On cyclically covering subspaces of $\mathbb{F}^n_q$
Yangcheng Li, Pingzhi Yuan, Shuang Li, Yuanpeng Zeng
TL;DR
The paper studies cyclically covering subspaces of $\mathbb{F}_q^n$ by characterizing when $h_q(n)=0$ using an isomorphism-based framework that links $\mathbb{F}_q^n$, $\mathbb{F}_{q^n}$, and $\mathbb{F}_q[x]/(x^n-1)$. It derives equivalent conditions via invariant subspaces, trace-orthogonal constructions, and direct-sum decompositions, and applies these to obtain zero-codimension results such as $h_q(\ell^t)=0$ when $q$ is a primitive root modulo $\ell^t$, as well as a doubling property $h_q(2n)=0$ for odd $n$ when $h_q(n)=0$. Additionally, the work connects coverings in $\mathbb{F}_{q^m}^n$ to those in $\mathbb{F}_q^{mn}$, yielding transfer criteria and bounds, and provides concrete examples (e.g., $h_3(11)=h_3(16)=1$) along with sufficient conditions for $h_{q^m}(n)=0$ in various arithmetic settings, including $h_4(n)=0$ for $n=3$ or $n=2^d$. Overall, the results broaden the toolkit for understanding cyclic coverings and give practical criteria for zero codimension across different finite-field regimes.
Abstract
For a prime power \( q \) and a positive integer \( n \), a subspace \( U \subseteq \mathbb{F}_q^n \) is called cyclically covering if the union of all its cyclic shifts covers the whole space \( \mathbb{F}_q^n \). Let \( h_q(n) \) denote the maximum possible codimension of such a subspace. This paper focuses on the case \( h_q(n) = 0 \). We provide necessary and sufficient conditions under which \( h_q(n) = 0 \) holds. As an application, we show that \( h_q(\ell^t) = 0 \) whenever \( q \) is a primitive root modulo \( \ell^t \). Moreover, we prove that if \( n \) is odd and \( h_q(n) = 0 \), then also \( h_q(2n) = 0 \). As an example, we show that \( h_3(11) =h_3(16) = 1 \). Furthermore, we investigate the relationship between the coverings of \(\mathbb{F}_{q^m}^n\) and \(\mathbb{F}_q^{mn}\), and obtain several sufficient conditions for \(h_{q^m}(n) = 0\). Specifically, we derive that if \(n = 3\) or \(n = 2^d\) (where \(d\) is a nonnegative integer), then \(h_4(n) = 0\).
