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My field of research is coding theory and I am working on cyclic codes. During my research, I tackled an algebraic problem. After some simple definitions, I asked my question. I will appreciate any helpful answer and comment.

Let ‎$‎‎F_{2^{2m}}‎$ ‎denote ‎the ‎finite ‎field ‎of ‎‎$‎{2^{2m}}‎$‎‏ ‎elements‎, where ‎$‎m‎$‎‏ ‎is a‎ ‎positive ‎integer. ‎Let‎ ‎$‎‎‎‎F_{2^{2m}}[x]‎$ ‎denote ‎the polynomial ring in indeterminate ‎$‎x‎$ ‎with ‎coefficients ‎from‎ ‎$‎‎F_{2^{2m}}‎$.‎ ‎

Suppose that ‎$‎f(x)‎$ ‎‎is a polynomial in $‎‎F_{2^{2m}}[x]‎$ ‎and‎ ‎‎$‎f(x)=f_0+f_1x+‎\cdots‎+f_kx^k‎$. We define the conjugate polynomial of ‎$‎f(x)‎$ ‎over‎ ‎$‎‎‎‎F_{2^{2m}}‎$ as follows:‎

‎$‎‎\overline{f(x)}‎={f_0}^{2^m}+f_1^{2^m}x+‎\cdots ‎+f_k^{2^m}x^k.$‎

In particular, if a polynomial is equal to its conjugate polynomial ‎over‎ ‎$‎‎F_{2^{2m}}‎$, then it is called self-conjugate polynomial.‎ ‎

Let ‎$‎n‎$ ‎be ‎an ‎odd ‎positive ‎integer. Since ‎‎$‎gcd(n,‎2^{2m})=1‎$‎, the polynomial ‏‎$‎‎‎x^n+1‎$ ‎can ‎be ‎factorized ‎into ‎distinct ‎irre‎ducible polynomials over ‎$‎‎F_{2^{2m}}‎$.‎ ‎‎

It ‎is ‎obvious ‎that ‎for ‎any ‎monic ‎irreducible ‎polynomial ‎dividing ‎‎$‎x^n+1‎$ ‎over‎‎ ‎$‎‎‎‎F_{2^{2m}}‎$, its conjugate polynomial ‎is ‎also a ‎monic ‎irreducible ‎polynomial ‎dividing ‎‎$‎x^n+1‎$ ‎over‎‎ ‎$‎‎‎‎F_{2^{2m}}‎$. ‎‎ ‎

For ‎example, let ‎‎$‎‎\omega‎‎$ be a‎ ‎primitive ‎element ‎of‎ $‎‎F_4‎$‎. ‎‎‎‎‎The factorization of ‎$‎‎x^5+1$ over $‎‎F_4‎$ is ‎

‎$‎‎x^5+1=(x+1)(x^2+‎\omega ‎x+1)(x^2+‎\omega‎^2x+1)‎‎$‎‎‎‎‎‎

It ‎is ‎obvious ‎that ‎$‎x+1‎$‎‏ ‎is a‎ ‎‎self-conjugate polynomial ‎and ‎‎$x^2+‎\omega ‎x+1‎$ ‎is ‎the‎ conjugate polynomial of ‎$x^2+‎\omega‎^2x+1‎$ ‎over $‎‎‎‎F_4‎$‎‎.‎ ‎‎

For another ‎example, let $‎‎\omega‎‎$ be a‎ ‎primitive ‎element ‎of‎ $‎‎F_{16}‎$‎. ‎The factorization of ‎$‎‎x^{11}+1$ over $‎‎‎‎F_{16}‎‎$ is‎

‎$‎‎x^{11}+1=(x+1)(x^5+‎\omega^5 ‎x^4+x^3+x^2+‎\omega‎^{10}x+1)(x^5+‎\omega^{10} ‎x^4+x^3+x^2+‎\omega‎^{5}x+1)‎‎$

It ‎is ‎obvious ‎that ‎$‎x+1‎$, ‎‎$x^5+‎\omega^5 ‎x^4+x^3+x^2+‎\omega‎^{10}x+1‎$ and ‎$x^5+‎\omega^{10} ‎x^4+x^3+x^2+‎\omega‎^{5}x+1‎$‎ ‎are‎ ‎‎self-conjugate polynomials over $‎‎‎‎F_{16}‎‎$.‎

Because of my researches I think that if $‎f(x)‎$ ‎‎is a self-conjugate monic ‎irreducible ‎polynomial ‎dividing ‎‎$‎x^n+1‎$ ‎over‎‎ ‎$‎‎F_{2^{2m}}‎‎$, then the degree of $‎f(x)‎$ is odd, ‎but I ‎‎could ‎not ‎prove ‎it.‎‎‎ ‎

Is this conjecture true in general? If the answer is no, please give me an example?

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1 Answer 1

up vote 4 down vote accepted

First, a polynomial $f$ is self-conjugate iff its coefficients belong to the fixed field of the Frobenius $x\mapsto x^{2^m}$, which is $F_{2^m}$.

Second, an $f\in F_{2^m}[x]$ which is irreducible over $F_{2^{2m}}$ must have an odd degree $d$: the splitting field of $f$ is the unique degree $d$ extension of $F_{2^m}$, which is $F_{2^{md}}$; if $d$ were even, then this field is a degree $d/2$ extension of $F_{2^{2m}}$, hence $f$ can’t be irreducible over the latter.

So, yes, any irreducible self-conjugate polynomial has an odd degree, $x^n+1$ has nothing to do with it.

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Thanks for your helpful answer. –  Zahra Jul 14 '12 at 7:48

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