Let $R$ be a commutative ring with unity, $e\in R$ is called and idempotent if $e^2=e$ and if $e\notin \{0,1\}$ then it is called a non-trivial idempotent.
want to show that $\text{Spec}R$ is not connected if and only if there exists a non trivial idempotent $e\in R$.
I have no idea how to solve this, need your help, thank you.
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i.a.m
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2Hint: If an idempotent $e$ exists, use $R=eR \oplus (1-e)R$ (or the fact that $V((e))$ and $V((1-e))$ are disjoint closed subsets) and if $X=SpecR$ is disconnected, then it's the union of two disjoint closed subsets $V(I), V(J)$ corresponding to ideals $I,J$ of $R.$ Can you conclude that $R=R/I \times R/J?$ if so, can you find a non-trivial idempotent? – Ehsan M. Kermani Mar 12 '13 at 02:15
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@EhsanM.Kermani, I'm a little lost, can you explain a little more, should I show that $R=R/I \times R/J$ or $X=R/I \times R/J$ – i.a.m Mar 12 '13 at 02:20
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$R$ is your ring not $X,$ right? in fact, $X=Spec(R/I) \amalg Spec(R/J).$ – Ehsan M. Kermani Mar 12 '13 at 02:35
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Dupliate of If $\mathop{\mathrm{Spec}}A$ is not connected then there is a nontrivial idempotent which was asked two days ago. – Martin Brandenburg Mar 12 '13 at 09:35
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@Ehsan: In your approach $I,J$ should be radical ideals. – Martin Brandenburg Mar 12 '13 at 09:36
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Dear @Martin, hmm... is it necessary for the closed subsets to be irreducible? – Ehsan M. Kermani Mar 13 '13 at 00:48