# lattice ideal

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

1. 1.

$I$ is a sublattice of $L$, and

2. 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. 1.

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

2. 2.

for any $a\in I$, if $b\leq a$, 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 $b\leq a$, then $b=a\wedge b\in I$. Conversely, since $a\wedge b\leq a\in I$, $a\wedge b\in I$ 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 $I\neq L$.

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

• $I$ 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 $S\neq\varnothing$ ($L\in S$), 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. 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. 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 $b\leq p\vee a$ for some $p\in I$. Then $b=b\wedge b\leq(p\vee a)\wedge b=(p\wedge b)\vee(a\wedge b)\in I$, which means $b\in I$. So $I$ is prime.

3. 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 $1\leq b\vee a\in J$, so $J=L$.

4. 4.

Combining the two results above, in a Boolean algebra, an ideal is prime iff it is maximal.

Examples. In the lattice $L$ below,

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

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 $a\leq a_{1}$. If $a_{1}$ is not maximal in $I$, repeat the process above so we get a chain $a\leq a_{1}\leq a_{2}\leq\ldots$ 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 $c\leq b$ 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