1. ## Ring

Let A be a ring with $\displaystyle 0 \not = 1$ and K be the set
$\displaystyle K =${$\displaystyle x\in A | x^2 - x + 1 = 0$}.
1) If $\displaystyle x\in K$ then $\displaystyle x$ is invertible and $\displaystyle x^{ - 1} = x^5$
2) If $\displaystyle K = A$ \ {0,1} then A is isomorphic with $\displaystyle Z_{3}$ or $\displaystyle 1 + 1 = 0$ .

2. Just to start you off on this, if $\displaystyle x^2-x+1=0$ then $\displaystyle x^2=x-1$, $\displaystyle x^3=x(x-1) = x^2-x = (x-1)-x = -1$.

3. Originally Posted by petter
Let A be a ring with $\displaystyle 0 \not = 1$ and K be the set
$\displaystyle K =${$\displaystyle x\in A | x^2 - x + 1 = 0$}.

2) If $\displaystyle K = A$ \ {0,1} then A is isomorphic with $\displaystyle Z_{3}$ or $\displaystyle 1 + 1 = 0$ .
suppose $\displaystyle 1+1 \neq 0.$ so $\displaystyle -1 \in A - \{0,1\}=K.$ thus: $\displaystyle 1+1+1=(-1)^2 + 1 + 1 =0.$ hence $\displaystyle 1+1=-1.$ in order to prove that $\displaystyle A \simeq \mathbb{Z}_3,$ we only need to show that $\displaystyle K=\{-1 \}.$

so suppose $\displaystyle -1 \neq x \in K.$ so $\displaystyle x^2-x+1=0.$ call this $\displaystyle (1).$ now since $\displaystyle x+1 \in A - \{0,1 \}=K,$ we also have: $\displaystyle (x+1)^2 - (x+1) + 1 = 0,$ which gives us: $\displaystyle x^2 + x + 1 = 0. \ \ (2)$

$\displaystyle (1)$ and $\displaystyle (2)$ give us: $\displaystyle -x=(1+1)x = 0,$ i.e. $\displaystyle x=0.$ contradiction! Q.E.D.