# Math Help - Ring

1. ## Ring

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

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

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

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

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

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