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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.

egreg
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Brandon
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    As stated this is not correct. $0$ is both a nilpotent and an idempotent, and $1$ is both a unit and an idempotent. – ronno May 20 '14 at 14:08
  • It follows that R has no nilpotent elements other than zero and no idempotent elements other than 0 and 1. R is a finite integral domain. – Brandon May 20 '14 at 14:21
  • @Brandon A.K.A. a finite field. – rschwieb May 20 '14 at 16:25
  • @Brandon are you assuming commutativity? – rschwieb May 20 '14 at 17:33
  • The statement, in that form, is incorrect, because $0$ is both nilpotent and idempotent. Also $1$ is both idempotent and a unit. And $-1$ is a unit and its square is idempotent. Fix: Every $x\in R$, $x\ne0$ and not a unit, is either nilpotent or $x^k$ is idempotent for some $k>0$. – egreg May 20 '14 at 20:44

1 Answers1

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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.

lhf
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  • ah that makes sense. do you think i should also show that if none of these cases are reached, a contradiction is made? – Brandon May 20 '14 at 14:27
  • I don't think that $x^t$ is idempotent though.. But you can prove by induction that $x^{kn}=x^{km}$ for all $k$, and that $x^{kn+jt}=x^{kn}=x^{km}$ for all $j$. From here I suspect you can create some $m_1,n_1$ with $m_1-n_1=gcd(m,n)=gcd(m_1,n_1)$ such that $x^{m_1}=x^{n_1}$, from where the conclusion is easy. – N. S. May 20 '14 at 16:08
  • @N.S., see my edited answer. – lhf May 20 '14 at 19:59
  • See also http://math.stackexchange.com/a/568093/589, which I've just found. – lhf May 21 '14 at 19:13