Consider the field $\mathbb{Z}_{5}$ and the polynomial ring $\mathbb{Z}_{5}[x]$ . $g(x) =x^{3}+x+1$ is an irreducible polynomial in $\mathbb{Z}_{5}[x]$ and hence $(g(x))$ is a prime and maximal idea. Thus $$\begin{align*} \ ^{\mathbb{Z}_{5}[x]}\!/\!_{(g(x))}=\left\{ ax^{2}+bx +c : a,b ,c \in \mathbb{Z}_{5} \right\} \end{align*}$$ is a finite field with order $125$
Since it is a finite field, I know that the "abelian group" part of a finite field is cyclic. How should I find the generator of $$\begin{align*} (\ ^{\mathbb{Z}_{5}[x]}\!/\!_{(g(x))})^{*} \end{align*}$$ I got stuck here. Any help? Thanks
I searched that by using the Frobenius map we can solve this problem. However, I don't know this, but there's a more elementary way to solve it
We have $$\begin{align*} \ ^{\mathbb{Z}_{5}[x]}\!/\!_{(g(x))}=\left\{ ax^{2}+bx+c: a, b , c \in \mathbb{Z}_{5} \right\}=S \end{align*}$$ since $\ ^{\mathbb{Z}[x]}\!/\!_{(g(x))}$ means all polynomials in $\mathbb{Z}_{5}[x]$ mod $x^{3}+x+1$ . That is all polynomials with degrees less or equal to $2$.
Then consider $x \in S$ , by Lagrange Theorem, we have $$\begin{align*} \text{ord}(x) \mid 124 \end{align*}$$ Thus $$\begin{align*} \text{ord}(x) = 2 \ or \ 4 \ or \ 31 \ or \ 62 \ or \ 124 \end{align*}$$ It's clear that $$\begin{align*} x^{2} \neq 1 \pmod{x^{3}+x+1} \\ x^{4} \neq 1 \pmod{ x^{3}+x+1} \end{align*}$$Thus $\text{ord}(x)=31 \ or \ 62 \ or 124$ . Then notice that $\text{ord}(2) =4$ in $S$ . Also $\langle 2 \rangle \cap \langle x \rangle =\left\{ 1 \right\}$ is trivial, and $2,x$ commute in $S$ . Thus $$\begin{align*} \text{ord}(2x) &= [\text{ord}(2),\text{ord}(x)] \\ &= [4, \text{ord}( x)] \end{align*}$$ No matter $\text{ord}(x)=31,62, \ or, 124$ , $\text{ord}(2x)=[4,\text{ord}(x)]=124$ . Thus $2x$ is the generator of $S$.