# The Unapologetic Mathematician

## The lattice of ideals

We know that the collection of all ideals of a given ring form a rig. In fact, they also form a lattice. We put the partial order of inclusion on ideals, so $I$ is below $J$ if $I\subseteq J$.

To show that this poset is a lattice we have to show that pairwise greatest lower bounds and least upper bounds exist. Lower bounds are easy: the intersection $I\cap J$ of two ideals is again an ideal. By definition, any ideal contained in both $I$ and $J$ is contained in $I\cap J$.

Upper bounds are a little trickier, since we can’t just take the union of two ideals. That would work for subsets of a given set, but in general the union of two ideals isn’t an ideal. Instead, we take their sum. Clearly $I\subseteq I+J$ and $J\subseteq I+J$. Also, if $K$ is another ideal containing both $I$ and $J$, then $K$ contains all linear combinations of elements of $I$ and $J$. But $I+J$ is the set of all such linear combinations. Thus $I+J\subseteq K$, and $I+J$ is the least upper bound of $I$ and $J$.

This lattice is related to the divisibility preorder. Given a commutative unital ring $R$ and two elements $a,b\in R$, recall that $a|b$ if there is an $x\in R$ so that $b=ax$. Then every multiple $bk$ of $b$ is also a multiple $axk$ of $a$. Thus we see that the principal ideal $(b)$ is contained in the principal ideal $(a)$. On the other hand, if $(b)\subseteq (a)$ we can see that $b=ax$ for some $x\in R$, so $a|b$. In particular, two elements are associated if and only if they generate the same principal ideal.

Notice that this correspondence reverses the direction of the order. If $a$ is below $b$ in the divisibility ordering, then $(a)$ is above $(b)$ in the ideal ordering. Thus the “greatest common divisor” of two ideals is actually now the least ideal containing both of them. The language of ideals, however, is far more general than that of divisibility. We now need to recast most of what we know about divisibility from our experience with natural numbers into these more general ring-theoretic terms.

May 15, 2007 - Posted by | Ring theory

1. Nice introduction!

But it felt too short, I was left wanting to know more about the lattice of ideals! Is it complete? What are its sublattices? These may be easy-to-answer questions, but it would be cool to see them added to this post anyways :-)

Regards!

Comment by Jose Brox | August 5, 2008 | Reply

2. Thanks, Jose. I might come back later and cover more when I get the chance. However, I’m really not a ring theorist, so I’m not an expert on this area. If there’s someone else who knows any general answers to these questions, I’d be glad to hear the answers.

Comment by John Armstrong | August 5, 2008 | Reply

3. Sure, the lattice of ideals is complete: any collection of ideals has an inf in this lattice, given by taking the intersection. There is a general result that any lattice or poset which admits arbitrary infs also admits arbitrary sups (see my post Lattices and Duality on my blog with Vishal, for instance).

I don’t know of any particularly nice way of characterizing the sublattices.

Comment by Todd Trimble | August 5, 2008 | Reply

4. There is one subset-which-forms-a-lattice of particular note*: the lattice of radical ideals (those closed under taking “nth roots” for all n > 0). One nice property of this is that it is isomorphic to the lattice of opens of a topological space (specifically, the Zariski spectrum of that ring); thus, it will be complete, though, again, an easier way to prove completeness would be by simply observing closure under arbitrary intersection. Furthermore, for any distributive lattice, the lattice of its (order) ideals will be isomorphic to the lattice of radical ideals of some ring, and vice versa.

*: It need not actually be a sublattice, as the joins within it may not coincide with the joins within the lattice of ideals. However, arbitrary meets within the two will coincide (being intersections in both cases); thus, it will be a sub-complete-meet-semilattice.

Comment by Sridhar Ramesh | August 5, 2008 | Reply

5. This is true, Sridhar. If I get around to some classical algebraic geometry, I’ll cover that topology. Or you can read more from Charles at Rigorous Trivialities.

Comment by John Armstrong | August 5, 2008 | Reply

6. what is the n.and s. conditions for the latiice of ideals of a ring to be distributive ?