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Let $R$ be a commutative ring of Krull dimension $0$.

Let $x\in R\setminus\{0\}$ such that $x$ is non-zero-divisor. Then, how do I prove that $x$ is a unit?

Rubertos
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  • Almost a duplicate of https://math.stackexchange.com/questions/518/why-does-a-minimal-prime-ideal-consist-of-zerodivisors – user26857 Sep 09 '17 at 18:43

2 Answers2

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The set of zero-divisors in a commutative ring is the union $$\bigcup_{\mathfrak p\in\operatorname{Ass}_f(A)}\mathfrak p,$$ where $\operatorname{Ass}_f(A)$ denotes the set of weakly associated prime ideals in $\operatorname{Spec}(A)$, and these sets are the same if $\dim A=0$.

Alternatively each non-zero divisor is a unit in each localisation $A_\mathfrak p$ since the maximal ideal of the latter is made up of nilpotent elements. Hence the image of a non-zero divisor in $\displaystyle\prod_{\mathfrak p\in\operatorname{Spec}(A)}\!\!\!\!A_\mathfrak p$ being a unit, is itself a unit by faithful flatness of $\displaystyle\prod_{\mathfrak p\in\operatorname{Spec}(A)}\!\!\!\!A_\mathfrak p$.

user26857
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Bernard
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It's known that such rings have the property that every chain of the form $xR\supseteq x^2R\supseteq x^3R\supseteq\ldots$ stabilizes after finitely many steps.

That being the case, eventually there will be an $n$ such that $x^n\in x^{n+1}R$, so that $x^n=x^{n+1}y$ for some $y\in R$. If $x$ isn't a zero divisor, you can repeatedly cancel it off the left until $1=xy$. Thus $x$ is a unit.

rschwieb
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