I've been trying to wrap my head around the proof that if $M$ is a finitely generated torsion-free $R$-module over a Dedekind domain $R$, then $M\cong R^n\oplus I$, where $I$ is an ideal of $R$. I'm trying to understand the uniqueness part of the proof, that if $M\cong R^n\oplus I\cong R^m\oplus J$, then $n=m$ and $I$ and $J$ are in the same ideal class.
Showing that $n=m$ I understand. According to proposition 6.5 of this source, we can then look at the determinant map to show $I\cong J$. Specifically, it mentions looking at $\wedge^{n+1}M$. Now I know $I$ can be generated by at most two elements. So it seems like $\wedge^{n+1}M\cong I\oplus\wedge^2I$. If $\wedge^2I$ was trivial, then I understand the isomorphism, but I can't prove that is always the case. What exactly is May getting at here? How does $\wedge^{n+1}M$ help show $I\cong J$?
You are using the theorem of exterior algebra that $${\wedge}^k(A\oplus B)\cong\bigoplus_{j=0}^k\left(\wedge^{k-j} A\right) \otimes\left(\wedge^j B\right)\tag1.$$ Take $B=I$. Then $\wedge ^0 I=R$ and $\wedge^1I=I$. But I claim $\wedge^2 I=0$. Let $I=Ru+Rv$. Then $\wedge^2 I=R(u\wedge v)$ is a cyclic $R$-module. Now $$u(u\wedge v)=u\wedge uv=v(u\wedge u)=0$$ and similarly $v(u\wedge v)=0$. So the annihilator of $\wedge^2 I$ contains $I$. But in any ideal class, there is an ideal coprime to any given ideal. So there is an ideal $J$ coprime to $J$ in the same ideal class as $I$. Then $I\cong J$ as modules, and $J$ annihilates $\wedge^2J\cong\wedge^2 I$. Thus $I+J=R$ annihilates $\wedge^2 I$, which therefore vanishes.
Now $(1)$ collapses to $${\wedge}^k(A\oplus I)\cong\left(\wedge^{k} A\right)\oplus \left(\left(\wedge^{k-1} A\right)\otimes I\right)\tag2.$$ Take $A=R^n$ and $k=n+1$.