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Recall that an ideal $I$ of a commutative ring with identity $R$ is called regular if it contains a regular element. An element of $R\setminus Z(R)$ is said to be regular, where $Z(R)$ denotes the set of zero divisors of $R$. Also, an ideal $I$ is faithful if $(0):I=(0)$.

There is a well-known fact that: In a Noetherian ring, every faithful ideal is regular. But I can't prove it.

Bernard
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M.Ramana
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1 Answers1

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By contrapositive:

If $I$ is not regular, $I\subset Z(R)$. Now $\;Z(R)=\mkern-12mu\bigcup\limits_{\mathfrak p\in\operatorname{Ass}R}\mkern-12mu\mathfrak p$, and in a noetherian ring, $\operatorname{Ass}R$ is finite.

Conclude with the avoidment lemma for prime ideals.

Bernard
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