$R$ is a ring which satisfies the following situation, $\exists n \ge 2; \forall x \in R; x^n = x$

Question

Suppose $R$ is a ring which satisfies the following situation:
$$\exists n \ge 2; \forall x \in R; x^n = x$$

1. $Nil(R)=\{0\}$
2. Every idempotent element of $R$ is in the center of $R$
3. $\forall x \in R , x^{n-1} \in Z(R)$ in which $Z(R)$ is the center of Ring

For the first one I know I have to show that $0$ is the only element in which you can find an integer to satisfy the Nilpotent element conditions. For the second one I have no clear clue.

Answer 1

For the first one, take $x\in Nil(R)$ and try to prove that $x=0$. To do that, think about these things:

• If $x\in Nil (R)$, there exists some $m\in \mathbb N$ such that $x^m=0$
• What is $x^{nm}$ equal to?
• $x^{mn} = (x^m)^n=(x^n)^m$

I won't answer the other questions because they should be asked separately - this website should have one question per post.

Answer 2

For (2), suppose $x^2=x$, and consider $(xyx-xy)^2$ and $(xyx-yx)^2$.

Answer 3

1. Suppose $x$ is nilpotent and $m$ is the smallest positive integer such that $x^m = 0$. Now $m \geq n$ is impossible because $x^m = x^{m-n} x^n = x^{m-n+1}$, and if $m \geq n$, then $m-n+1$ is a smaller positive number with $x^{m-n+1}=0$. So $m < n$. But then $x = x^n = x^m x^{m-n} = 0 x^{m-n} = 0$.

2. See other answers.

3. We can show $x^{n-1} \in Z(R)$ by showing that $x^{n-1}$ is idempotent. We have $$ (x^{n-1})^2 = x^{2n-2} = x^n x^{n-2} = x^1 x^{n-2} = x^{n-1}. $$