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 $H^c = H$.

Thus, one has $(0 : (0 : J)) = J$ if and only if $J$ is $r$-closed. 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 $R$.

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.

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.

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 $H^c = H$.

Thus, one has $(0 : (0 : J)) = J$ if and only if $J$ is $r$-closed. 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 an intersection of the form $\bigcap_{H \in X}(0:H)$ for some set $X$ of ideals of $R$ is $r$-closed. Conversely, any $r$-closed ideal of $R$ has this form. The reason is that that the collection $Y$ of all such ideals $J$ is closed under colons by ideals $H$ and is also closed under intersections, so the map $s: I \longmapsto \bigcap\{J \supseteq I: J \in Y\}$ is a semiprime operation on $R$ with $r \leq s$, and since $(0)^s \subseteq (0: R) = (0)$, one has $r = s$ by maximality of $r$. Thus, the $r$-closed ideals of $R$ are precisely the ideals that can be expressed as an intersection of annihilator ideals $(0:H)$

Now, one has $(*)$ $J^r = J$ for all ideals $J$ 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. Indeed, 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.

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 $H^c = H$.

Thus, one has $(0 : (0 : J)) = J$ if and only if $J$ is $r$-closed. 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 $R$.

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.