I have the following two definitions:
- If $K$ is an extension field of $F$ and $K = F(a)$ for some $a \in K$, then $a$ is a primitive element of $K$.
- If $K$ is a finite field and $a$ is a generator for its multiplicative group $K^*$, then $a$ is a primitive element of $K$.
It's clear that if $a$ is a primitive element (def2) of the finite field $K$ (characteristic $p$), then we can write $K = Z_p(a)$ (because $K^* = \langle a \rangle $). So in this case $a$ is also primitive (in the sense of def1).
However, if I'm not wrong, it is possible for $K = Z_p(a)$ to be a finite field, without $a$ being a generator for $K^*$. Can someone provide a counterexample?
Sure. Let $p\equiv 3 \ [4]$ be a prime number. Then $X^2+1\in\mathbb{F}_p[X]$ is irreducible. Take an element $\alpha$ in an algebraic closure of $\mathbb{F}_p$ satisfying $\alpha^2=-1$.
Set $K=\mathbb{F}_p(\alpha)$. Then $K$ is a finite field with $p^2$ elements, and $\alpha\in K$ is a primitive element in the sense of $1$, by definition.
Now $\alpha^2=-1$ and $\alpha^4=1\in K$, so $\alpha$ has order $4$ in $K^*$, while $K^*$ has order $p^2-1\geq 8$. Hence $\alpha$ is not a primitive element in the sense of $2$.
For $p=3$, one may check that $\alpha+1$ is a primitive element in sense of $2$.