Lang's elementary divisors theorem is wrong?

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?

• I took a look at Lang's proof of this statement, and it's puzzling to me, to say the least. I can follow everything up to the point that the decomposition exists, but then he justifies the uniqueness of the decomposition by appealing to a localization argument, where he claims that the result follows from well-known uniqueness theorems for principal ideal rings. But... as your post point out, the situation seems anything but clear even over $\mathbb{Z}$! And it seems to me that localizing could only recover information up to isomorphism, anyway. Commented Jul 14 at 7:15
• It seems to me that Lang's statement is incomplete with respect to his proof. Lang uses A SPECIFIC CONSTRUCTION, which (I also did not check the localization argument yet) if you follow it, you will find the "maximal decomposition", in the sense that you cannot replace any ideal by a bigger one. In your example of $\Bbb Z^2$, Lang's "algorithm" yields $(1) \oplus (1)$, but it is true that if one forgets about the algorithm and just looks at the statement, $(1) \oplus (2)$ should also work. Commented Jul 14 at 7:19
• By Steinitz's isomorphism theorem, we have for any nonzero ideals $I, J$ that $I \oplus J \cong A \oplus IJ$, so we can always reduce to a sum of copies of $A$ and then a product of ideals. IIRC, you can get a unique decomposition this way, in the sense that the rank is unique and then the two ideals you get at the end are isomorphic (and in fact belong to the same class in the ideal class group, I think?). Otherwise, I have no idea what Lang meant here, and the statement of the proposition seems wrong to me. (Incidentally, I usually see "elementary divisors" in the context of... Commented Jul 14 at 7:20
• ...the torsion part of a f.g. module over a PID, where you do get a uniqueness division statement. But here, things are torsion free, so I'm not really sure what's going on. Caveat: it's late for me, and I'm rusty besides, so maybe I'm saying something stupid above!) Commented Jul 14 at 7:22
• The uniqueness statement is for the choice (up to an inverse) of a sequence $a_0 | a_1 | \dotsc | a_r$. Commented Jul 14 at 12:25

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

The uniqueness follows by localizing at primes $$\mathfrak{p}$$, and invoking the corresponding uniqueness over principal rings, which is part of standard algebra.

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.