Degrees of irreducible polynomials over binary field

TL;DR

This paper investigates the degrees of polynomials over binary fields related to their associated matrices and irreducibility, providing insights into polynomial factorization over finite fields.

Contribution

It introduces a study of the degrees of polynomials over binary fields linked to their matrices and irreducibility, expanding understanding of polynomial structure.

Findings

Analysis of degrees of polynomials over binary fields

Relationship between matrices and polynomial irreducibility

Insights into polynomial factorization over finite fields

Abstract

An algorithm for factoring polynomials over finite fields is given by Berlekamp in 1967. The main tool was the matrix Q corresponding to each polynomial. This paper studies the degrees of polynomials over binary field that associated with their corresponding matrices Q and irreducibility.

Equations

$$x^{2i}\equiv \sum _{j=0}^{m-1}{Q}_{i+1,j+1}{x}^j(\text{mod}f)$$

$$G^{o(G)}=I_m.$$

$$o(G)=\text{lcm}(d_1,d_2,\dots ,d_r).$$

$$\text{lcm}(k{n}_1,k{n}_2,\dots ,k{n}_{\ell })=k\cdot \text{lcm}(n_1,n_2,\dots ,n_{\ell }).$$

$${\mathcal{P}}_1:G^m=I_m\text{if and only if}f(x)\text{is irreducible}$$

$$m\text{is an odd prime or 9}.$$

$${\mathcal{P}}_1^{\prime }:f(x)\text{divides}{x}^{2^m}-x\text{if and only if}f(x)\text{is irreducible.}$$

$$x^{2^m}-x=\prod _{d\mid m}(\text{Irreducible polynomials of degree}d),$$

$${\mathcal{P}}_1^{\prime \prime }:\text{If}f(x)\text{divides}{x}^{2^m}-x\text{then}f(x)\text{is irreducible.}$$

$$N(\ell)\left\{\begin{array}[]{ll}=2&\text{if}~{}\ell=1\\ =\ell-1&\text{if}~{}\ell=2,3,4\\ \geq\ell&\text{if}~{}\ell>4\end{array}\right..$$

$$x^{2^m}-x=x(x+1)\prod \text{(Irreducible polynomials of degree m)},$$

$$x^{2^9}-x=x(x+1)(x^3+x+1)(x^3+x^2+1)\prod \text{(Irreducible polynomials of degree 9)},$$

$${\mathcal{P}}_2:o(G)=m\text{if and only if}f(x)\text{is irreducible}$$

$$m=p^i\text{or}m=p^{i}q$$

$$p_1{p}_2{p}_3=1\cdot p_1{p}_3+(p_2-p_1)\cdot p_2+(p_1-1)\cdot p_2{p}_3,$$

$$p^i{q}^j=(p-1)\cdot p^{i-1}{q}^j+1\cdot p^i{q}^{j-1}+(q-p)\cdot p^{i-1}{q}^{j-1},$$

$$x^{2^m}-x=(x^{2^{p^i}}-x)\prod _{j=1}^i\left(\text{Irreducible polynomials of degree}qj\right),$$

$$\{uq(\text{mod}{p}^i)\mid u=1,2,\dots ,p^i-1\}=\{1,2,\dots ,p^i-1\}.$$

Full text

Keywords: Irreducible polynomial, binary field, Berlekamp matrix.

MSC2010: 11T06.

1 Introduction

Let $F=\mathbb{F}_{2}=\{0,1\}$ be the binary finite field. Then for each polynomial $f(x)\in F[x]$ of degree $m,$ the Berlekamp matrix $Q$ proposed in [1] of $f(x)$ is the $m\times m$ matrix over $F$ whose $i$th row represents $x^{2(i-1)}$ reduced modulo $f(x).$ Specifically,

[TABLE]

for $i=0,1,\ldots,m-1.$

2 Preliminary

Let $G$ be the Berlekamp matrix with respect to the polynomial $f(x)$ of degree $m$ over $F.$ It is not difficult to show that $G^{m}=I_{m}$ if and only if $f(x)$ has no square factors, where $I_{m}$ denotes the identity matrix of order $m.$ If $f(x)$ has no square factors then the order $o(f(x))=o(G)$ of $f(x)$ is defined to be the least positive integer such that

[TABLE]

Lemma 2.1**.**

Let $f(x)=\prod_{i=1}^{r}g_{i}(x)$ where $g_{i}(x)$ are distinct polynomials of order $d_{i}$ over $F$ for $1\leq i\leq r.$ Then

[TABLE]

It is not hard to have the following observation for least common multiple.

Lemma 2.2**.**

For positive integers $n_{1},n_{2},\ldots,n_{\ell}$ and $k,$ the least common multiple

[TABLE]

Proof.

