Editing Modular lattice (section)

== Properties ==
Every [[distributive lattice]] is modular.<ref>{{Cite book |chapter-url=https://link.springer.com/book/10.1007%2Fb139095 |title=Lattices and Ordered Algebraic Structures |last=Blyth |first=T. S. |publisher=Springer |year=2005 |isbn=978-1-85233-905-0 |series=Universitext |location=London |page=65 |chapter=Modular lattices |doi=10.1007/1-84628-127-X_4}}</ref><ref>In a distributive lattice, the following holds: <math>(a\wedge b)\vee (x\wedge b) = ((a\wedge b)\vee x)\wedge((a\wedge b)\vee b)</math>. Moreover, the absorption law, <math>(a\wedge b)\vee b = b</math>, is true for any lattice. Substituting this for the second conjunct of the right-hand side of the former equation yields the Modular Identity.</ref>

{{harvtxt|Dilworth|1954}} proved that, in every finite modular lattice, the number of join-irreducible elements equals the number of meet-irreducible elements. More generally, for every {{mvar|k}}, the number of elements of the lattice that cover exactly {{mvar|k}} other elements equals the number that are covered by exactly {{mvar|k}} other elements.<ref>{{citation
 | last = Dilworth | first = R. P. | authorlink = Robert P. Dilworth
 | doi = 10.2307/1969639
 | journal = [[Annals of Mathematics]]
 | mr = 0063348
 | pages = 359–364
 | series = Second Series
 | title = Proof of a conjecture on finite modular lattices
 | volume = 60
 | issue = 2 | year = 1954| jstor = 1969639 }}. Reprinted in {{citation
 | editor1-last = Bogart | editor1-first = Kenneth P.
 | editor2-last = Freese | editor2-first = Ralph
 | editor3-last = Kung | editor3-first = Joseph P. S.
 | doi = 10.1007/978-1-4899-3558-8_21
 | location = Boston
 | pages = 219–224
 | publisher = Birkhäuser
 | series = Contemporary Mathematicians
 | title = The Dilworth Theorems: Selected Papers of Robert P. Dilworth
 | year = 1990| chapter = Proof of a Conjecture on Finite Modular Lattices
 | isbn = 978-1-4899-3560-1
 }}</ref>

A useful property to show that a lattice is not modular is as follows:

: A lattice {{Mvar|G}} is modular if and only if, for any {{Math|''a'', ''b'', ''c'' &isin; ''G''}},
::<math>\Big((c\leq a)\text{ and }(a\wedge b=c\wedge b)\text{ and }(a\vee b=c\vee b)\Big)\Rightarrow(a=c)</math>

Sketch of proof: Let G be modular, and let the premise of the implication hold. Then using absorption and modular identity:

: ''c'' = (''c''∧''b'') ∨ ''c'' = (''a''∧''b'') ∨ ''c'' = ''a'' ∧ (''b''∨''c'') = ''a'' ∧ (''b''∨''a'') = ''a''

For the other direction, let the implication of the theorem hold in G. Let ''a'',''b'',''c'' be any elements in G, such that ''c'' ≤ ''a''. Let ''x'' = (''a''∧''b'') ∨ ''c'', ''y'' = ''a'' ∧ (''b''∨''c''). From the modular inequality immediately follows that ''x'' ≤ ''y''. If we show that ''x''∧''b'' = ''y''∧''b'', ''x''∨''b'' = ''y''∨''b'', then using the assumption ''x'' = ''y'' must hold. The rest of the proof is routine manipulation with infima, suprema and inequalities.{{Citation needed|date=September 2018}}