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I have the following two definitions:

  1. If $K$ is an extension field of $F$ and $K = F(a)$ for some $a \in K$, then $a$ is a primitive element of $K$.
  2. 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?

Strata771
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    The difference between these two definitions is a recurring source of headache on this site. the latter is standard in the context of finite fields, the former in the context of general field extensions. I included an explanation into the tag wiki. See this CW post of mine for more discussion. +1 to y'all. – Jyrki Lahtonen Jun 16 '20 at 10:30

1 Answers1

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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 $2$.

GreginGre
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