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 $R$. \\\