lattice ideal
Let $L$ be a lattice^{}. An ideal $I$ of $L$ is a nonempty 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 nontrivial 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,
$$\text{xymatrix}\mathrm{\&}1\text{ar}\mathrm{@}[ld]\text{ar}\mathrm{@}[rd]a\text{ar}\mathrm{@}[rd]\mathrm{\&}\mathrm{\&}b\text{ar}\mathrm{@}[ld]\mathrm{\&}c\text{ar}\mathrm{@}[d]\mathrm{\&}\mathrm{\&}d\text{ar}\mathrm{@}[ld]\text{ar}\mathrm{@}[rd]\mathrm{\&}e\text{ar}\mathrm{@}[rd]\mathrm{\&}\mathrm{\&}f\text{ar}\mathrm{@}[ld]\mathrm{\&}0$$ 
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 $a\le {a}_{1}$. 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  20130322 15:48:58 
Last modified on  20130322 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  nontrivial ideal 