I'm supposed to prove that, in a ring, if $a$ is an element such that the set $\{a^n; n=0,1,...\}$ is finite, then $a$ is either invertible or a zero divisor. I don't understand how $a$ can not be invertible, since if the set is finite then there has to be an $i$ such that $a^i=1$.
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1What happens for $a=0$? – asdq Nov 21 '18 at 09:17
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Though not an exact dupe, many answers in the dupe apply here too. – Bill Dubuque Nov 21 '18 at 15:15
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e.g. In $,\Bbb Z/8,$ the positive powers of $2$ being $,2,4,0,0,\ldots$ are not $\equiv 1\pmod{!8},$ being even $= 2^n !+ 8k$ $\ \ \ \ \ \ \ \ \ \ $ – Bill Dubuque Nov 21 '18 at 15:24
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Since the set is finite there exist $m,n$ such that $a^m=a^n$. Without loss of generality assume that $m>n$. If $a$ is not a zero divisor then $0=a^m-a^n=a^n(a^{m-n}-1)$ implies that $a^{m-n}=1$ hence $a$ is invertible.
If you consider the ring $k[x]/(x^2)$ for example, the powers of $x$ are finite and $x$ is a zero divisor.
Levent
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It is not true that if that set is finite, then $a^i=1$ for some $i\in\mathbb N$. Take, for instance, the ring of all $2\times2$ rational matrices. Then each matrix $M$ of the form$$M=\begin{bmatrix}1-d&\frac{d-d^2}c\\c&d\end{bmatrix},$$($c\in\mathbb{Q}\setminus\{0\}$, $d\in\mathbb Q$) is such that $M^2=M$. So, the set $\{M^n\,|\,n\in\mathbb{Z}^+\}$ is surely finite. However, there is no natural $i$ such that $M^i=\operatorname{Id}$.
Of course, since $M^2=M$, you have $M\times(M-\operatorname{Id})=0$ and therefore $M$ is a zero divisor.
José Carlos Santos
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