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 $rm = (z_1σ+z_2)(a,b) = (z_1a+z_2a,2z_1a-z_1b+z_2b)$. Note that $σ(σ(a,b)) = (a,b)$ so that it satisfies action associativity.
Observe$$(a,b) \wedge (c,d) = ( (a,0)+(0,b) ) \wedge ( (c,0) + (0,d) ) = (ad+bc)(e_1 \wedge e_2)$$so that $\bigwedge^2 M$ is generated by $e_1 \wedge e_2$ as an $R$-module. As well, we have $σ(e_1 \wedge e_2) = (σe_1) \wedge e_2 = (e_1+2e_2) \wedge e_2 = e_1 \wedge e_2$ and $σ(e_1 \wedge e_2) = e_1 \wedge (σe_2) = e_1 \wedge -e_2 = -(e_1 \wedge e_2)$ so that $(z_1σ+z_2)(e_1 \wedge e_2) = (\overline{z_1+z_2})(e_1 \wedge e_2)$ where the overline denotes reduction modulo 2, implying $\bigwedge^2 M$ is either of order 1 or 2. To prove the latter, we shall construct a bilinear alternating form $φ$ on $M$ such that $φ(e_1,e_2) \neq 0$.
First, define $ψ : R → \mathbb{Z}/2\mathbb{Z}$ by $ψ(z_1σ+z_2)=\overline{z_1+z_2}$ and prove it is a ring homomorphism$$ψ(z_1σ+z_2+z_1'σ+z_2')=\overline{z_1+z_2+z_1'+z_2'}=ψ(z_1σ+z_2)+ψ(z_1'σ+z_2')$$$$ψ((z_1σ+z_2)(z_1'σ+z_2'))=ψ((z_1z_2'+z_2z_1')σ+(z_1z_1'+z_2z_2'))=$$$$\overline{z_1z_2'+z_2z_1'+z_1z_1'+z_2z_2'}=\overline{(z_1+z_2)(z_1'+z_2')}=ψ(z_1σ+z_2)ψ(z_1'σ+z_2')$$Now define $φ : M × M → \mathbb{Z}/2\mathbb{Z}$ by $φ((a,b),(c,d))=ψ(ad-bc)$. It alternates and is 1 on $(e_1,e_2)$ so all that remains to show is that it is bilinear by being additive in its components$$φ((a,b)+(a',b'),(c,d))=ψ((a+a')d-(b+b')c))=$$$$ψ(ad-bc)+ψ(a'd-b'c)=φ((a,b),(c,d))+φ((a',b'),(c,d))$$and linear over $R$$$φ((z_1σ+z_2)(a,b),(c,d))=φ((z_1a+z_2a,2z_1a+z_2b-z_1b),(c,d))=$$$$\overline{z_1ad+z_1bc+z_2ad-z_2bc}=\overline{z_1ad-z_1bc+z_2ad-z_2bc}=$$$$\overline{z_1(ad-bc)+z_2(ad-bc)}=(z_1σ+z_2)φ((a,b),(c,d))~\square$$
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