X. Kai and S. Zhu in the paper "On cyclic self-dual codes", AAECC, vol. 19, pp. 509-525, 2008, at page 510 in line 6 said that, It is well Known that there are no cyclic self-dual codes over $F\_q$ when $q$ is odd.


I know that W. Cary Huffman and V. Pless in the book "Fundamentals of error correcting codes" proved that there are no self-dual cyclic codes of length $n$ over $F\_q$ when $gcd(n,q)=1$.


Also Y. Jia, S. Ling and C. Xing in the paper with name "On self-dual cyclic codes over finite fields" at 2011 proved that there exist at least one self-dual cyclic code of length $n$ over $F\_q$ if and only if $q$ is a power of $2$ and $n$ is even.


My questions are:


1) Where can I find the proof of the claim that introduced in my first paragraph?


2) Generally, how can I find some good sources about self-dual (self-orthogonal) cyclic codes, precisely about the existence of these codes over finite fields?



request sources about self-dual cyclic codes
I think (1) is straightforward: suppose $C$ is a cyclic code of length $n$ with generator polynomial $f(x) \in F\_q[x]$. Let $C'$ be the code with generator polynomial $g(x) = (x^n-1)/f(x)$. As I understand the definition of duality for cyclic codes (following page 84 of van Lint, Introduction to Coding Theory, 3rd edition), the dual of $C$ is $C'$. Hence, if $C$ is self-dual then $f(x) = g(x)$ and so $x^n-1 = f(x)^2$. This is impossible when $q$ is odd. 


I can't help with (2), except to say that a quick Google search found several papers that look relevant, e.g. [Self-dual cyclic codes](http://www2.research.att.com/~njas/doc/cyclic.pdf) by N.J.A Sloane and J.G. Thompson and [On the Minimal Distance of Binary Self-Dual Cyclic Codes](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.157.9295&rep=rep1&type=pdf) by Bas Heijne and Jaap Top.


**Edit:** as Zahra's comment says, with the other definition of duality for cyclic codes, the generator polynomial for $C^\bot$ is a multiple of $h(x) = g(x^{-1})x^{\deg g}$. The roots of $h(x)$ are the reciprocals of the roots of $g(x)$, counted with multiplicities. Suppose that $1$ is a root of $x^n-1$ with multiplicity $m$. If $1$ has multiplicity $r$ as a root of $f(x)$, then $1$ has multiplicity $m-r$ as a root of $g(x)$ and $h(x)$. Hence, if $C$ is self-dual then $f(x) = h(x)$, and $m-r = r$. Therefore $m$ is even, and again this is impossible when $q$ is odd.


