LCM decomposition of linear differential operators in positive characteristic
Raphaël Pagès
TL;DR
This work tackles the problem of computing LCLM-decompositions of linear differential operators in positive characteristic. It leverages the $p$-curvature and its Frobenius normal form to determine the feasible LCLM-shape, then constructs a small representative operator $L^*$ in the same equivalence class with a known LCLM, and finally propagates the decomposition to the original operator via an isomorphism of differential modules. The proposed algorithms run in time polynomial in the order $r$, the coefficient degree $d$, and the characteristic parameter $p$, with factor degrees bounded polynomially in $r$ and $d$ and practical evidence for near-quasi-linear dependence on $p$. This provides a practical framework for decomposing linear differential operators in characteristic $p$, with potential applications to arithmetic geometry and the study of solution spaces.
Abstract
We present an algorithm to compute $\mathrm{LCLM}$-decompositions for linear differentials operators with coefficients in the rational function field of characteristic $p$, $\mathbb{F}_{p^n}(t)$. We show that for an operator $L$ of order $r$ with coefficients of degree $d$, it finishes in polynomial time in $r$, $d$ and $p$. This algorithm proceeds in three steps. We begin by showing that the ''shape'' of the factorisation of $L$ can be easily obtained from the Frobenius normal form of its $p$-curvature, which can be efficiently computed an algorithm from Bostan, Caruso and Schost. Using results from the thesis of the author, we are then able to construct an operator $L^*$ in the same equivalence class as $L$ for which an $\mathrm{LCLM}$-decomposition is known. Finally, by computing an isomorphism between the quotient modules $\mathbb{F}_q(t)\langle\partial\rangle/\mathbb{F}_q(t)\langle\partial\rangle L^*$ and $\mathbb{F}_q(t)\langle\partial\rangle/\mathbb{F}_q(t)\langle\partial\rangle L$, we find a corresponding $\mathrm{LCLM}$-decomposition of $L$.