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Homelattice ideal
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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 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 $b\leq a$, 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 nontrivial 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. 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 $b\leq p\vee a$ 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. 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. 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 $\{b,c,d,e,0\}$.
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