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Let $I$ and $J$ be two ideal of a ring $R$ (commutative with $1$) such that $I\subseteq Ann_R(Ann_R(J))$ and $I$ is a principal ideal. Is there any conditions on $I$ or $J$ or both of them under which we can deduce that $I\subseteq J$?

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  • $\begingroup$ Is the existence of some $R$-module supposed to be implicit here? What is $J$ annihilating things in? I am not an expert in commutative algebra by any means, so forgive me if this is a common convention. $\endgroup$ – Alec Rhea Feb 14 '18 at 6:11
  • $\begingroup$ @AlecRhea An ideal is an $R$-module. $\endgroup$ – André 3000 Feb 14 '18 at 6:27
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    $\begingroup$ I haven't been able to think of any nontrivial conditions (e.g. when $R$ is a dual ring, and $Ann(Ann(J))=J$ for all $J$.) Is there some motivation for a connection you could share? $\endgroup$ – rschwieb Feb 14 '18 at 19:12
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    $\begingroup$ I don't think there is any non-trivial condition that gives you this. On one hand you always have $J\subseteq \mathrm{Ann}(\mathrm{Ann}(J))$. On the other hand if $J\subsetneq \mathrm{Ann}(\mathrm{Ann}(J))$, then let $t\in \mathrm{Ann}(\mathrm{Ann}(J))\setminus J$ and $I=(t)$. Then this $I$ satisfies the condition, but the desired statement fails. The only way to guarantee this is that $J= \mathrm{Ann}(\mathrm{Ann}(J))$, but that's a trivial condition. $\endgroup$ – Sándor Kovács Feb 15 '18 at 2:52
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A semiprime operation on a ring $R$ (commutative with identity) is a closure operation $c: I \longmapsto I^c$ on the lattice of all ideals of $R$ such that $I^c J^c \subseteq (IJ)^c$ for all ideals $I$ and $J$ of $R$.

I think it's useful to observe that the operation $r: J \longmapsto (0 : (0 : J))$ on the ideals of $R$ is the largest semiprime operation $c$ on $R$ such that $(0)^c = (0)$. More generally, for any fixed ideal $H$ of $R$, the operation $J \longmapsto (H : (H : J))$ is the largest semiprime operation $c$ on $R$ such that $H^c = H$.

By definition of $r$, one has $(0 : (0 : J)) = J$ if and only if $J$ is $r$-closed, that is, if and only if $J^r = J$. Moreover, if $J$ is $r$-closed, then $(J : H)$ is $r$-closed for any ideal $H$ (which allows you to get many more $r$-closed ideals from a single one), and the intersection of any family of $r$-closed ideals is $r$-closed. This is because $r$ is a semiprime operation. Thus, for example, any ideal that $J$ that is of the form $(0:H)$ for some ideal $H$ of $R$ is $r$-closed. Conversely, any $r$-closed ideal of $R$ has this form, since $J = (0:H)$ for $H = (0:J)$ if $J$ is $r$-closed. Thus, the $r$-closed ideals are equivalently the ideals that are the annihilator of some ideal. You can also see this by observing that $(0: (0 : (0 : J))) = (0:J)$ for any ideal $J$ of a ring $R$ (which implies that $r$ is indeed a closure operation).

Now, one has $(*)$ $J^r = J$ for all ideals $J$ of $R$, if and only if any ideal of $R$ is an annihilator of some ideal of $R$, if and only if $r$ acts trivially on all ideals of $R$, if and only if the only semiprime operation $c$ on $R$ such that $(0)^c = (0)$ is the trivial semiprime operation $J \longmapsto J$. This is a very strong condition. It implies that $R$ can have no proper ideals containing a non-zerodivisor. In fact, suppose that $R$ is a reduced ring satisfying $(*)$. Then the radical operation $J \longmapsto \sqrt{J}$ is a semiprime operation on $R$ with $\sqrt{(0)} = (0)$. Therefore one must have $\sqrt{J} = J$ for all ideals $J$ of $R$, which holds if and only if $R$ is von Neumann regular (or equivalently, reduced and of Krull dimension zero). Therefore, a reduced ring satisfying $(*)$ must be von Neumann regular. However, there probably exist von Neumann regular rings that don't satisfy $(*)$, although I can't verify an example at the moment. Certainly a finite direct product of fields satisfies $(*)$, as does any finite direct product of rings satisfying $(*)$. Does an arbitrary direct product of fields satisfy $(*)$? I doubt it (you have to look at filters to check this), but an arbitrary direct product of fields is von Neumann regular, so that could be an example.

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  • $\begingroup$ Nice answer but partial: OP asks about conditions both on $a\in R$ and/or $J$ which would ensure that $a$ cannot lie in $J^r\setminus J$. $\endgroup$ – მამუკა ჯიბლაძე Feb 24 '18 at 14:58
  • $\begingroup$ Agreed it's partial, but it was too long to include just as a comment. I characterized the $J$ for which there can be no $a$ in $J^r \backslash J$. $\endgroup$ – Jesse Elliott Feb 24 '18 at 22:44

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