… as promised in my Quora answer here. Also, here’s a PDF version of the post: Classification of finitely generated modules over Dedekind domains, with and without projective modules, and reconciliation of approaches.

In this post (specifically in Section 4) we give a proof of the classification of finitely generated (f.g.) modules over a Dedekind domain avoiding the notion of projective modules. (Instead we rely on pure submodules, which give a slightly more transparent/natural/direct approach to the classification problem at hand, at the expense of the greater generality afforded by the projective module approach. Along the way we also explain why these approaches are really not so different after all.)

We gradually build up from the more familiar special cases when is a principal ideal domain (PID) or even just a field. Along the way we discuss some relevant abstractions, such as the splitting lemma (Section 2).

I would like to thank Profs. Yifeng Liu and Bjorn Poonen for teaching me in 18.705 and 18.785, respectively, this past (Fall 2014) semester at MIT. Thanks also to Alison Miller for catching a typo in Exercise 3.

**1. Preliminaries **

First we recall the simpler case when is a PID. In fact, we start with the simplest case of fields:

Exercise 1 (Structure theorem for f.g. modules over a field (i.e. vector spaces), “right splitting” approach via free modules)Let be the field, the module (i.e. -vector space).Let be a -submodule (i.e. subspace) of , and suppose is free over . Lift a basis of (arbitrarily) to prove that .

(The key point is that if some -linear combination , i.e. , then by linear independence in we must have .)

Conclude by induction that the f.g. modules over are precisely the free modules over , i.e. those isomorphic to for some .

Remark 1It’s a separate (but obviously important) issue to show that the -dimension is well-defined, i.e. if and only if . For (PIDs and) Dedekind domains there is a similar notion of rank in the torsion-free case, but we won’t focus on these sorts of issues throughout the post.

Exercise 2 (Structure theorem for f.g. modules over a PID, algorithmic generators and relations approach)Let be the PID, the module.

- Take a finite set of generators of , i.e. a surjection , and look at the kernel (so ), which describes the “relations” among the set of generators. Show that is f.g. over . (In other words, is finitely presented.)
- Take a finite presentation (here ), which corresponds to an (possibly “overdetermined”) “relations matrix” . Mimicking Gaussian elimination (but using only -row/column operations), show that can be “diagonalized”, i.e. put in a Smith normal form. This corresponds to a suitable -linear change of generators of , combined with an alternative -linear change of the “description” of the relations among the generators. (This shouldn’t be confused with diagonalization of linear operators. It’s really more or less just solving a system of -linear equations.)

Exercise 3 (F.g. torsion-free modules over PIDs are free, “right splitting” approach via free modules)Let be the PID, the torsion-free f.g. module.Mimicking Exercise 1, prove that is a free -module. (Hint: Take a “saturated” submodule of , in the sense that if for some then , or equivalently, that is torsion-free. Now show that .)

Exercise 4 (Torsion part (of f.g. module over PID) breaks off, “right splitting” approach via free modules)Let be the PID, the f.g. module, its torsion submodule.Using and then mimicking Exercise 3, prove that .

Remark 2Exercise 2 is the first proof on Wikipedia, as well as the proof in Artin’s “Algebra”.Another approach, with key steps outlined in Exercises 3 and 4, (which doesn’t explicitly reduce or otherwise work with generators/relations) is to (less explicitly) find a special -submodule of and try to prove that is a direct summand of . (The whole point is that this is generally sutble when isn’t a vector space—this is one of the main problems of representation theory.) For instance, see the second proof on Wikipedia, or Emerton’s answer at the discussion on MathOverflow. (These are effectively done using the fact that free modules are projective, which is also the standard approach to the general Dedekind domain problem.)

Nonetheless, these two approaches have the same overall flavor: intuitively, the fully “saturated” submodules are the ones that break off (as direct summands), whether we identify them algorithmically (as in the first approach) or through some characterization (as in the second approach).

**2. Starting from scratch: (canonical) direct sum decomposition **

In general, say we have an -submodule (or slightly more generally, an injection , though importantly, this injection may not be unique), and we wish to prove is a direct summand of , i.e. for some .

Slightly more generally, let’s look at (all the terms of) the whole short exact sequence (sometimes it’s better to think in terms of , so that , and sometimes it’s better to think in terms of , so that ). What does it take to get a “canonical” splitting? (We roughly follow the terminology in the linked Wikipedia article.)