Let $A=\text{lcm}(kn_{1},kn_{2},\ldots,kn_{\ell})$ and $B=\text{lcm}(n_{1},n_{2},\ldots,n_{\ell}).$ For each $i=1,2,\ldots,\ell,$ one has $kn_{i}|kB$ since $n_{i}|B,$ and hence $A|kB.$ On the other hand, $kn_{i}|A$ implies $n_{i}|A/k,$ and thus $B|A/k.$ The result follows. ∎

3 Main results

Let $G$ be the Berlekamp matrix with respect to the polynomial $f(x)$ of degree $m$ over $F.$ The property $\mathcal{P}_{1}$ is defined as

[TABLE]

where $I_{m}$ is the identity matrix of order $m.$

Theorem 3.1**.**

Let $f(x)$ be a polynomial over $F$ of degree $m\geq 2.$ Then $f(x)$ has the property $\mathcal{P}_{1}$ if and only if

[TABLE]

Proof.

Note that $G^{m}=I_{m}$ if and only if $f(x)$ divides $x^{2^{m}}-x.$ Then $\mathcal{P}_{1}$ is also realized as

[TABLE]

Moreover, since

[TABLE]

the property $\mathcal{P}_{1}^{\prime}$ is equivalent to

[TABLE]

To prove the sufficiency, suppose that $f(x)$ has the property $\mathcal{P}_{1}^{\prime\prime}.$ Note that the number of irreducible polynomials $N(\ell)$ over $F$ of degree $\ell$ is

[TABLE]

Hence the degree $m$ of $f(x)$ can not be written as $m=n\ell$ for some positive integers $1<n\leq\ell$ and $\ell\geq 5,$ or otherwise $N(\ell)\geq n$ and a product of $n$ irreducible polynomials of degree $\ell$ does not have $\mathcal{P}_{1}^{\prime\prime}.$ For the some reason, $m$ can not be written as $m=n\ell$ for some positive integers $1<n<\ell,$ either. Furthermore, $m\neq 2$ and $m\neq 4$ since both of $x(x+1)$ and $x(x+1)(x^{2}+x+1)$ do not have $\mathcal{P}_{1}^{\prime\prime}.$ To conclude the above argument, $m$ is an odd prime or $m=3\cdot 3=9.$

For the necessity, assume $m$ is an odd prime or $9.$ If $m$ is an odd prime, then

[TABLE]

and hence $f(x)$ is an irreducible polynomial of degree $m$ whenever $f(x)$ divides $x^{2^{m}}-x.$ Besides, if $m=9$ then

[TABLE]

and thus $f(x)$ is an irreducible polynomial of degree $9$ provided that $f(x)$ divides $x^{2^{9}}-x.$ It says that $f(x)$ has $\mathcal{P}_{1}^{\prime\prime},$ and the proof is completed. ∎

Let $G$ be the Berlekamp matrix with respect to the polynomial $f(x)$ of degree $m$ over $F.$ The property $\mathcal{P}_{2}$ is defined as

[TABLE]

where $o(G)$ is the order of $G.$

Theorem 3.2**.**

Let $f(x)$ be a polynomial over $F$ of degree $m\geq 2.$ If $f(x)$ has the property $\mathcal{P}_{2}$ then $m$ can be written as

[TABLE]

for primes $p<q$ and positive integer $i.$

Proof.

Suppose to the contrary that $m$ can be written as a product of three pairwise coprime factors that more than $1,$ or $m=p^{i}q^{j}$ for primes $p<q$ and positive integers $i,j$ with $j\geq 2.$

Assume that $m$ can be written as $m=p_{1}p_{2}p_{3},$ where positive integers $1<p_{1}<p_{2}<p_{3}$ are pairwise coprime. Since

[TABLE]

a product of $1,$ $p_{2}-p_{1},$ and $p_{1}-1$ irreducible polynomials respectively of degrees $p_{1}p_{3},$ $p_{2},$ and $p_{2}p_{3}$ does not have the property $\mathcal{P}_{2},$ which is a contradiction. (It is quick to check that the number $N_{2}(\ell)$ of irreducible polynomials over $F$ of degree $\ell$ satisfies $N_{2}(\ell)\geq\ell-1$ for each positive integer $\ell,$ and $\text{lcm}(p_{1}p_{3},p_{2},p_{2}p_{3})=p_{1}p_{2}p_{3}=m.$)

Next, suppose that $m$ can be written as $m=p^{i}q^{j}$ for primes $p<q$ and positive integers $i,j$ with $j\geq 2.$ Since

[TABLE]

a product of $p-1,$ $1,$ and $q-p$ irreducible polynomials respectively of degrees $p^{i-1}q^{j},$ $p^{i}q^{j-1},$ and $p^{i-1}q^{j-1}$ does not have the property $\mathcal{P}_{2},$ which is a contradiction. (It is immediate to check that $\text{lcm}(p^{i-1}q^{j},p^{i}q^{j-1},p^{i-1}q^{j-1})=p^{i}q^{j}=m.$) The result follows. ∎

The case will be trivial if the degree $m$ is a prime power. It may be quick to show that $f(x)\in F[x]$ has $\mathcal{P}_{2}$ if its degree $m=p^{i}$ for some prime $p$ and positive integer $i.$ Now, focus on the case $m=p^{i}q$ for primes $p<q$ and positive integer $i.$

Note that the order $o(f(x))$ of $f(x)$ is the least common multiple (l.c.m.) of the degrees of factors in $f(x).$ Then a quick observation is given below.

