an:06695038 Zbl 1372.11111 Tuxanidy, Aleksandr; Wang, Qiang Compositional inverses and complete mappings over finite fields EN Discrete Appl. Math. 217, Part 2, 318-329 (2017). 00364177 2017
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11T06 05B15 permutation polynomials; complete mappings; compositional inverse; linearized polynomials; finite fields; binomial; trace Let $$\mathbb {F}_{q}$$ be a finite field with $$q=p^m$$ elements and $$f\in \mathbb {F}_{q}[x]$$ a univariate polynomial. If $$f$$ induces a permutation of $$\mathbb {F}_{q}$$ under evaluation then there exists a unique $$f^{-1}\in \mathbb {F}_{q}[x]$$ of degree less than $$q$$ satisfying $$f(f^{-1}(x)) \equiv f^{-1}(f(x)) \equiv x \pmod{x^q-x}$$. The paper under review studies this compositional inverse in case $$f$$ is a linearized binomial permuting the kernel of the trace map $$T_{q^n|q^s}: \mathbb {F}_{q^n} \longrightarrow \mathbb {F}_{q^s}$$, where $$s$$ is a positive divisor of $$n$$. In Theorem 2.4 it is shown that for any positive integer $$r$$ such that $$d:=\gcd (n,r) =\gcd (r,s)$$ and any $$c\in \mathbb {F}_{q^s}$$ whose norm $$N^{q^s|q^d}(c)$$ equals 1, the binomial $$L_{c,r}(x) :=x^{q^r}-cx\in \mathbb {F}_{q^n} [x]$$ induces a permutation of $$\ker (T_{q^n|q^s})$$ if and only if $$n/s$$ is not divisible by the characteristic $$p$$. Theorem 2.4 also contains an explicit formula for $$L_{c,r}^{-1}$$ when it exists. This result is used in order to construct a class of complete mappings (i.e., permutation polynomials $$f\in \mathbb{F}_q [x]$$ such that $$f(x)+x$$ also permutes $$\mathbb {F}_{q}$$) for which the computational inverse is again explicitly obtained. A recursive construction of a set of complete mappings with the property that the difference of any two distinct elements permutes $$\mathbb {F}_{q}$$ is also given. Mihai Cipu (Bucure??ti)