On primitive polynomials over finite fields (Q908964)

Let \(F= \mathrm{GF}(q)\) and \(E=\mathrm{GF}(q^k)\) be the finite fields with \(q\) and \(q^k\) elements, respectively. It is shown that given \(a\in F^*\) there exists a primitive element \(\omega\in E\) satisfying \(T_{E| F}(\omega)=a\) for all but finitely many pairs \((q,k)\). Here \(T_{E| F}(\omega)\) is the trace function of \(E\) with respect to \(F\). Equivalently, the coefficient of \(x^{k-1}\) in a primitive polynomial of degree \(k\) over \(F\) may be arbitrarily selected from \(F^*\). For the case \(k=2\) it is shown the only exceptions to the theorem are for \(q\le 3,847,271\) and there are 147 of them. Some comments on the distribution of the traces of primitive elements are given and in particular the following asymptotic result is shown: if \(q\) sufficiently large then for any \(a\in F^*\) there are at least \(q^{(k-1)(1-\delta)}\) primitive elements of \(E\) with trace a where \(\delta >0\) is arbitrary. For elements of trace zero the following is shown: if \(k\ge 3\) for all but finitely many pairs there exists a primitive element of \(E\) with trace zero over \(F\).