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lattice ideal (Definition)

Let $ L$ be a lattice. An ideal $ I$ of $ L$ is a non-empty subset of $ L$ such that

  1. $ I$ is a sublattice of $ L$, and
  2. for any $ a\in I$ and $ b\in L$, $ a\wedge b\in I$.

Note the similarity between this definition and the definition of an ideal in a ring (except in a ring with 1, an ideal is almost never a subring)

Since the fact that $ a\wedge b\in I$ for $ a,b\in I$ in the first condition is already implied by the second condition, we can replace the first condition by a weaker one:

for any $ a,b\in I$, $ a\vee b\in I$.

Another equivalent characterization of an ideal $ I$ in a lattice $ L$ is

  1. for any $ a,b\in I$, $ a\vee b\in I$, and
  2. for any $ a\in I$, if $ b\le a$, then $ b\in I$.

Here's a quick proof. In fact, all we need to show is that the two second conditions are equivalent for $ I$. First assume that for any $ a\in I$ and $ b\in L$, $ a\wedge b\in I$. If $ b\le a$, then $ b=a\wedge b\in I$. Conversely, since $ a\wedge b\le a\in I$, $ a\wedge b\in I$ as well.

Special Ideals. Let $ I$ be an ideal of a lattice $ L$. Below are some common types of ideals found in lattice theory.

  • $ I$ is a proper ideal if $ I\ne L$, and, if $ L$ contains 0, $ I\ne 0$.
  • $ I$ is a prime ideal if it is proper, and for any $ a\wedge b\in I$, either $ a\in I$ or $ b\in I$.
  • $ I$ is a maximal ideal of $ L$ if $ I$ is proper and the only ideal having $ I$ as a proper subset is $ L$.
  • ideal generated by a set. Let $ X$ be a subset of a lattice $ L$. Let $ S$ be the set of all ideals of $ L$ containing $ X$. Since $ S\ne\varnothing$ ($ L\in S$), the intersection $ M$ of all elements in $ S$, is also an ideal of $ L$ that contains $ X$. $ M$ is called the ideal generated by $ X$, written $ (X]$. If $ X$ is a singleton $ \lbrace x\rbrace$, then $ M$ is said to be a principal ideal generated by $ x$, written $ (x]$. (Note that this construction can be easily carried over to the construction of a sublattice generated by a subset of a lattice).
Remarks. Let $ L$ be a lattice.
  1. Given any subset $ X\subset L$, let $ X'$ be the set consisting of all finite joins of elements of $ X$, which is clearly a directed set. Then $ \downarrow\!\!X'$, the down set of $ X'$, is $ (X]$. Any element of $ (X]$ is less than or equal to a finite join of elements of $ X$.
  2. If $ L$ is a distributive lattice, every maximal ideal is prime. Suppose $ I\subseteq L$ is maximal and $ a\wedge b\in I$ with $ a\notin I$. Then $ L$ is generated by $ I$ and $ a$, so that $ b\le p\vee a$ for some $ p\in I$. Then $ b=b\wedge b\le (p\vee a)\wedge b=(p\wedge b)\vee (a\wedge b)\in I$, which means $ b\in I$. So $ I$ is prime.
  3. If $ L$ is a complemented lattice, every prime ideal is maximal. Suppose $ I\subseteq L$ is prime and $ a\notin I$. Let $ b$ be a complement of $ a$, then $ b\in I$, for otherwise, $ 0=a\wedge b\notin I$, a contradiction. Let $ J$ be the ideal generated by $ I$ and $ a$, then $ 1\le b\vee a\in J$, so $ J=L$.
  4. Combining the two results above, in a Boolean algebra, an ideal is prime iff it is maximal.

Examples. In the lattice $ L$ below,

$\displaystyle \xymatrix{ & 1 \ar@{-}[ld] \ar@{-}[rd] \\ a \ar@{-}[rd] & & b \ar... ... & \\ & d \ar@{-}[ld] \ar@{-}[rd] & \\ e \ar@{-}[rd] & & f \ar@{-}[ld] \\ & 0 }$    

Besides $ L$ and $ \lbrace 0\rbrace$, below are all proper ideals of $ L$:

  • $ M=\lbrace a, c, d, e, f, 0\rbrace$,
  • $ N=\lbrace b, c, d, e, f, 0\rbrace$,
  • $ R=\lbrace c, d, e, f, 0\rbrace$,
  • $ S=\lbrace d, e, f, 0\rbrace$,
  • $ T=\lbrace e, 0\rbrace$, and
  • $ U=\lbrace f, 0\rbrace$.
Out of these, $ M,N,S,T,U$ are prime, and $ M,N$ are maximal. The ideal generated by, say $ \lbrace c,e\rbrace$, is $ R$. Looking more closely, we see that $ R$ can actually be generated by $ c$, and so is principal. In fact, all ideals in $ L$ are principal, generated by their maximal elements. It is not hard to see, that in a lattice $ L$ with acc (ascending chain condition), all ideals are principal:
Proof. . First, let's show that an ideal $ I$ in a lattice $ L$ with acc has at least one maximal element. Suppose $ a\in I$. If $ a$ is not maximal in $ I$, there is a $ a_1\in I$ such that $ a\le a_1$. If $ a_1$ is not maximal in $ I$, repeat the process above so we get a chain $ a\le a_1\le a_2\le\ldots$ in $ I$. Eventually this chain terminates $ a_n=a_{n+1}=\cdots$. Thus $ b=a_n$ is maximal in $ I$. Next, suppose that $ I$ has two distinct maximal elements. Then their join is again in $ I$, contradicting maximality. So $ b$ is unique and all elements $ c$ such that $ c\le b$ must be in $ I$. Therefore, $ I=(b]$. $ \qedsymbol$

Finally, an example of a sublattice that is not an ideal is the subset $ \lbrace b, c, d, e, 0\rbrace$.



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See Also: lattice filter, upper set, order ideal, lattice of ideals

Other names:  prime lattice ideal, maximal lattice ideal
Also defines:  ideal, proper ideal, prime ideal, sublattice generated by, ideal generated by, principal ideal, maximal ideal
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Cross-references: eventually, chain, ACC, maximal elements, iff, Boolean algebra, contradiction, complement, complemented lattice, prime, maximal ideal is prime, distributive lattice, down set, directed set, joins, finite, generated by, singleton, intersection, proper subset, contains, theory, types, characterization, equivalent, subring, ring, similarity, sublattice, subset, lattice
There are 48 references to this entry.

This is version 13 of lattice ideal, born on 2006-03-27, modified 2007-05-05.
Object id is 7781, canonical name is LatticeIdeal.
Accessed 5950 times total.

Classification:
AMS MSC: 06B10 (Order, lattices, ordered algebraic structures :: Lattices :: Ideals, congruence relations)
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