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  "body": "\n## Statement\n\nLet $L$ be a [[lattice|WH4SFAACAS]].\nLet $N_5$ be the [[pentagon lattice|DHC8TZXGYS]], with elements $0 \u003c a \u003c c \u003c 1$ and an element $b$ that is incomparable to $a$ and $c$, with $b \\wedge c = 0$ and $a \\vee b = 1$.\nThen $L$ is a [[modular lattice|QX7WZ92QAK]] if and only if $L$ has no [[sublattice|PV90EW00XK]] isomorphic to $N_5$.\n\n## Remarks\n\nOne direction is easy: $N_5$ is not modular, and the modular law holds in every sublattice of a modular lattice.\nFor the other direction, suppose $a \\le b$ and $x$ violate the law, and put $u = a \\vee (x \\wedge b)$ and $v = (a \\vee x) \\wedge b$, so $u \u003c v$.\nA direct computation gives $x \\wedge u = x \\wedge v = x \\wedge b$ and $x \\vee u = x \\vee v = a \\vee x$.\nSo the five elements $x \\wedge b$, $u$, $v$, $x$ and $a \\vee x$ form a sublattice isomorphic to $N_5$.\n\nExample: the lattice of subspaces of a vector space is modular, so it contains no copy of $N_5$.\nTogether with the [[M3–N5 characterization of distributivity|C6GH4VHAMQ]], it shows that the pentagon and the diamond are the only obstructions to distributivity.\n\n## Explanations\n\n### L1 human\n\nThe modular law is an equation that must hold in all of a lattice.\nThe theorem says that it fails only when a five-element picture, the pentagon, appears inside the lattice.\nSo to test modularity, one looks for a pentagon.\n\n### L2 human\n\nThe theorem turns an equational condition into a forbidden-substructure condition.\nIt is used to show that a given lattice is not modular by exhibiting a pentagon, and to prove modularity of lattices built from modular pieces.\nThe same style of result, with the diamond $M_3$ added, characterizes [[distributive lattices|0HM0HQCK6N]].\nThe proof is short and gives the pentagon explicitly from any failure of the modular law.\n",
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