Let $R=\mathbb{Z}[\sqrt{5}]$ and consider $I=(2,\, 1+\sqrt{5})$ as an $R$-module. I'm struggling to prove that the element $2\wedge (1+\sqrt{5})$ of the exterior power $\Lambda^2(I)$ is non-zero. I should construct some alternating bilinear map from $I\times I$ in $\mathbb{Z}$ (regarded in some way as an $R$-module?) not sending $(2,\, 1+\sqrt{5})$ in $0$: I'm thinking of something similar to the determinant since each element $u\in I$ can be expressed as $2a+b(1+\sqrt{5})$, but it seems it doesn't work so well... Any idea about how to do this?
See also Remark 4.7 here: https://kconrad.math.uconn.edu/blurbs/linmultialg/extmod.pdf
We have an exact sequence of $R$-modules $$K\hookrightarrow R^2\xrightarrow{[2,1+\sqrt 5]}I\to\{0\}$$ where $$K=\{(1+\sqrt 5)b-2d,(1+\sqrt 5)d-2(b+d)):b,d\in\Bbb Z\}\subseteq I^2$$ By this, we get a commutative diagram of $R$-module homomorphisms with exact rows $\require{AMScd}$: \begin{CD} K\otimes_R\Lambda_R^1(R^2)@>>>\Lambda_R^2(R^2)@>>>\Lambda_R^2(I)@>>>\{0\}\\ @V\sim VV@V\sim VV@|\\ K\otimes_RR^2@>>\varphi>R@>>\psi>\Lambda_R^2(I)@>>>\{0\} \end{CD} where $\varphi((h,k)\otimes(u,v))=hv-ku$ and $\psi(1)=2\wedge(1+\sqrt 5)$. If $\pi:R\to R/I$ denote the canonical projection, then $\pi\circ\varphi=0$, hence $\pi$ factors through $\psi$ giving rise to a non-zero $R$-module homomorphism $\Lambda_R^2(I)\to R/I$.