If $m,n>1$ are integers such that $g.c.d.(m,n)=1$ then is it true that there are at-least four elements in $\mathbb Z_{mn}$ such that $x^2=x$ ( i.e. idempotent ) ?
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Do you know the Chinese remainder theorem? – Tobias Kildetoft Dec 12 '14 at 11:02
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@TobiasKildetoft: yes ... – Dec 12 '14 at 11:04
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1In general, given two rings $R$ and $S$ (with units), can you find at least $4$ distinct idempotents in $R\times S$? – Tobias Kildetoft Dec 12 '14 at 11:05
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Using the Chinese remainder theorem, $\Bbb Z_{mn}$ has the same structure as $\Bbb Z_{m}\times \Bbb Z_{n}$.
In this algebraic structure, there are the idempotents: $$ (0,0), (0,1), (1,0), (1,1) $$
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