Proof: Since there is a unique representation for r∈R and m∈M we may define the action of R on M from their mutual bases' actions to obtain rm=(z1σ+z2)(a,b)=(z1a+z2a,2z1a−z1b+z2b). Note that σ(σ(a,b))=(a,b) so that it satisfies action associativity.
Observe(a,b)∧(c,d)=((a,0)+(0,b))∧((c,0)+(0,d))=(ad+bc)(e1∧e2)so that ⋀2M is generated by e1∧e2 as an R-module. As well, we have σ(e1∧e2)=(σe1)∧e2=(e1+2e2)∧e2=e1∧e2 and σ(e1∧e2)=e1∧(σe2)=e1∧−e2=−(e1∧e2) so that (z1σ+z2)(e1∧e2)=(¯z1+z2)(e1∧e2) where the overline denotes reduction modulo 2, implying ⋀2M is either of order 1 or 2. To prove the latter, we shall construct a bilinear alternating form φ on M such that φ(e1,e2)≠0.
First, define ψ:R→Z/2Z by ψ(z1σ+z2)=¯z1+z2 and prove it is a ring homomorphismψ(z1σ+z2+z′1σ+z′2)=¯z1+z2+z′1+z′2=ψ(z1σ+z2)+ψ(z′1σ+z′2)ψ((z1σ+z2)(z′1σ+z′2))=ψ((z1z′2+z2z′1)σ+(z1z′1+z2z′2))=¯z1z′2+z2z′1+z1z′1+z2z′2=¯(z1+z2)(z′1+z′2)=ψ(z1σ+z2)ψ(z′1σ+z′2)Now define φ:M×M→Z/2Z by φ((a,b),(c,d))=ψ(ad−bc). It alternates and is 1 on (e1,e2) so all that remains to show is that it is bilinear by being additive in its componentsφ((a,b)+(a′,b′),(c,d))=ψ((a+a′)d−(b+b′)c))=ψ(ad−bc)+ψ(a′d−b′c)=φ((a,b),(c,d))+φ((a′,b′),(c,d))and linear over Rφ((z1σ+z2)(a,b),(c,d))=φ((z1a+z2a,2z1a+z2b−z1b),(c,d))=¯z1ad+z1bc+z2ad−z2bc=¯z1ad−z1bc+z2ad−z2bc=¯z1(ad−bc)+z2(ad−bc)=(z1σ+z2)φ((a,b),(c,d)) ◻
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