Primitivity, freeness, norm and trace
Consider the finite field \(GF(q^n)\) as a vector space over its subfield \(GF(q)\). A basis for this vector space of the form \(\{w, w^q, \cdots, w^{q^{n-1}}\}\) is called a normal basis, and such an element \(w \in GF(q^n)\) is called free over \(GF(q)\). If, in addition, \(w\) is a primitive element of \(GF(q^n)\), the resulting basis is called a primitive normal basis. In [\textit{H. W. Lenstra jun.} and \textit{R. J. Schoof}, Math Comput. 48, 217-231 (1987; Zbl 0615.12023)] it is proven that primitive normal bases always exist. In the paper under review the authors prove that for any primitive element \(b\) of \(GF(q)\), there is a primitive element of \(GF(q^n)\) which is free over \(GF(q)\) and whose \(GF(q^n)/GF(q)\)-norm is equal to \(b\). Furthermore, if \((q,n)\not=(3,2)\), it is shown that for any nonzero element \(b\) in \(GF(q)\), there is an element of \(GF(q^n)\) which is free and whose \(GF(q^n)/GF(q)\)-norm is \(b\). Finally, if \((q,n) \not \in \{(89,8), (41,8), (25,8), (17,8), (13,8), (7,8), (64,7), (4,7)\}\), it is shown that for any primitive element \(b\) of \(GF(q)\) and any nonzero element \(a\) of \(GF(q)\), there is a primitive element of \(GF(q^n)\) which is free over \(GF(q)\), whose \(GF(q^n)/GF(q)\)-norm is equal to \(b\), and whose \(GF(q^n)/GF(q)\)-trace is equal to \(a\). The philosophy in tackling these problems is to exploit non-counting theoretical arguments (such as links between various problems) rather than direct counting arguments using various character sums.
scientific article; zbMATH DE number 1423738
scientific article