\overset{n-r}{\sum }}{q}^{n-r-m}{h}_{m}$.

$\underset{f\in {Μ}_{n}}{\sum }{\tau }^{\ast \ast }\left(f\right)={q}^{n}\underset{r=0}{\overset{n}{\sum }}\underset{m=0}{\overset{n-r}{\sum }}{q}^{-m}{h}_{m}$.(3.3)

${h}_{0}=1$ 我们有

$\underset{r=0}{\overset{n}{\sum }}\underset{m=0}{\overset{n-r}{\sum }}{q}^{-m}{h}_{m}=n+1+O\left(\underset{r=0}{\overset{n-1}{\sum }}\underset{m=1}{\overset{n-r}{\sum }}{q}^{-m}{h}_{m}\right)=n+1+O\left(n\underset{m=1}{\overset{n}{\sum }}{q}^{-m}{h}_{m}\right)$.(3.4)

$|n\underset{m=1}{\overset{n}{\sum }}{q}^{-m}{h}_{m}|\le n\underset{m=1}{\overset{n}{\sum }}{q}^{-m}\left(C\left(\epsilon \right){q}^{m\left(1+\epsilon \right)/2}\right)\le C\left(\epsilon \right)n{q}^{\left(\epsilon -1\right)/2}\underset{m=0}{\overset{n-1}{\sum }}{q}^{m\left(\epsilon -1\right)/2}$. (3.5)

$\underset{m=0}{\overset{n-1}{\sum }}{q}^{m\left(\epsilon -1\right)/2}\le \underset{m=0}{\overset{\infty }{\sum }}{q}^{-m/4}\le \underset{m=0}{\overset{\infty }{\sum }}{2}^{-m/4}$,

$\underset{m=0}{\overset{n-1}{\sum }}{q}^{m\left(\epsilon -1\right)/2}=O\left(1\right)$. (3.6)

$n\underset{m=1}{\overset{n}{\sum }}{q}^{-m}{h}_{m}={O}_{\epsilon }\left(n{q}^{\left(\epsilon -1\right)/2}\right)$. (3.7)

$\underset{r=0}{\overset{n}{\sum }}\underset{m=0}{\overset{n-r}{\sum }}{q}^{-m}{h}_{m}=n+1+{O}_{\epsilon }\left(n{q}^{\left(\epsilon -1\right)/2}\right)$. (3.8)

$\underset{f\in {Μ}_{n}}{\sum }{\tau }^{\ast \ast }\left(f\right)=\left(n+1\right){q}^{n}+{O}_{\epsilon }\left(n{q}^{n+\left(\epsilon -1\right)/2}\right)$.

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