# What kind of algebra is the lattice of ideals of a ring?

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.

• A first remark is that this is not an $R$-module, because there is no additive inverse for ideals. Also as you said your $\neg a$ is not very interesting: it's just $\neg a = R$ unless $a=R$ and then $\neg R=0$. Commented Dec 9, 2022 at 6:42
• Multiplication by a fixed element $r$ can shrink the ideal, I think. If I take $3 \in \mathbb{Z}$ and use it to scale $2\mathbb{Z}$, I get back $6\mathbb{Z}$ ... it's an ideal but a different one. Commented Dec 9, 2022 at 6:53
• Note that lattice you've defined isn't distributive in general. It is, for instance, in a Dedekind domain, but it's not hard to find examples in which it's not distributive. Commented Dec 9, 2022 at 7:45
• encyclopediaofmath.org/wiki/Multiplicative_lattice Commented Dec 10, 2022 at 0:57

What kind of algebra is the lattice of ideals of a ring?

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:

Gondran, Michel, and Michel Minoux. Graphs, dioids and semirings: new models and algorithms. Vol. 41. Springer Science & Business Media, 2008.

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.