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 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 $S\ne\varnothing$ ($L\in S$ ), 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).

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 $\down X'$ , 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 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,

• $M=\lbrace a, c, d, e, f, 0\rbrace$ ,
• $N=\lbrace b, c, d, e, f, 0\rbrace$ ,
• $R=\lbrace c, d, e, f, 0\rbrace$ ,
• $S=\lbrace d, e, f, 0\rbrace$ ,
• $T=\lbrace e, 0\rbrace$ , and
• $U=\lbrace f, 0\rbrace$ .

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, $I=(b]$ .

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