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Let $q = p^s$ and $r = q^m$, where $p$ is a prime, $s$ and $m$ are positive integers. Let $N>1$ be an integer dividing $r - 1$, and put $n = (r - 1)/N$.

Let $\alpha$ be a primitive element of $\mathbb{F}_r$, $\theta = \alpha^N$, and $Tr_{r/q}$ be the trace function from $\mathbb{F}_r$ to $\mathbb{F}_q$. The set

$$C(r,N) = \{ (Tr_{r/q}(\beta), Tr_{r/q} (\beta \theta),\dots, Tr_{r/q}(\beta \theta^{n-1})) : \beta \in \mathbb{F}_r\}$$

is an irreducible cyclice code of length $n$ over $\mathbb{F}_q$.

Let's define the set

$$Z(r,a) = \#\{x \in \mathbb{F}_r: \text{Tr}_{r/q}(x) = 0 \}.$$

The paper "The weight distribution of some irreducible cyclic codes" by Cunsheng Ding makes the following statement:

Hence, for any $\beta \in \mathbb{F}_r^*$ the Hamming weight of the codeword

$$c(\beta) = (Tr_{r/q}(\beta), Tr_{r/q} (\beta \theta),\dots, Tr_{r/q}(\beta \theta^{n-1}))$$

of the code $C(r,N)$ is equal to

$$n - \frac{Z(r, \beta) - 1}{N}.$$

Can anyone provide any proof for this statement?

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  • $\begingroup$ That's surely $Z(r,a)$ isn't it? $\endgroup$
    – kodlu
    Mar 14, 2023 at 16:49
  • $\begingroup$ I meant $Z(r,a)=\{x:Tr(x)=a\}$ isn't it? $\endgroup$
    – kodlu
    Mar 14, 2023 at 17:04
  • $\begingroup$ did you see my answer? $\endgroup$
    – kodlu
    Mar 20, 2023 at 14:07

1 Answer 1

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The codeword $c(\beta)$ in your question is defined on a set via the trace map that does not include the zero element. Thus its number of nonzero elements is simply the length minus the number of $k\in \{0,1,\ldots,n-1\}$ which give $Tr(\beta \theta^k)=0.$

From the equidistribution properties of the trace function and the fact that we are considering a subfield of index $N$ the number of zeroes of the trace function minus 1, in the extension field, needs to be divided by $N$ when projected into the subfield.

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