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We have a cyclic code $C$ of length $n$ on a finite field $\mathbb{F}_q$ and $g$ is the generating polynomial of $C$. We define another code $\bar{C}=\{c\in C\mid\sum\limits_{i=0}^{n-1} c_i=0 \}$. We identify a word $c$ with his polynomial $c(x)=c_0+...+c_{n-1}x^{n-1}$.

We want to find a generating polynomial for $\bar{C}$ using $g$.

First I prove that $\bar{C}$ is a cyclic code. Let $(c_0,...,c_{n-1})\in \bar{C}$. Using the fact that $C$ is cyclic we have $(c_{n-1},c_0,...,c_{n-2})\in C$ and the sum is $0$ because we keep the same coefficient (just the order has changed). So $(c_{n-1},c_0,...,c_{n-2})\in \bar{C}$.

By the identification of $c$ to $c(x)$ we notice that if $c\in \bar{C}$ then $c(1)=0$ and $g$ must divide $c$ by definition of $g$. Which means that $(x-1)$ and $g$ divide $c$.

So if we have $\gcd{(x-1,g)}=1$ we take $(x-1)g$ as the generating polynomial of $\bar{C}$. But for the other case I meet difficulties.

Thanks in advance !

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    If $g(x)$ is divisible by $(x-1)$, then $\tilde{C} = C$ because _every_ codeword in $C$ satisfies $\sum_i c_i = 0$ and so the generator polynomial of $\tilde{C}$ is $g(x)$. – 2017-02-17
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    @DilipSarwate Thank you but I still don't understand why it's not $(x-1)$ which divides $g$ ? – 2017-02-17
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    If $\gcd(x-1,g(x) \neq 1$, then it must be that $\gcd(x-1,g(x)=x-1$, that is, $x-1$ is already a divisor of $g(x)$. In this case, _every_ codeword in $C$ has even weight, and so $\tilde{C} = C$ and $\tilde{C}$ has generator polynomial $g(x)$. Not $g(x)/(x-1)$ because the code generated by $g(x)/(x-1)$ is of _larger_ dimension than $C$ whereas $\tilde{C}$ is a subcode of $C$ and so has dimension no larger than the dimension of $C$. – 2017-02-18
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    @DilipSarwate Is $(x-1)$ a codeword of $C$ ? – 2017-02-18

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All linear binary codes $C$ have the property that either all the codewords have even Hamming weight, or half the codewords have even Hamming weight while half have odd Hamming weight. In the latter case, the codewords of even Hamming weight are a linear subcode of $C$. Note that the codewords of even Hamming weight have the property that $\sum_{i=0}^{n-1} c_i = 0$.

If $C$ is a linear binary cyclic code of length $n$ ($n$ odd) with generator polynomial $g(x)$ and parity check polynomial $h(x)$, then $g(x)h(x) = x^n-1$ and since $(x-1) \vert (x^n-1)$, $(x-1)$ is a divisor of either $g(x)$ or $h(x)$ but not both. Note that since all codeword polynomials are divisible by $g(x)$, if all the codewords of $C$ have even Hamming weight, then it must be that $g(x)$, which is a codeword polynomial, has even weight. Hence, $g(1) = \sum_i g_i = 0$ and $x-1$ is a divisor of $g(x)$. If $C$ has codewords of odd Hamming weight, then $g(1)=0$ would imply that $c(1)=0$ for all codewords, and so it must be that $g(1)=1$ and so $x-1$ is a divisor of $h(x)$.

So, if all the codewords of $C$ have even Hamming weight, then $C^\prime = C$ and is a cyclic code with generator polynomial $g(x)$.

If $C$ has codewords of odd Hamming weight, then $C^\prime$ is the subcode of codewords of even Hamming weight. Each $c(x)$ is divisible by $g(x)$ (which does not have $x-1$ as a factor), but $c(1)=0$ meaning that $c(x)$ is also divisible by $x-1$. Hence, each $c(x)$ is divisible by $(x-1)g(x)$, not $g(x)/(x-1)$ as the OP insists is correct. Let's check dimensions. If $C$ is a $[n,k]$ code, then $\deg g(x) = n-k$. $C^\prime$ is a $[n,k-1]$ code and its generator polynomial must have degree $n-(k-1)=n-k+1 = \deg [(x-1)g(x)]$.