Group of units in a finite field Let $f(x) \in \mathbb{Z}/3\mathbb{Z}[x]$ be a cubic irreducible polynomial and let $F = \mathbb{Z}/3\mathbb{Z}[x]/(f(x))$. I want to show that either $x$ or $2x$ generates the group of units of $F$.
My only thoughts so far are that $F$ is a finite field of order $27$, so its group of units has order $26$. Clearly both $x,2x$ do not have order $2$, so they must have order $13$ or $26$. But how can I narrow it down more?
I could try to find all cubic irreducibles, or I could use that $F$ is a splitting field of $x^{27}-x$, or use that the group of units is cyclic, but I'm not sure. Those strategies all seem to computational or don't lead anywhere.
 A: The $f$ is a red herring. The procedure is equivalent to picking $x \in \Bbb F_{27} \setminus \Bbb F_3$.
Let $\zeta$ be a primitive $26$th root of unity, so $\Bbb F_3 = \{0, 1, \zeta^{13}\}$. Let $x = \zeta^e$ with $0 < e < 26$, so $e \ne 13$, and $2x = \zeta^{e+13}$.
It suffices to show that one of $e$ or $e+13$ is coprime to $26 = 2 \times 13$.
Since $0 < e < 26$ and $e \ne 13$, we know that $e$ is not divisible by $13$, so $e+13$ is also not divisible by $13$.
One of $e$ or $e+13$ is odd, i.e. not divisible by $2$ either.
So one of $e$ or $e+13$ is coprime to $26 = 2 \times 13$.
So one of $x = \zeta^e$ or $2x = \zeta^{e+13}$ generates the unit group.
A: You’ve noted that $x$ has order $13$ or $26$. If it has order $26$ you are done. If $x$ has order $13$, then just note that since $2=-1$ in $F$, then $2$ has order $2$. So $\gcd(o(2),o(x))=1$, and since the multiplicative group is abelian, it follows that $o(2x)=\mathrm{lcm}(o(2),o(x)) = 26$.
So if $x$ does not generate, then $2x$ certainly does.
