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?



Special polynomials over finite fields
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.


