# Weight of a codeword in a cyclic code as a function of the number of solutions of an equation involving the trace function

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?

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

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.