Finding indempotents in a quotient ring 
I am trying to find the nontrivial indempotents in the ring $\mathbb{Z_3}[x]/(x^2+x+1)$. 

We can clearly see that $0,1$ are indempotents. I want to prove they are the only ones. Thus I am wondering if there is a way other than just brute force to show this. Currently I am down to 9 possibilities but would like to avoid checking each one individually
 A: Hint $\ {\rm mod}\ 3\!:\ x^2\!+x+1 = (x\!-\!1)^2,\,$ and $\,(x\!-\!1)^2\!\mid f(1\!-\!f)\!\iff \!(x\!-\!1)^2\!\mid\color{#c00}{ f}\ $ or $\ (x\!-\!1)^2\!\mid \color{#0a0}{1\!-\!f}\ $ since $\,x\!-\!1\,$ is prime, but $f,\,1\!-\!f$ are coprime. So  in $\,\Bbb Z_3[x]/(x\!-\!1)^2$ only $\,f = \color{#c00}0,\,\color{#0a0}1$ are idempotent. 
A: I think the other two solutions are relatively straightforward, but here is an alternative with different flavor. 
After you factor $x^2+x+1$ over the field of three elements and discover it is $(x+2)^2$, you should see right away that the only proper ideal above this ideal is $(x+2)$. So this ring has exactly one nontrivial ideal.
If you had a nontrivial idempotent e, then both $eR$ and $(1-e)R$ would be distinct nontrivial ideals.

Actually, that line of reasoning does not depend on counting the number of ideals. The main point is that there is only one maximal ideal. In such rings, there can only be trivial idempotents. The ideals  $eR$ and $(1-e)R$ Cannot lie in the same maximal ideal, and yet they must each fall into some maximal ideal.
A: Any element of the ring can be written (uniquely) in the form $ax+b$.  If such an element is an idempotent, then $(ax+b)^2=ax+b$.  Expand this, use the relation $x^2 = - x -1$, and equate coefficients to get a system of two equations for $a$ and $b$.  If that system has a solution in $\mathbb{Z}_3$ then you have found an idempotent; if it has no solution, then you have shown no idempotent exists.
