This question might be below the level of MO, so apologies in advance. I posted the [same question](https://math.stackexchange.com/questions/4503962/does-the-set-of-ideals-whose-jacobson-radical-nilradicals-coincide-form-a-la) in MS about a week ago without an answer so far.


Let $R$ be a unital commutative ring and $L(R)$ denote the [lattice of ideals](https://planetmath.org/latticeofideals) of $R$. Let $\mathcal{S}\subseteq L(R)$ be the set of all ideals whose Jacobson radicals and nilradicals coincide.


**Q1:** For which classes of rings $\mathcal{S}$ is a sublattice (respectively, $\sigma$-complete or complete sublattice) of $L(R)$?


**Q2:** For which classes of rings $\mathcal{S}$ is a lattice? Specifically, we could define $I\vee\_S J$ as the *smallest* ideal in $\mathcal{S}$ containing $I+J$ for a given pair $I,J\in\mathcal{S}$. For which classes of rings $I\vee\_S J$ exists for any $I,J\in\mathcal{S}$?


For example, $\mathcal{S}=L(R)$ iff $R$ is a [Hilbert-Jacobson ring](https://en.wikipedia.org/wiki/Jacobson_ring), so $\mathcal{S}$ is trivially a complete lattice.




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**edit (Aug 9, 2022):** Professor Keith Kearnes' reply below indeed answers the question in the title by providing a counterexample. I appreciate his detailed response. I'm carrying the two questions in the body of the post to a subsequent [separate question](https://mathoverflow.net/questions/428175/set-of-ideals-whose-jacobson-radical-nilradicals-coincide-ii) that look for an affirmative answer.



Does the set of ideals, whose Jacobson radical & nilradicals coincide, form a sublattice?
Let's call an ideal $I\lhd R$ *Jacobson* if $J(I)=\sqrt{I}$.
I will answer the question in the title by constructing, in stages, an example of a unital ring $R$ where the set of Jacobson ideals
is not a sublattice of the ideal lattice of $R$. It will turn out that more is true about this example: its ordered set of Jacobson ideals does not form a lattice at all.
Since the formation of Jacobson radical or nilradical commutes with intersection, it follows that if $A, B\lhd R$ are Jacobson, then $J(A\cap B)=J(A)\cap J(B)=\sqrt{A}\cap \sqrt{B}=\sqrt{A\cap B}$, so $A\cap B$ is also Jacobson. This means that the example I want to construct should have Jacobson ideals $A$ and $B$ where $A+B$ is not Jacobson.
**Stage 1.** Let $\mathbb Q$ be the field of rational numbers and let $L$ be the subring of $\mathbb Q$ consisting of fractions $m/n$ with odd denominator.
The key fact here is that $L$ is a local integral domain with field of fractions equal to $\mathbb Q$. 
**Stage 2.**
Let $S$ be the subring of
$\mathbb Q^{\omega}$ consisting of those tuples
$\textbf{q}=(q\_0,q\_1,\ldots)\in \mathbb Q^{\omega}$
which are eventually constant and which satisfy the condition that the limit
$q\_{\infty}:=\lim\_{n\to\infty} q\_n$ belongs in the subring $L\leq \mathbb Q$. Saying that $\textbf{q}\in S$
means the same thing as saying that
$\textbf{q}\in \mathbb Q^{\omega}$ and
all but finitely many entries of $\textbf{q}$ are equal to
some fixed $q\_{\infty}\in L$.


The key facts here are that  



* $S$ has trivial Jacobson radical. This is because the $n$th
coordinate projection $\pi\_n\colon S\to \mathbb Q\colon \textbf{q}\mapsto q\_n$
is surjective. Since the image of $\pi\_n$ is a field we get that $\ker(\pi\_n)\lhd S$ is a maximal ideal of $S$.
This ideal contains exactly those
elements of $S$ that vanish in the $n$th coordinate.
If one intersects the maximal ideals of this type, one is left with
$\{\textbf{0}\}=\{(0,0,\ldots)\}$ only. Thus, if one intersects all maximal ideals of $S$,
one must get the zero ideal of $S$.
* $S$ has trivial nilradical. This is because the nilradical is contained
in the Jacobson radical.
* $S$ has a retraction onto a
subring $L'$ that is isomorphic to the $L$ from Stage 1.
Here, the retraction is the map $(q\_0,q\_1,q\_2, \ldots)\mapsto (q\_{\infty},q\_{\infty},q\_{\infty},\ldots)$ where $q\_{\infty}$ is $\lim\_{n\to\infty} q\_n$.