** 2.1. (“Canonical”) left splitting **

Viewing as a submodule of , it suffices to get a “compatible -projection” , i.e. a (surjective) map such that , where is the inclusion map. (Then will be a projection in the usual linear algebra sense.)

For instance, when is a group algebra, one can prove Maschke’s theorem by constructing a -projection (specifically, by “averaging” a -projection into a -projection).

( is called an injective module if and only if we have such a “left splitting” whenever is a module containing .)

** 2.2. (“Canonical”) right splitting **

Let denote the surjection . It suffices to get a “compatible -inclusion” , i.e. an (injective) map such that .

( is called a projective module if and only if we have such a “right splitting” whenever is a module with a surjection onto .

For instance, when is a Dedekind domain, see Exercise 6.)

** 2.3. Briefly, the relation between splittings and the direct sum decomposition **

In the case (for definiteness), the and are related as follows: for any .

Exercise 5 (based on Exercise 3.38 from Pete Clark’s aforementioned commutative algebra notes)Let be a ring. Prove that [every -module is injective] if and only if [every -module is projective].For example, prove that if for a field and finite group , then both statements hold for the group algebra.

**3. Sketch of (classical?) projective module approach **

We now return to the general case of Dedekind domains . Let be the fraction field of .

Exercise 6 (F.g. torsion-free modules over Dedekind domains are projective, “right splitting” approach via projective modules)Let be the Dedekind domain, the torsion-free module.For a detailed treatment of projective modules, we defer to Pete Clark’s notes. (Incidentally, he also started an interesting MathOverflow discussion on the difficulties of teaching free, projective, and flat modules.)

- The
rankof is defined as the (clearly finite) dimension of , or equivalently the localization , as a vector space over . (In other words, we’re embedding the -module inside a vector space in the only reasonable way.)- (Key step.) Prove that rank- modules are just (up to -module isomorphism) the fractional ideals of , and show that these are projective.
- Prove that direct sums of projective modules are projective, and conclude by induction that the f.g. torsion-free modules over are (projective and) precisely (up to isomorphism) the finite direct sums of fractional ideals of .
- Give a direct proof (not using that direct sums of projective modules are projective) that f.g. torsion-free modules over are projective, by mimicking the proof for rank- modules. Of course, by induction, this further induces an alternative proof of the classification of f.g. torsion-free modules.

Remark 3Exercise 6 is the heart of the (classical?—I believe it’s Steinitz’ original proof) projective module approach to breaking off submodules (as direct summands). The use of localization (or similar Bezout’s identity arguments) seems pretty fundamental in general; we use it as well to verify the the pure submodule criterion below (in the alternative approach: see the key Theorem 3). Exercise 7 further clarifies the similarity. Here are some references:

- hilbertthm90‘s sequence of three blog posts, which I believe roughly follow May’s notes on Dedekind domains.
- Chapter III, section 22, page 144 of “Representation theory of finite groups and associative algebras” by C. Curtis and I. Reiner.
- Pete Clark’s commutative algebra notes, Chapter 20.6. Finitely generated modules over a Dedekind domain, page 290.
- The first chapter of Henri Cohen’s “Advanced Topics in Computational Number Theory” gives an interesting proof of the projectivity: , so is a direct summand of a free module, hence projective.
- This was the approach given in 18.785 (MIT Fall 2014, taught by Prof. Bjorn Poonen) Problem Set 4.

Remark 4It seems difficult to apply the generators and relations approach (from Exercise 2) to the Dedekind domain problem. For instance, it doesn’t seem easy to describe an analogous Smith normal form—in the final decomposition into (torsion) and (torsion-free fractional ideals), the fractional ideals are not principal (though they are generated by at most elements). That said, since , all but one component can be taken principal (in general), so maybe there’s some hope. Also maybe “Hermite and Smith normal form algorithms over Dedekind domains” by Henri Cohen is relevant.

**4. Alternative pure submodule approach **

** 4.1. Pure submodule criterion, in the spirit of generators and relations **

Let be a ring, a submodule. The usefulness of the following definition will soon become clear. Overall the idea is to refine the “right splitting approach” from Exercises 3 and 4 in the direction of the algorithmic generators-and-relations ideas of Exercise 2, rather than the projective module direction from Exercise 6.

