Consider the field $E:= \frac{\mathbb{Z}_3[X]}{\langle x^2 + x + 2\rangle}$.
If I'm right the elements of the quotient ring can be found as:
$$a_0 + a_1x + \langle x^2 + x + 2\rangle.$$
So we got the possibilities in $\mathbb{Z}_3$:
$$\{0,1,2,\beta, 1+\beta , 2+\beta, 2\beta, 1+2\beta ,2+2\beta \}.$$
Here $\beta = \overline{x} = x + \langle x^2 + x + 2\rangle$ is a root of $x^2 + x+2$.
(Correct me if my notation is wrong.)
So how do we get the elements of unit of $E^{\times},\cdot$. I assume $1$ is in it, but don't know how to calculate the other elements. With the elements, what would be the Cayley table of $E^{\times},\cdot$?
Other little question: we know that $\beta$ is a solution of $x^2 + x+2$, what is the other root?
Best Answer
After I figured out how to proper multiplicate in a quotient ring via: Constructing a multiplication table for a finite field, I managed to find the unit elements by calculating every possible combination. I found for instance: \begin{split} \beta(1+\beta)& = x^2 + x + \langle x^2 + x + 2\rangle \\ & =x^2 + x + \langle x^2 + x + 2\rangle + (0 + \langle x^2 + x + 2\rangle)\\ &= x^2 + x + \langle x^2 + x + 2\rangle + 2x^2+2x+4+ \langle x^2 + x + 2\rangle\\ &= 3x^2+ 3x +4 +\langle x^2 + x + 2\rangle\\ &=0+0+1+\langle x^2 + x + 2\rangle\\ &=1 \end{split} If I do this for the other elements, I find that $(2+\beta)(1+2\beta)=1$ and $(2\beta)(2+2\beta)=1$.
So the elements of unit become: $E^{\times},\cdot = \{1,\beta,1+\beta,2+\beta,1+2\beta,2\beta,2+2\beta\}$. The Cayley table is found by multiplying all the elements with each other. They are calculated similar as above.