Linear Network Coding over Rings, Part II: Vector Codes and Non-Commutative Alphabets

TL;DR

This paper explores the conditions under which networks are linearly solvable over various algebraic structures, establishing equivalences and bounds involving fields, rings, and modules, including non-commutative rings.

Contribution

It characterizes the relationships between scalar and vector linear solvability over different rings and modules, and identifies minimal non-commutative rings necessary for certain network solvability.

Findings

Equivalence of vector linear solvability over fields and scalar linear solvability over rings.

Existence of networks solvable over non-commutative rings but not over commutative rings, with size bounds.

Scalar solvability over rings of size p^k implies k-dimensional vector solvability over GF(p) for 2 ≤ k ≤ 6.

Abstract

We prove the following results regarding the linear solvability of networks over various alphabets. For any network, the following are equivalent: (i) vector linear solvability over some finite field, (ii) scalar linear solvability over some ring, (iii) linear solvability over some module. Analogously, the following are equivalent: (a) scalar linear solvability over some finite field, (b) scalar linear solvability over some commutative ring, (c) linear solvability over some module whose ring is commutative. Whenever any network is linearly solvable over a module, a smallest such module arises in a vector linear solution for that network over a field. If a network is linearly solvable over some non-commutative ring but not over any commutative ring, then such a non-commutative ring must have size at least $16$, and for some networks, this bound is achieved. An infinite family of networks is demonstrated, each of which is scalar linearly solvable over some non-commutative ring but not over any commutative ring. Whenever $p$ is prime and $2 \le k \le 6$, if a network is scalar linearly solvable over some ring of size $p^k$, then it is also $k$-dimensional vector linearly solvable over the field $GF(p)$, but the converse does not necessarily hold. This result is extended to all $k\ge 2$ when the ring is commutative.