Ternary parallelogram scales are MOS substitution: Difference between revisions
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Assume without loss of generality that | Assume without loss of generality that | ||
'''u'''<sub>'''x'''</sub> = (''t'', 0), ''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 ==== | ==== 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}.}} | 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}.}} | ||
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Claim: ''k''<sub>'''w'''</sub> has order ''n'' in {{nowrap|ℤ/''mn''ℤ.}} | Claim: ''k''<sub>'''w'''</sub> has order ''n'' in {{nowrap|ℤ/''mn''ℤ.}} | ||
Proof: The order cannot be less than ''n'', lest we have {{nowrap|φ((0, ''uk''<sub>'''w'''</sub>)) {{=}} 0}} for some {{nowrap|0 < ''u'' < ''n'',}} contradicting injectivity of φ within a fundamental domain (following from previous lemma). | Proof: The order cannot be less than ''n'', lest we have {{nowrap|φ((0, ''uk''<sub>'''w'''</sub>)) {{=}} 0}} for some {{nowrap|0 < ''u'' < ''n'',}} contradicting injectivity of φ within a fundamental domain (following from previous lemma). If the order is ''N'' > ''n'', then by Lemma 2, {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 1)''a''}}, where ''a'' = ψ(''k''<sub>'''w'''</sub>), are not disjoint, contradicting injectivity. | ||
==== Template word is MOS ==== | ==== Template word is MOS ==== | ||