# If $R$ is a Dedekind domain and $I, J \triangleleft R$, then $R^d \oplus I \simeq R^{d'} \oplus J$ implies $d = d'$ and $I \simeq J$

I'm trying to figure out the following observation I have written on my notes on Dedekind domains,

If $R$ is a Dedekind domain, $I, J$ two integral ideals and $d, d' \in \mathbb{N}$, such that $R^d \oplus I \simeq R^{d'} \oplus J$, then we have that $d = d'$ and $I \simeq J$.

It seems quite simple, but given my lack of familiarity with abstract algebra, I haven't made much progress. I've tried composing the given isomorphism $\phi: R^d \oplus I \longrightarrow R^{d'} \oplus J$ with the projections to both coordinates to get each summand isomorphic to $R^d \oplus I$ divided by the preimages of $R^{d'} \oplus 0$ and $0 \oplus J$ respectively, but it is not obvious to me that (if) we indeed have $\phi(R^d \oplus 0) = R^{d'} \oplus 0$ and $\phi(0 \oplus I) = 0 \oplus J$.

Would you mind providing any hints? Thanks in advance.

• First, $d=d'$ since $rank_R(R^d\oplus I)=d+1$ since ideals have rank 1, and two isomorphic finitely generated projective modules have the same rank. – xarles Jul 24 '18 at 15:09
• Second, you can try to do some "operation" $\lambda$ with $R$-modules such that $\lambda(R^d\oplus I)\cong I$.... – xarles Jul 24 '18 at 15:19
• @xarles would you mind providing an answer? I've thought about this a good deal of time and I've come up with nothing. I can easily see how to prove this for PIDs, but we had not defined rank for other rings other than PIDs. Neither have I come up with an operation that 'isolates' the ideals out of the sum, either I'm not seeing it or maybe I have not encountered such machinery in class. In either case, I'd appreciate if you could clarify these steps. Thanks in advance. – Guido A. Jul 25 '18 at 10:59

## 1 Answer

To understand the answer you need to know some properties of Dedekind domains. First of all, finitely generated torsion free modules over a Dedekind domain has a well defined rank, which can be either defined as the rank of the base change to the quotient field, or the maximum number of linearly independent elements (but not the minimum number of generators). Ideals, and more generally, fractional ideals, have rank 1 (but need more than 1 generator if they are not principal): in fact, fractional ideals are the (torsion free) rank 1 $R$-modules.

Two isomorphic finitely generated torsion free modules (=projective) have the same rank, but the reverse is not the true. For example, ideals are isomorphic (as $R$-modules) if and only if the represent the same class in the class group.

Now, for any finitely generated torsion free $R$-module $M$ of rank $d$, its wedge product $\lambda(M):=\Lambda^dM$ (this notation is not standard, I am not sure if there is a standard notation) is a torsion free $R$-module of rank 1, so a fractional ideal. If $M=R^{d-1}\oplus I$, then $\lambda(M)\cong I$, and hence $M=R^{d-1}\oplus I\cong R^{d-1}\oplus I'=M'$, then $\lambda(M)\cong \lambda(M')$, so $I\cong I'$.

In fact, something more is true: if $M$ and $M'$ are torsion free finitely generated $R$ modules, they are isomorphic if and only if $rank_R(M)=rank_R(M')$ and $\lambda(M)\cong \lambda(M')$, or, equivalently, they represent the same element in the class group of $K$.