Editing Ideal (order theory) (section)

==Maximal ideals==

An ideal {{mvar|I}} is a {{em|{{visible anchor|maximal ideal}}}} if it is proper and there is no ''proper'' ideal ''J'' that is a strict superset of {{mvar|I}}. Likewise, a filter ''F'' is maximal if it is proper and there is no proper filter that is a strict superset.

When a poset is a [[distributive lattice]], maximal ideals and filters are necessarily prime, while the converse of this statement is false in general.

Maximal filters are sometimes called [[ultrafilter]]s, but this terminology is often reserved for Boolean algebras, where a maximal filter (ideal) is a filter (ideal) that contains exactly one of the elements {''a'', ¬''a''}, for each element ''a'' of the Boolean algebra. In Boolean algebras, the terms ''prime ideal'' and ''maximal ideal'' coincide, as do the terms ''prime filter'' and ''maximal filter''.

There is another interesting notion of maximality of ideals: Consider an ideal {{mvar|I}} and a filter ''F'' such that {{mvar|I}} is [[Disjoint sets|disjoint]] from ''F''. We are interested in an ideal ''M'' that is maximal among all ideals that contain {{mvar|I}} and are disjoint from ''F''. In the case of distributive lattices such an ''M'' is always a prime ideal. A proof of this statement follows.

{{math proof|1=Assume the ideal ''M'' is maximal with respect to disjointness from the filter ''F''. Suppose for a contradiction that ''M'' is not prime, i.e. there exists a pair of elements ''a'' and ''b'' such that {{math|''a'' &and; ''b''}} in ''M'' but neither ''a'' nor ''b'' are in ''M''. Consider the case that for all ''m'' in ''M'', {{math|''m'' &or; ''a''}} is not in ''F''. One can construct an ideal ''N'' by taking the downward closure of the set of all binary joins of this form, i.e. {{math|''N'' {{=}} {{mset| ''x'' | ''x'' ≤ ''m'' &or; ''a'' for some ''m'' &isin; ''M''}}}}. It is readily checked that ''N'' is indeed an ideal disjoint from ''F'' which is strictly greater than ''M''. But this contradicts the maximality of ''M'' and thus the assumption that ''M'' is not prime.

For the other case, assume that there is some ''m'' in ''M'' with {{math|''m'' &or; ''a''}} in ''F''. Now if any element ''n'' in ''M'' is such that {{math|''n'' &or; ''b''}} is in ''F'', one finds that {{math|(''m'' &or; ''n'') &or; ''b''}} and {{math|(''m'' &or; ''n'') &or; ''a''}} are both in ''F''. But then their meet is in ''F'' and, by distributivity, {{math|(''m'' &or; ''n'') &or; (''a'' &and; ''b'')}} is in ''F'' too. On the other hand, this finite join of elements of ''M'' is clearly in ''M'', such that the assumed existence of ''n'' contradicts the disjointness of the two sets. Hence all elements ''n'' of ''M'' have a join with ''b'' that is not in ''F''. Consequently one can apply the above construction with ''b'' in place of ''a'' to obtain an ideal that is strictly greater than ''M'' while being disjoint from ''F''. This finishes the proof.}}

However, in general it is not clear whether there exists any ideal ''M'' that is maximal in this sense. Yet, if we assume the [[axiom of choice]] in our set theory, then the existence of ''M'' for every disjoint filter–ideal-pair can be shown. In the special case that the considered order is a [[Boolean algebra (structure)|Boolean algebra]], this theorem is called the [[Boolean prime ideal theorem]]. It is strictly weaker than the axiom of choice and it turns out that nothing more is needed for many order-theoretic applications of ideals.