Is there a cyclic code with the following parameters?

Question

Is there exist a cyclic code over $F_{2^6}$ with parameters $[31,27,\geq5]$? I tried to construct such code but I didn't succeeded?

Answer 1

No. That's because there do not exist ANY cyclic codes over $K=\Bbb{F}_{64}$ with $(n,k)=(31,27)$.

Over the field $K$ the polynomial $p(x)=x^{31}-1$ factors as $$ p(x)=(x-1)\prod_{j=1}^6 q_j(x),\qquad(*) $$ where the factors $q_j(x), j=1,2,\ldots,6$, all have degree five. They are the six irreducible quintic factors of $x^{31}-1$ over the prime field $\Bbb{F}_2$. As $\gcd(5,6)=1$ an irreducible quintic remains irreducible over the degree six extension $K$.

A cyclic code of length $31$ has a generator polynomial $g(x)\in K[x]$ that is a factor of $p(x)$. If a rank $27$ cyclic code of length $31$ existed, we would need $$\deg g(x)=31-17=4.$$

But the factorization $(*)$ implies that $$ \deg g(x)\equiv \text{$0$ or $1$}\pmod 5, $$ so this is impossible.

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However, it is easy to construct a $(31,27,5)$-code with alphabet $K$. That will be an MDS-code, so we can simply shorten a Reed-Solomon code. If $\alpha$ is a generator of the multiplicative group of $K$, we can, for example, use $g(x)=(x-\alpha)(x-\alpha^2)(x-\alpha^3)(x-\alpha^4)\in K[x]$. It would generate a cyclic code of length $63=|K|-1$. If we restrict the message polynomials $m(x)$ to degrees $\le 26$, we get the space $$ \mathcal{C}=\{g(x)m(x)\mid m(x)\in K_{\le 26}[x]\}, $$ and that will work. Alternatively (and more or less equivalently), we can select some subset $S\subset K$ of $31$ elements. If $m(x)\in K[x]$, let's denote by $m(S)$ the vector $(m(s))_{s\in S}\in K^{31}$ gotten by evaluating the polynomial $m$ at all the points of $S$. Then the space $$ \mathcal{C}=\{m(S)\mid m(x)\in K[x], \deg m(x)\le 26\} $$ is a $27$-dimensional subspace of $K^{31}$. Because a degree $26$ polynomial can have at most $26$ zeros in $K$ (hence in $S$), the minimum Hamming weight of $\Bbb{C}$ is five.