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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:

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

$x ‎\longmapsto‎ x^q$

Suppose that we define Hermitian inner product over $F_{q^2}$ as follow:

$(x,y)=x.\bar{y}$

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 $g(x)$?

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# The generator polynomial of cyclic code

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:

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

$x ‎\longmapsto‎ x^q$

Suppose that we define Hermitian inner product over $F_{q^2}$ as follow:

$(x,y)=x.\bar{y}$

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 $g(x)$?