lattice ideal

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 (http://planetmath.org/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 proper if $I\ne L$.

• If $L$ contains $0$, $I$ is said to be non-trivial if $I\ne 0$.

• $I$ 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 $S\ne \mathrm{\varnothing }$ ($L\in S$), 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 $\{x\}$, 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 $X^{\prime }$ be the set consisting of all finite joins of elements of $X$, which is clearly a directed set. Then $\downarrow X^{\prime }$, the down set of $X^{\prime }$, 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 the ideal generated by $I$ and $a$ must be $L$, 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,

Besides $L$ and $\{0\}$, below are all proper ideals of $L$:

• $M=\{a,c,d,e,f,0\}$,

• $N=\{b,c,d,e,f,0\}$,

• $R=\{c,d,e,f,0\}$,

• $S=\{d,e,f,0\}$,

• $T=\{e,0\}$, and

• $U=\{f,0\}$.

Out of these, $M,N,S,T,U$ are prime, and $M,N$ are maximal. The ideal generated by, say $\{c,e\}$, 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 $a\le a_1\le a_2\le \mathrm{\dots }$ in $I$. Eventually this chain terminates $a_n=a_{n+1}=\mathrm{\cdots }$. 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, $I=(b]$.∎

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

Title	lattice ideal
Canonical name	LatticeIdeal
Date of creation	2013-03-22 15:48:58
Last modified on	2013-03-22 15:48:58
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	17
Author	CWoo (3771)
Entry type	Definition
Classification	msc 06B10
Synonym	prime lattice ideal
Synonym	maximal lattice ideal
Related topic	LatticeFilter
Related topic	UpperSet
Related topic	OrderIdeal
Related topic	LatticeOfIdeals
Defines	ideal
Defines	proper ideal
Defines	prime ideal
Defines	sublattice generated by
Defines	ideal generated by
Defines	principal ideal
Defines	maximal ideal
Defines	non-trivial ideal