| Popis: |
It is well known that if $\mathbb{F}$ is a finite field then $\mathbb{F^{*}}$, the set of non zero elements of $\mathbb{F}$, is a cyclic group. In this paper we will assume $\mathbb{F}=\mathbb{F}_{p}$ (the finite field with p elements, p a prime) and $\mathbb{\mathbb{F}}_{p^{2}}$ is a quadratic extension of $\mathbb{F}_{p}$. In this case, the groups $\mathbb{F}_{p}^{*}$ and $\mathbb{F}_{p^{2}}^{*}$ have orders $p-1$ and $p^{2}-1$ respectively. We will provide necessary and sufficient conditions for an element $u\in\mathbb{F}_{p^{2}}^{*}$ to be a generator. Specifically, we will prove $u$ is a generator of $\mathbb{F}_{p^{2}}^{*}$ if and only if $N(u)$ generates $\mathbb{F}_{p}^{*}$ and $\frac{u^{2}}{N(u)}$ generates Ker$\,N$, where $N:\mathbb{F}_{p^{2}}^{*}\rightarrow\mathbb{F}_{p}^{*}$ denotes the norm map. We will also provide a method for determining if $u$ is not a generator of Ker$\,N$. |
