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Let $R$ be a commutative ring with unity. I want a proof of the fact that

$R$ is zero-dimensional (in the sense that all prime ideals are maximal) if and only if $R/J(R)$ is von Neumann regular and $J(R)$ is nil, where $J(R)$ is the Jacobson radical of $R$.

If $J(R)$ is nil then it coincides with the nil radical $\sqrt 0$ of $R$. So, if $R/J(R)$ is regular we have $R/\sqrt 0$ regular which implies that $R$ is zero-dimensional by an easy argument. Now, the reverse implication is at stake. Thanks for any suggestion,

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  • $\begingroup$ The result is also mentioned (without proof) in this answer. $\endgroup$ – user26857 Feb 27 '16 at 9:22
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"$\Rightarrow$" Since $\dim R=0$ we have $J(R)=N(R)$, so $J(R)$ is nil and $R/J(R)$ is reduced. But a zero-dimensional reduced ring is von Neumann regular hence $R/J(R)$ is von Neumann regular.

(One can easily show that a commutative ring is von Neumann regular iff all its localizations are fields.)

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