  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$\,N$.
