I have been messing around with rings and ideals for the past week or so in an attempt to prove Nakayama's lemma as an exercise. I completely failed to prove the lemma, but I did notice something interesting about the lattice of ideals of a fixed ring.
They seem to form an interesting structure that's simultaneously a lattice and an $R$-module. I'm wondering if they have a real name.
We have a few interesting binary operations on ideals. We seem to have two families of operations, one from the lattice structure, one from the ring structure, and one bonus operation that gives us the zeroes.
We have the lattice operations, note that the intersection of two ideals is an ideal.
$$a \land b = a \cap b \\
a \lor b = \langle a \cup b \rangle \\
\lnot a = \langle a^c \rangle $$
We also have the ring operations.
$$ ra = \{rx : x \in a\} \\
a+b = \langle \{x + y : x \in a \land y \in b\} \rangle \\
a*b = \langle \{x * y : x \in a \land y \in b\} \rangle $$
Proof that scalar multiplication sends an ideal to an ideal: Suppose $x = rx'$ is in $ra$ and $y = ry'$ is in $ra$, then $x – y = rx' – ry' = r(x' – y')$ is also in $ra$. Suppose $x = rx'$ is in $ra$, then $sx$ is $srx'$ and $r(sx')$ is in $ra$.
And we have one bonus operation $Z(a)$ which can be defined as $\{r : ra = 0\}$. Suppose $x$ and $y$ are in in $Z(a)$, then $xa=0$ and $ya=0$, thus $xa-ya = 0$ thus $(x-y)a = 0$ thus $x-y$ is in $Z(a)$. Suppose $x$ is in $Z(a)$, then $xa=0$, thus $r(xa) = 0$, thus $(rx)a = 0$ thus $rx$ is in $Z(a)$.
We can also define $\bot$ as $\langle 0 \rangle$ and $\top$ as $\langle R \rangle$ for good measure.
So, this gives us a kind of strange algebraic structure in the signature $(\bot, \top, \land, \lor, \lnot, +, *, R, Z)$ where $R$ denotes scalar multiplication by all ring elements.
So, this thing is a distributive lattice, but not a Boolean lattice ($\lnot$ in general is a super weird operation here).
Multiplication also distributes over addition which is nice: let $\oplus$ and $\otimes$ denote elementwise versions of $+$ and $*$ which are allowed to fail to produce an ideal.
$$(a+b) * c = \langle\langle a \oplus b \rangle \otimes c \rangle = \langle \langle a \otimes c \rangle \oplus \langle b \otimes c \rangle \rangle = ac+bc $$
It's also entirely possible that this kind of structure just isn't interesting. It is essentially just $2^R$ equipped with a closure operator $\langle \rangle$ and the operations taken from $R$ applied elementwise.
Best Answer
It's well-known that the set of ideals of a ring forms what is called a semiring and indeed also carries its natural partial order (inclusion of ideals), and that the partial order is always a modular lattice.
I've seen one place where semirings with partial orders are discussed:
In particular the semiring of ideals has idempotent addition, which I think in their terms this qualifies as an "idempotent dioid," but that is not a very widespread term, I think.
Complementation is covered but it has nothing to do with the set-complement of the ideal in the ring (we've already seen the problems with that in the comments. There are much better candidates for pairing elements.)
You do indeed get an action of sorts of $R$ on the semiring of ideals: you could just consider $r\cdot I:=(r)I$. I think this makes the set of ideals an $R$ semimodule over $R$ considered as a semiring.