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arXiv:2611.13044 · v1

Cyclically covering subspaces and additive perfect codes

首次提交 2026-11-09 20:35
版本更新 2026-11-09 20:35
雷达观测 2026-11-11 11:05
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摘要

Let $h_q(n)$ be the largest codimension of a subspace of $\mathbb{F}_q^n$ whose cyclic shifts cover the whole space. For $r>0$, we characterize the inequality $h_q(n)\ge r$ in terms of cyclic $\mathbb{F}_q$-linear codes over vector alphabets with covering radius one, no nonzero word of block weight one, and a scalar dual generated by the shifts of one word. For $k,m\ge2$, let $I_q(j)$ count the monic irreducible polynomials of degree $j$ over $\mathbb{F}_q$. If $k$ is prime or $m\ge2\sum_{j=1}^{k-1}I_q(j)+2$, we prove that a cyclic $\mathbb{F}_q$ linear perfect code of length $\frac{q^{km}-1}{q^k-1}$ over $\mathbb{F}_q^k$ exists if and only if $\gcd(m,q-1)=1$. When this condition holds, we also obtain \(h_q\!\left(\frac{q^{km}-1}{q^k-1}\right)=k(m-1).\) Restriction of scalars gives the same criterion for every $m\ge2$ when $q$ is odd, $q+1$ is a power of two, and $k=2p$ with $p$ prime. We also show that $h_q(q^k+1)=k$ exactly when $q$ is even, for every $k\ge1$.

作者备注:20 pages

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v1 · 2026-11-09 20:35 · 最新观测版本

Cyclically covering subspaces and additive perfect codes

Let $h_q(n)$ be the largest codimension of a subspace of $\mathbb{F}_q^n$ whose cyclic shifts cover the whole space. For $r>0$, we characterize the inequality $h_q(n)\ge r$ in terms of cyclic $\mathbb{F}_q$-linear codes over vector alphabets with covering radius one, no nonzero word of block weight one, and a scalar dual generated by the shifts of one word. For $k,m\ge2$, let $I_q(j)$ count the monic irreducible polynomials of degree $j$ over $\mathbb{F}_q$. If $k$ is prime or $m\ge2\sum_{j=1}^{k-1}I_q(j)+2$, we prove that a cyclic $\mathbb{F}_q$ linear perfect code of length $\frac{q^{km}-1}{q^k-1}$ over $\mathbb{F}_q^k$ exists if and only if $\gcd(m,q-1)=1$. When this condition holds, we also obtain \(h_q\!\left(\frac{q^{km}-1}{q^k-1}\right)=k(m-1).\) Restriction of scalars gives the same criterion for every $m\ge2$ when $q$ is odd, $q+1$ is a power of two, and $k=2p$ with $p$ prime. We also show that $h_q(q^k+1)=k$ exactly when $q$ is even, for every $k\ge1$.

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