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How can I prove it?

Principal ideal ring if and only if it is both a Noetherian ring and a Bezout ring. I think that The fact that a Noetherian Bezout ring is a principal ideal ring follows by a single line of logic based on the above definitions. The fact that every principal ideal ring is Bezout is again direct from the definitions. The fact that it is Noetherian follows from the fact that a principal ideal is, by definition, finitely generated.

But I need something better explained than mine.

Alan Simonin
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johnsaa
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2 Answers2

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Let $R$ be a commutative ring. Suppose that $R$ is a principal ring. Then for each ideal $I \subset R$ we have that $I=(a)$ for some $a \in R$ so $R$ is noetherian, since ideal is $1$-generated. Similarly $R$ must be Bezout because the sum of any two principal ideals is again principal. Conversely if $R$ is Bezout and Noetherian then for any ideal $I \subset R$ we have that $I$ is finitely generated. Since $R$ is Bezout any finitely generated ideal is principal thereby $I$ is principal. But $I$ was an arbitrary ideal, so $R$ is a principal ring.

JSchlather
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This all seems clear... Nothing to add or change.

Perhaps you should elaborate that the Noetherian property means that all ideals are finitely generated.

Then, using the definition of Bezout ring: all finitely generated ideal is principal, we can see the equivalence.

Berci
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