**Stage 3.** The ring that I have been aiming for is $R:=\{(\textbf{u},\textbf{v})\in S\times S\;|\;u\_{\infty}=v\_{\infty}\}$. To make this clear, let me repeat the definition using more words than symbols: $R$ is the subring of $S\times S$ consisting of those pairs of tuples that have the same limit.


The key facts here are that  



* $R\leq S\times S$ is a subdirect embedding. That is, $R$ projects onto $S$ in each of the two factors.
This means that the composition of the embedding $R\leq S\times S$
with either projection
$\pi\_i\colon S\times S\to S$, $i=1,2$, is surjective.
Let the kernels of these two projections be $A, B\lhd R$, so that
  

$A=\ker(\pi\_1)=\{\{(\textbf{0},\textbf{v})\in S\times S\;|\;v\_{\infty}=0\}$
and
$B=\ker(\pi\_2)=\{\{(\textbf{u},\textbf{0})\in S\times S\;|\;u\_{\infty}=0\}$.   

Notice that any pair of tuples in $A$ or $B$ is zero in all but finitely
many coordinates.

* $A$ and $B$ are Jacobson ideals of $R$.
This is because $R/A\cong S$, $R/B\cong S$,
and $S$ has trivial Jacobson radical.
* Let $L''$ be the subring of $R$ consisting of all pairs
$(\textbf{u},\textbf{v})=((\ell,\ell,\ell,\ldots),(\ell,\ell,\ell,\ldots))$ for some $\ell\in L$.
I will refer to $L''$ as the 'diagonal copy' of $L$ inside $R$.
It is clear that $L''\cong L$.
* Let $C=A+B$. This is the ideal of $R$ consisting of exactly those pairs of tuples in $R$ that are zero almost everywhere.  
 
I claim that $C$ is the kernel of a retraction of $R$
onto $L''$. To see this, first consider the intersection $C\cap L''$:
since pairs of tuples in $C$ are zero almost everywhere and $L''$
is the diagonal copy of $L$ in $R$, the only common pair
must be the diagonal pair with all coordinates zero. Thus $C\cap L'' = \{(\textbf{0},\textbf{0})\}$.  

Now I want to show that $R=C+L''$. Choose any
$x=(\textbf{u},\textbf{v})\in R$
and define $\ell:=u\_{\infty}=v\_{\infty}$. The pair
$d:=((\ell,\ell,\ldots),(\ell,\ell,\ldots))$ belongs to $L''$
while the difference $x-d\in R$ is zero almost everywhere,
hence belongs to $C$. Thus, $x=(x-d)+d\in C+L''$. This completes the proof that
$C$ is the kernel of a retraction of $R$ onto $L''\cong L$.
(That is, $C$ is an ideal of $R$, $L''$ is a subring of $R$, and the conditions $C\cap L'' = \{(\textbf{0},\textbf{0})\}, C+L''=R$ guarantee that $L''$ is a complete irredundant set of coset representatives for $C$ in $R$. The map that sends an element of $R$ to its $C$-coset representative in $L''$ is the retraction.)


We explained above why $A, B\lhd R$ are Jacobson ideals.
Now I argue that $C:=A+B$ is not Jacobson,
i.e., that $J(C)\neq \sqrt{C}$. For this we work in
$R/C\cong L''\cong L$. It suffices to show that
the zero ideal of this quotient is not Jacobson.
Equivalently, we must show that
the $J(0)\neq \sqrt{0}$ in $L$.
But, $L$ is a local domain that is not a field,
so $J(0)\neq 0=\sqrt{0}$.
The final observation I want to make is that
$A$ and $B$ do not have a join in the ordered set of Jacobson ideals
of $R$, so the Jacobson ideals of $R$ do not form a lattice.
For this it suffices to note that the ideals of $L$ are
$$
L\supsetneq (2) \supsetneq (4) \supsetneq (8) \supsetneq \cdots \supsetneq (0),
$$
that $J(L)=\sqrt{L}=L$, and $J((2^k))=\sqrt{(2^k)}=(2)$ for $k>0$.
This shows that every ideal of $L$ is Jacobson except the zero ideal.
In particular, since there is no least nonzero ideal of $L$, there is no least Jacobson ideal of $L$. Now $L\cong R/C$ for $C=A+B$. We can translate this information back to $R$ to conclude that every ideal above $C$
is Jacobson except $C$ itself, and there is no least ideal properly above $C$. Writing this in terms of $A$ and $B$, this means that every ideal of $R$ containing both $A$ and $B$ is Jacobson except $A+B$, and there is no least element in the set of ideals properly above $A+B$. Hence there is no join of $A$ and $B$ in the ordered set of Jacobson ideals of $R$.
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