Lemma 3.3**.**

Assume that $f(x)\in F[x]$ is of degree $m=p^{i}q$ for primes $p<q$ and positive integer $i.$ If the order $o(f(x))=m$ then there exists a factor of $f(x)$ that of degree $p^{i}$ or $p^{i}q.$

∎

Corollary 3.4**.**

Let $f(x)\in F[x]$ of degree $m=p^{i}q$ for primes $p<q$ and positive integer $i.$ Then the following properties follow.

• (i)

If $f(x)$ has $\mathcal{P}_{2}$ then a polynomial $\tilde{f}(x)\in F[x]$ of degree $\tilde{m}=p^{\tilde{i}}q$ with positive integer $\tilde{i}\leq i$ also has $\mathcal{P}_{2}.$

• (ii)

If $q>2^{p^{i}}$ then $f(x)$ has $\mathcal{P}_{2}.$

• (iii)

If $p^{i}=2$ then $q>4$ if and only if $f(x)$ has $\mathcal{P}_{2}.$

• (iv)

If $q>p^{i}>2$ and $f(x)$ has $\mathcal{P}_{2},$ then $(p^{i}-2)q>2^{p^{i}}-2^{p^{i-1}}+1.$ If $q<p^{i}$ and $f(x)$ has $\mathcal{P}_{2},$ then $(q-2)p^{i}>2^{q}.$

Proof.

(i) is direct from Lemma 2.2.

To prove (ii), suppose $q>2^{p^{i}}$ and the order $o(f(x))$ of $f(x)$ equals $m=p^{i}q.$ Since $f(x)$ is of degree $m$ and divides

[TABLE]

the degree of each factor of $f(x)$ is a multiple of $q.$ Hence by Lemma 3.3 there is a factor in $f(x)$ that of degree $p^{i}q,$ which means that $f(x)$ is composed of exactly one factor polynomial and is irreducible.

The sufficiency of (iii) is straightforward from (ii). Then the necessity of (iii) is only to examine the case $q=3,$ i.e., $m=2\cdot 3=6.$ One can see that the reducible polynomial $f(x)=x(x^{2}+x+1)(x^{3}+x+1)$ is of order $o(f(x))=6,$ and thus does not have $\mathcal{P}_{2}.$ The proof of (iii) is completed.

To prove the first part of (iv), suppose to the contrary that $q>p^{i}>2$ and $f(x)$ has $\mathcal{P}_{2},$ but $(p^{i}-2)q\leq 2^{p^{i}}-2^{p^{i-1}}+1.$ Since $p<q$ are two distinct primes, one has $\gcd(p^{i},q)=1,$ and thus

[TABLE]

By Pigeon Hole Principle, there exists positive integer $\hat{u}$ with $1\leq u\leq p^{i}-2$ such that $\hat{u}q\equiv 1~{}\text{or}~{}2~{}(\text{mod}~{}p^{i}).$ If $\hat{u}q\equiv 1~{}(\text{mod}~{}p^{i}),$ say, $\hat{u}q=\ell p^{i}+1,$ then the product of $x$ and $\ell$ irreducible polynomials of degree $p^{i}$ and $p^{i}-\hat{u}$ irreducible polynomials of degree $q$ does not has $\mathcal{P}_{2},$ which is a contradiction. (Note that the fact $N(p^{i})=(2^{p^{i}}-2^{p^{i-1}})/p^{i}$ and $(p^{i}-2)q\leq 2^{p^{i}}-2^{p^{i-1}}+1$ implies $\ell$ exists.) On the other hand, if $\hat{u}q\equiv 2~{}(\text{mod}~{}p^{i}),$ say, $\hat{u}q=\ell p^{i}+2,$ then the product of $x(x+1)$ and $\ell$ irreducible polynomials of degree $p^{i}$ and $p^{i}-\hat{u}$ irreducible polynomials of degree $q$ does not has $\mathcal{P}_{2},$ which makes a contradiction. Similarly, to show the second part of (iv), suppose to the contrary that $q<p^{i}$ and $f(x)$ has $\mathcal{P}_{2},$ but $(q-2)p^{i}\leq 2^{q}.$ An analogue version of the above contradiction will occur by exchanging the positions of $p^{i}$ and $q.$ The result follows. ∎