Ring subsets

Let $R$ be a ring. Then the subsets of $R$ can be classified by which operations and relations they are closed under. This produces an ontology of Moore families.

Every single one of these Moore families produces a type of lattice: lattices of additive subgroups, lattices of subrings,lattices of ideals, etc. Subfamilies in the inclusion ordering form lattice-ordered suborders of one another as lattices. These may or may not be sublattices depending upon the families in consideration.

Lattice of ideals:
Theorem. the lattice of ideals forms a sublattice of the lattice of additive subgroups

Proof.
Meet closure:
Ideals and additive subgroups are both Moore families, so they both preserve intersection as their meet operation. Therefore, the lattice of ideals is a meet subsemilattice of the lattice of additive subgroups.

Join closure:
Let $R$ be a ring, then the additive group of the ring $(R,+)$ is an commutative group. The join of two groups is equal to the set of all sums of sequences formed by elements of the two groups, but with commutativity this reduces to all sums of two elements with one chosen from each group. Let $A,B$ be subgroups of $(R,+)$ then $A \vee B$ is equal to the set {a+b} of sums of elements from each group.This forms a subgroup, so the only thing to remain is to show that this set {a+b} is also an ideal.

Let $x$ be an element of the ring $R$. Then let $a+b$ be an element of $A \vee B$. The product of $x$ and $a+b$ is equal to $x(a+b)$. By the distributive law, this is equal to $xa+xb$. The set $A$ is an ideal so there exists some element of $A$ denoted $a_0$ which is equal to the product $xa$ and since $B$ is an ideal there is some element $b_0$ equal to the product $xb$. By substitution $xa+xb = a_0+b_0$, but now $a_0+b_0 \in {a+b} = A \vee B$ and therefore the product is in the original set. Ideals therefore form a join subsemilattice of the lattice of additive subgroups.

Remarks. the join of two ideals in the lattice of ideals is typically referred to as $A+B$ rather then $A \vee B$. The fact that ideals are a sublattice of additive subgroups puts this in context. Therefore, from one now one the join of ideals can be referred to as $A+B$.

Corollary. the lattice of ideals is a sublattice of $Sub(R)$.

Proof. Additive subgroups, subrings, and ideals form a chain of lattices. Ideals are a sublattice of additive subgroups, and therefore they must be a sublattice of the intermediate lattice of subrings. To see this, consider the join of two ideals $A,B$ in the lattice of subrings. This is the smallest set that is both an additive subgroup and multiplicatively closed which contains both of them, but the smallest additively closed set is also multiplicatively closed. Therefore, the subring join of ideals is the additive subgroup join which is an ideal. Therefore, the join of ideals is an ideal. The meet of ideals is an ideal because both subrings and ideals are Moore families.

Overview. ideals form a union-free modular Moore family

Lattice of radical ideals
Another lattice of ring subsets is the lattice of radical ideals. This does not necessarily form a sublattice of the lattice of ideals. The closure operation associated with this family is the radical of an ideal $\sqrt{I}$. Prime ideals are the intersection-irreducible members of this family, therefore by considering the lattice of radical ideals we can give the prime ideals a lattice-theoretic perspective.