Specific irreducible polynomials with linearly independent roots over finite fields (Q677139)

If \(F=\text{GF}(q)\) is a finite field and \(n\geq 1\) an integer, then it is one of the major problems in the theory of finite fields to (explicitly) construct an irreducible polynomial of degree \(n\) over \(F\) (possibly satisfying additional properties such as normality, primitivity or prescribed coefficients). The authors determine three families of so-called completely normal polynomials. (A monic polynomial \(f \in\) GF\((q)\) of degree \(n\) is completely normal if \(f\) is irreducible and if every root \(\alpha\) of \(f\) simultaneously generates a normal basis for GF\((q^n)\) over every intermediate field of GF\((q^n)\) over GF\((q)\).) Completely normal (or completely free) elements have attained much interest. In this respect, we would like to mention the following references (the second one is a monograph on simultaneous module structures over finite fields). [1] \textit{R. Chapman} [Finite Fields Appl. 3, 1-10 (1997);]. [2] \textit{D. Hachenberger}, Finite fields: Normal bases and completely free elements, Kluwer Academic Publishers, Boston (1997; Zbl 0864.11065). The degrees of the polynomials determined in the paper under review depend strongly on properties of the given ground field. The first construction works for all \((n,q)\) in which there exists an irreducible binomial of degree \(n\) over GF\((q)\) (in fact binomials play a crucial role for the construction). The second construction works for all pairs \((2^{k+1},q)\), \(k \geq 0\), \(q-3\) divisible by \(4\), while the third family has parameters \((p^{k+1},p)\), \(k \geq 0\), \(p\) prime. Compared to the first and the third family, the proof concerning the second one is very involved and demands more than half of the paper. In fact it is the normality over the ground field rather than the completeness which causes the trouble. The (iterative) construction in [1] of a family of completely normal polynomials with parameters \((2^{k+1},q)\), \(k \geq 0\), \(q-3\) divisible by \(4\) allows a simpler proof. In fact, due to the content of Section 27 of [2], it is very likely that both approaches (of the paper under review and that of [1]) can be handled within a unified framework.