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
from Hot Weekly Questions - Mathematics Stack Exchange
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