Let $R$ be a commutative ring with unity such that for each ideal $I$ of $R$ there exists an ideal $J$ of $R$ such that $I=J^2$. Is there any characterizations for such rings? Clearly, regular rings have this property, but the reverse is probably not true (but I couldn't prove it).
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5$\begingroup$ I don't think this holds for regular rings. For example, it fails for any discrete valuation ring. $\endgroup$– Brian ShinCommented yesterday
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5$\begingroup$ Suppose that $R$ is a Noetherian ring with such property. Then for every prime ideal $\mathfrak p$, you pick an ideal $\mathfrak q$ such that $\mathfrak p=\mathfrak q^2$, which forces $\mathfrak q=\mathfrak p$, and in particular, $\mathfrak p=\mathfrak p^2$, and thus by Nakayama's lemma, $\mathfrak p$ vanishes in the localization $R_{\mathfrak p}$. In other words, every local ring of $R$ is necessarily a field, which forces that $R$ is a reduced Artinian ring, i.e. a finite product of fields. $\endgroup$– Z. MCommented yesterday
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5$\begingroup$ The op might mean von Neumann regular rings, in which case all ideals are idempotent. $\endgroup$– Benjamin SteinbergCommented yesterday
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2$\begingroup$ @rschwieb Yes, as long as the value group is $2$-divisible, it works, since ideals of valuation rings one-to-one corresponds to ideals of the value group, [SP, Tag 00IH]. And these valuation rings are not von Neumann regular. $\endgroup$– Z. MCommented yesterday
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2$\begingroup$ @Z.M I think this should be understood in reverse, i.e. that the interesting examples are non-Noetherian. Think about, say, the ring of all algebraic integers. $\endgroup$– Kevin CastoCommented 22 hours ago
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Another class of rings: the Jaffard-Ohm-Kaplansky construction allows you to take any linearly ordered Abelian group and to produce a valuation domain whose ideals reflect part of the order.
If you choose your group to be $2$-divisible (like, say $(\mathbb Q,+)$ or $\mathbb R,+)$, then you can ensure that you can always extract a "square root" of an ideal.
In particular, since such a valuation domain is not a field, it cannot be von Neumann regular.
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$\begingroup$ Thanks to Z.M. for confirming that this idea worked like I thought it did... $\endgroup$– rschwiebCommented yesterday