How to prove the following?
$R$ is an associative ring with identity. $R$ contains element $r$. The element is not invertible on the right and is not a left divisor of zero. Then the ring $R$ cannot be Artinian on the right.
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Suppose $R$ is right Artinian. Then $rR \supset r^2 R \supset r^3 R \supset \ldots $ is a descending sequence of right ideals. Thus for some $n$, $r^n R = r^{n+1} R$. Use this fact to show that either $r$ is right invertible, or $r$ is a left divisor of zero.
Since $R$ has an identity 1, from $r^nR=r^{n+1}R$ it follows that $r^n1=r^{n+1}s$, or rewritten: $r^n=r^n(rs)$. Since $r$ is not a left divisor of $0$, $r^n\neq0$ and $r^{n+1}s\neq 0$, so it follows that $1=rs$. Then $s$ is a right inverse of $r$, but it contradicts with the statement. Hence $R$ is not Artinian.
This is a generalization of the commutative ring theorem that every Artinian integral domain is a field (with essentially the same proof).
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I took the liberty of fixing your usage of $n$ instead of $r$; also I changed $m$ into $s$ to make clearer what set it belongs to. – egreg Jun 24 '14 at 09:02
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+1 also proves that strongly $\pi$-regular rings have the property, which is called the right cohopfian property. Actually such rings are both right and left cohopfian. – rschwieb Jun 24 '14 at 11:26