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Given that the ring R with unity has finite number of idempotents. How do we show that the number of idempotents is even?

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up vote 1 down vote accepted

If $e$ is an idempotent, so is $1-e$. This pairs the idempotents off, and so you have an even number of them.

I don't know if you want any more ideas for a formal proof. Basically you could just find a maximal set $S$ of idempotents such that if $e\in S$, and $1-e\notin S$. If $X$ is the full set of idempotents, then you can show there is a bijection between $S$ and $X\setminus S$, and since these are finite sets, it implies $X$ is of order $2n$ where $n$ is the size of $S$.

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The map $e\mapsto 1-e$ is its own inverse, you'll notice. – rschwieb Oct 23 '12 at 19:51
Wonderful.Bad that I didnt notice it at once.Thank you ! – p.s Oct 23 '12 at 19:53
@Pilot Another trick about idempotents to have in your pocket: $(1-2e)^2=1$! – rschwieb Oct 23 '12 at 19:54
To show the bijection between $S$ and $X\setminus S$ it helps to know $e \neq 1-e$. This is easily shown from $e(1-e) = 0$ but $ee=e$ (and $e=0$ doesn't work unless $R=\{0\}$, which is a counter-example, but presumably is supposed to be ignored). – Jack Schmidt Oct 23 '12 at 19:57
It is required that R has unity as well,that means R has at least 0 and 1 in it. – p.s Oct 23 '12 at 20:01

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