The problem comes from Serge Lang's Algebraic Number Theory:
Proposition 27 Elementary divisors theorem. Let $M$ be a non-zero finitely generated projective module over a Dedekind ring $A$. Then there exists ideals $a_i, i= 1, \dots, r$ such that
$$
M \cong \oplus_{i=1}^{r} a_i
$$
These ideals can be so chosen that $a_i | a_{i+1}$ for all i, and are then uniquely
determined.
This theorem seems wrong. A counterexample is $A = \mathbb{Z}$, the integer ring. Then $(2) \oplus (3) \cong (5) \oplus(7)$, since every non-zero ideal is isomorphic to $\mathbb{Z}$ as modules. Is the theorem wrong?
Best Answer
You are correct, the existence holds, but the uniqueness statement is false in several regards.
First of all, it is true that $\Bbb Z^2$ as a $\Bbb Z$-module can be expressed as, for instance $(1) \oplus (1)$ as well as $(1) \oplus (2)$. More generally, you could observe that the statement contradicts its analogue for PID, which says that finitely generated projective modules over a PID $A$ are free, i.e. isomorphic to direct sums of the unit ideal: take any non-unit ideal of $A$ as a counterexample for the uniqueness of Lang's statement.
Seeing this argument, you might think that perhaps there is a way to salvage uniqueness, by imposing that the ideals $\mathfrak{a}$ are taken to be as big as possible: any principal ideal of $A$ is isomorphic (as an $A$-module) to the unit ideal and is contained inside of it ; if we replace principal ideals by the unit ideals in our argument for PID's, we do find uniqueness.
Unfortunately, the answer is still no, and here is a counterexample. Consider $A = \Bbb Z[\sqrt{-5}]$, which is known to be a non-principal Dedekind domain. Let $M=\mathfrak{p} := (2,1+\sqrt{-5})$, let $N = \mathfrak{q} := (3,1-\sqrt{-5})$. These are distinct maximal ideals of $A$, so neither is contained in the other. However, the map $$m \mapsto \frac{m \times 3}{1+\sqrt{-5}}$$ defines an isomorphism of $A$-modules from $\mathfrak{p}$ to $\mathfrak{q}$ (it suffices to check the images of generators).
You can stop here if you just wanted a definitive counterexample. I will now carry on to show where Lang's proof fails.
Lang shows existence as follows. Pick any nonzero $\lambda \in \operatorname{Hom}_A(M,A)$, which exists (pick a nonzero linear map $M \otimes_A K \rightarrow K$, where $K= \operatorname{Frac}(A)$, and multiply by the common denominator of the images of the generators of $M$). Denote $\mathfrak{a}_1 := \lambda(M)$, it is a finitely generated torsion-free $A$-module, hence projective (Lang Proposition 26), so the following short exact sequence splits: $$0 \rightarrow M_1 := \ker \lambda \rightarrow M \rightarrow \mathfrak{a}_1 \rightarrow 0.$$ Hence we can write $M \simeq \mathfrak{a}_1 \oplus M_1$. By tensoring with $K$, one sees that the rank of $M_1$ is less than the rank of $M$, so one would be tempted to carry on by induction. But doing so, we would have no way of imposing the division relation between the ideals. Instead, if we first pick $\lambda$ as above so that the image is as large as possible (I'm avoiding the term "maximal" because the image need not be a maximal ideal), then we are guaranteed that the remaining ideals are contained in $\mathfrak{a}_1$. To see this, assume that $M_1 \simeq \mathfrak{a}_2 \oplus M_2$ with $\mathfrak{a}_2$ not contained in $\mathfrak{a}_1$, then $M \simeq \mathfrak{a}_1 \oplus \mathfrak{a}_2 \oplus M_2$, from which we find a map $\lambda':M \rightarrow A$ whose image contains both $\mathfrak{a}_1$ and $\mathfrak{a}_2$, contradicting the maximality of $\lambda$. This concludes the existence.
For uniqueness, Lang simply says
However, this does not have the desired effect. If we write $$M = \bigoplus_{s=1}^t \mathfrak{p_1}^{e_{1,s}}...\mathfrak{p_m}^{e_{m,s}}$$ where each $(e_{j,s})_s$ is a nondecreasing sequence of nonnegative integers (this is a reformulation of the existence), localizing at (the complement of) $\mathfrak{p}$ gives: $$M_{(\mathfrak{p})} = \bigoplus_{s=1}^t A_{(\mathfrak{p})}\mathfrak{p_1}^{e_{1,s}}...\mathfrak{p_m}^{e_{m,s}}.$$ Now each term of the sum is an ideal of $A_{(\mathfrak{p})}$ (the unit ideal if $\mathfrak{p}$ is not one of the factors), hence principal, hence isomorphic to a free module of rank 1, and we have effectively lost all information about which ideals were there, and to which power. Only the rank of $M$ remains.
In the proof of existence, it might happen that there exists two choices for a maximal $\lambda$ which map to distinct, incomparable ideals $\alpha$, in which case uniqueness would fail. This is how I found the counterexample above.