Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For what $(n,k)$ there exists a polynomial $p(x) \in F_2[x]$ s.t. $\deg(p)=k$ and $p$ divides $x^n-1$?

Motivation: if exists $p(x)$, then ideal generated by $p(x)$ is "cyclic error correcting code". It seems to me not for all $n,k$ it exists, but MatLab generates some polynoms for cyclic codes for all $n,k$ I tried. So something is wrong with my understanding.

The question probably elementary, sorry.

share|improve this question

2 Answers 2

One version of this problem has been studied very, very recently by Lola Thompson (arXiv) in her PhD thesis.

A value of $n$ for which $x^n-1$ admits divisors (over $\mathbb Q$) of all degrees $k\le n$ is called $\phi$-practical. By earlier work of Thompson, these are about as common as the primes: the number of $\phi$-practical numbers up to $x$ has order of magnitude $x/\log x$.

So for most $n$, there is some $k \le n$ such that $x^n-1$ does not have a divisor of degree $k$. Of course, working over $\mathbb F_2[x]$, there are more divisors available. Empirically, this does not seem to change things by more than a constant factor (there are tables of counts for $\mathbb F_2$, $\mathbb F_3$, $\mathbb F_5$ in the above preprint). However, a proof of zero density is only known under the assumption of GRH.

Thompson and Pollack have a conditional result (also on GRH) for the converse problem: for a fixed $k \ge 3$, the number of $n \le x$ for which $x^n-1$ has a divisor of degree $k$ over $\mathbb F_2$ is $\ll x/(\log k)^{2/35}$. So for very large $k$ there are somewhat fewer pairs $(n,k)$ with the given property.

share|improve this answer

To answer the question, we need to know something about the factorization of $x^n-1$. Let us begin by writing $n=2^\ell m$ with $m$ odd. Then $$ (x^n-1)=(x^m+1)^{2^\ell}, $$ and the problem is reduced to the case of an odd exponent.

I have given the answer to that in different disguise here, and the answer is that degree of the factors of $x^m+1$ are in 1-1 correspondence with the sizes of $2$-cyclotomic cosets modulo $m$. The ($2$-) cyclotomic coset modulo $m$ of an integer $a$ consists of the residue classes of $2^ja,j=0,1,\ldots$. If $g$ is a primitive root of unity of order $m$, then the cyclotomic coset $[a]$ of $a$ corresponds to the factor $$ f_a(x)=\prod_{j\in [a]}(x-g^j), $$ that clearly is a factor of $x^m-1$ over $\mathbb{F}_2[g]$, but actually has coefficients in $\mathbb{F}_2$, because its zeros are all conjugates under powers of the Frobenius automorphism.

As an example let us consider $n=31$. Here there is a single cyclotomic coset of size $1$ namely $[0]$ corresponding to the obvious factor $x+1$. The other cyclotomic cosets have all 5 elements, so $x^{31}+1$ is a product of a linear factor and six irreducible quintic polynomials. Therefores all its factors have degrees congruent to either zero or one modulo 5, and all those values ($\le 31$) occur.

I don't know how Matlab handles it. Engineers are in the habit of shortening cyclic codes, because you can use the generator polynomial even if you shorten the code at the end. Obviously you will destroy cyclicity, if you do that. Anyway, you are right. Cyclic codes do not exist for all combinations of $n$ and $k$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.