Ternary parallelogram scales are MOS substitution: Difference between revisions

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 Assume without loss of generality that
-'''u'''<sub>'''x'''</sub> = (''t'', 0) (parallel to '''v''').
+'''u'''<sub>'''x'''</sub> = (''t'', 0), ''t'' > 0 (parallel to '''v''').
 ==== The two non-axial step vectors differ by (0, ''n'') if the axial step is parallel to '''v''' and by (''m'', 0) otherwise ====
+Under the above assumption, since {{nowrap|φ(''t'''''v''') {{=}} 1}} we have that {{nowrap|φ('''v''') {{=}} ''k''<sub>'''v'''</sub>}} is a cyclic generator of the group {{nowrap|ℤ/''mn''ℤ,}} such that {{nowrap|''tk''<sub>'''v'''</sub> {{=}} 1 mod ''mn''.}} Multiplication by ''t'' is a group automorphism ψ of {{nowrap|ℤ/''mn''ℤ}} such that ψ(''k''<sub>'''v'''</sub>) = 1, so the first row {{nowrap|[0 : ''m''] × {0}}} is mapped as {{nowrap|ψφ((''i'', 0)) {{=}} ''i''}}; so the image of the first row under ψφ is {{nowrap|{0, ..., ''m'' - 1}.}}
 ==== Template word is MOS ====
 ==== Filling word is MOS ====
 [[Category:Pages with proofs]]