Show that $[x]$ generates the set of units for $R = \mathbb F_3[x]/\langle x^2+x+2\rangle$

Question

Given $R = \mathbb F_3[x]/\langle x^2+x+2\rangle$, which I have previously shown is a field with $9$ elements, I now have to show that $[x]$ generates $R^*$, the set of units of $R.$ In other words, I have to show that $\langle [x]\rangle = R^*.$

My logic shows that I have misunderstood something fundamental:

Since $R$ is a field, we must have $R^* = R \setminus \{[0]\}.$

Since $\langle [x]\rangle = \{ k[x] \ | \ k \in R \}$ we must have $[0][x] = [0x] = [0] \in \langle [x]\rangle.$

But this is in contradiction with $\langle [x]\rangle = R^* = R \setminus \{[0]\}. $

What am I misunderstanding and how can I show that $\langle [x]\rangle = R^*?$

Answer 1

What you missed is that $R^*$ is a multiplicative group, and the question is about the multiplicative subgroup generated by the coset $[x]$.

Therefore $$\langle [x]\rangle=\{[x^t]\mid t\in\Bbb{Z}\}.$$ When proving that $\langle[x]\rangle=R^*$ it is beneficial to observe that $R^*$ has exactly eight elements (do you why?). Therefore it suffices to check that $[x^4]\neq[1]$. Do you see why?

Answer 2

$\langle[x]\rangle$ is not to be understood as the "ideal" in the field (which would coincide with the whole field, of course). It should be just understood as a subgroup of $R^*$ generated by $[x]$, i.e. $\{[x]^n\mid n\in\mathbb Z\}=\{[x^n]\mid n\in\mathbb Z\}$

For that, because the order of $R^*$ is $8$, you just need to prove that the order of $[x]$ is $8$. Do it by proving that it is not $1$, $2$ or $4$, i.e. that $[x]\ne[1]$, $[x^2]=[2x+1]\ne[1]$ and $[x^4]=[(2x+1)^2]=[x^2+x+1]=[2]\ne [1]$.