Suppose a finite ring $R$. Show that each $x \in R$ is exactly one of a unit, nilpotent, or $x^k$ is idempotent.
I know I must show this in cases.
Case 1: Suppose $x$ is a unit. Then there exists a $y \in R$ such that $xy=1=yx$. Can I then say this implies $1/y$ times $1/y$ cannot equal $0$ therefore $x$ is not nilpotent ($x^n=0$ for some $n\ge0$). And also that $1/(y^k)$ times $1/(y^k)$ cannot equal $1/(y^k)$ therefore $x^k$ is not idempotent ($x^k\cdot x^k = x^k$ for some $k\ge0$ such that $x^k\notin\{0,1\}$)
(I may be completely overlooking something here)
Case 2: Suppose $x$ is nilpotent... I am stuck here
Case 3: Suppose $x^k$ is idempotent $x^k\cdot x^k = x^k$ for some $k\ge1$, therefore $x^n$ cannot equal $0$; $x$ is not nilpotent I am not sure how to show this is not a unit
also do I have to prove a fourth case where I suppose $x$ is none of the above and show this is a contradiction? Therefore it is exactly one of a unit, nilpotent, or $x^k$ idempotent.
Consider the powers of $x$: $1$, $x$, $x^2$, … . Since the ring is finite, these powers must repeat.
If you reach $1$, then $x$ is a unit.
If you reach $0$, then $x$ is nilpotent.
Otherwise, you must have $x^{n+t}=x^n$ for some $n,t$ with $t\ge1$.
Then $x^{e+t}=x^e$, for every $e\ge n$, and so $x^{e+tq}=x^e$, for every $q\ge 1$.
Take $q$ such that $e=tq\ge n$. Then $x^{2e}=x^{e+tq}=x^e$. Thus, $x^{tq}$ is idempotent.