Every strict sum of cubes in $\mathbb F_4[t]$ is a strict sum of 6 cubes

Luis H. Gallardo

doi:10.4171/pm/1807

JournalspmVol. 65, No. 2pp. 227–236

Every strict sum of cubes in $F_{4} [t]$ is a strict sum of 6 cubes

Luis H. Gallardo
Université de Brest, France

$Every strict sum of cubes in $\mathbb F_4[t]$ is a strict sum of 6 cubes cover$

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Abstract

It is easy to see that an element $P (t) \in F_{4} [t]$ is a strict sum of cubes if and only if $P (t) \in M (4)$ where

M (4) = {P (t) \in F_{4} [t] ∣ P (r) \in {0, 1} for all r \in F_{4} and such that either 3 does not divide deg (P (t)), or 3 does divide deg (P (t)) and P (t) is monic} .

We say that $P (t)$ is a “strict” sum of cubes $A_{1} (t)^{3} + \cdot\cdot\cdot + A_{g} (t)^{3}$ if $deg (A_{i}^{3}) < deg (P) + 3$ for each $i$ , and we define $g (3, F_{4} [t])$ as the least $g$ such that every element of $M (4)$ is a strict sum of $g$ cubes. The main result is that

g (3, F_{4} [t]) \leq 6.

This improves an earlier result of the author that $g (3, F_{4} [t]) \leq 9$ .

Cite this article

Luis H. Gallardo, Every strict sum of cubes in $F_{4} [t]$ is a strict sum of 6 cubes. Port. Math. 65 (2008), no. 2, pp. 227–236

DOI 10.4171/PM/1807