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Let $q$ be a power of prime number $p$ and let $F_{q^2}$ be a finite field of order $q^2$. Suppose that "-" be a conjugation operation that is defined as follow:

$-:F_{q^2} ‎\longrightarrow‎ ‎F_{q^2}$

$x ‎\longmapsto‎ x^q$

Let $C$ be a cyclic code of length n over $F_{q^2}$ with the generator polynomial $g(x)$ and let $\bar{C}=\lbrace \bar{c} : c \in C \rbrace$ be the conjugate code of $C$.

It is obvious that $\bar{C}$ is also a cyclic code.

Is it possible to determine the generator polynomial of $\bar{C}$ from $g(x)$?

edited Dec 16 2011 at 18:07

asked Dec 16 2011 at 17:44

Zahra
179●1●1●7

2

The obvious choice works. – Felipe Voloch Dec 16 2011 at 18:33

Thank you for your helpful answer. – Zahra Dec 17 2011 at 19:20

Shahrooz · Accepted Answer · 2011-12-16 23:25:55Z UTC

2

The generator polynomial of $\bar{C}$ is $‎\overline{g(x)}‎$. Because the conjugation operation is distributive over summation and multiplication.

answered Dec 16 2011 at 23:25

Shahrooz
962●1●2●13

The generator polynomial of cyclic code

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