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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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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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    @Pilot Jack's referring to the case of $0=1$ which some people insist on taking seriously :)2012-10-23