Definition 1We define a submodule (of a fixed -module ) to bepureif the following condition holds: if we choose finitely many , and we can write the as -linear combinations of some , then the can in fact be written as corresponding -linear combinationswith the taken inside. In other words, we can find such that for all .Of course, we can rephrase this condition concisely in terms of matrices: for and , we have if and only if .

Example 1If is a direct summand of , i.e. for some (internal direct sum), then the projection allows us to simply take for all in any such scenario. So is a pure submodule of .

Observation 5 (Equivalent definition of purity)is pure if and only if following condition holds: for any finite choice of and with , there exist with .

Theorem 2 (Pure submodule criterion (for direct summands), 18.705 (MIT Fall 2014, taught by Prof. Yifeng Liu) Midterm, Problem 3.4)Let be a ring, a submodule. Further suppose is finitely presented (e.g. if is noetherian and is f.g.).If is a pure submodule of , then is a direct summand of , i.e.

internallyfor some submodule (in particular, the direct sum here must be canonical). In other words, when is finitely presented, the converse of Example 1 holds.

*Proof:* Everything will be canonical here, i.e. we’ll prove that the short exact sequence splits (in the sense of the splitting lemma from Section 2).

Observation 6If () generate , then a right splitting (induced by) is just a well-defined map sending each to some lift/pre-image (i.e. the must lie in ).It’s easy to see that a choice of works (i.e. the corresponding construction of is consistent) if and only if we have whenever (because the latter is equivalent to ). (And if is well-defined, then the corresponding projection satisfies .)

This closely resembles the condition for purity given in Observation 5, modulo a final technical point—since is finitely presented, the generators can be chosen such that **we only have to worry about finitely many -linear conditions** (since in a finite presentation, the set of -linear conditions, viewed via corresponding “coefficient vectors” as a submodule of , is f.g.). So we’re done.

Remark 7Example 1 and Theorem 2 together show that when is finitely presented, purity is more or less a “relative (to )” version of injective modules.)

Exercise 7 (Connection between projective module and pure submodule approaches)Prove that afinitely presented-module is projective if and only if the following condition holds: for any -surjection , the kernel is a pure submodule of .Given the “equational” nature of Definition 1, is this connection related to the dual basis lemma, or another “equational” characterization of projective modules? (I haven’t thought through this much myself.)

Remark 8If we only assume to be f.g., then as long as we tweak Definition 1 to includeinfinitesystems of equations, the natural analog of Theorem 2 holds, with essentially the same proof. This is probably still useful; for instance, I believe the proof of the Dedekind module classification below would still go through. Cf. Greg Kuperberg’s related comments in the aforementioned MathOverflow discussion.(I believe this is more or less a “relative (to )” version of algebraically compact (or pure-injective) modules.)

Although we won’t need it for the application below, here’s the rest of Problem 3:

Exercise 8 (Based on 18.705 (MIT Fall 2014, taught by Prof. Yifeng Liu) Midterm, Problem 3)In this exercise we donotassume that is finitely presented.

- Show that is pure if and only if the induced -map is injective for every finitely presented -module . (This is the usual first definition.)

Remark 9It may or may not help to use the equational criterion for vanishing of elements of tensor products; I haven’t thought about this too carefully myself.- Prove Example 1 using the tensor product formulation.
- When is a PID (e.g. take ), show that a submodule of is pure if and only if for all nonzero . (In other words, the condition in Definition 1 is “enough” over PIDs!)

Remark 10Hint: Probably the easiest solution is to combine the tensor product formulation with the structure theorem over PIDs. It would be interesting to find a more conceptual way to prove the result without the structure theorem, which might even lead to an alternative proof of the structure theorem.- Does the previous part still work for Dedekind domains ? If not, what’s the best analog (e.g. how many pairs are necessary in Definition 1)? (I haven’t actually worked this out myself, but I imagine there’s a reasonable answer. Feel free to take the structure theorem over Dedekind domains for granted.)

** 4.2. Applying pure submodule criterion to original problem **

The following is the key result.

Theorem 3 (Saturated submodules (of f.g. modules over Dedekind domains) break off as direct summands, “right splitting approach” via pure submodule criterion)Let be a Dedekind domain, a f.g. module over . Let be an -saturated submodule of , i.e. such that if for some nonzero and , then , or equivalently, is torsion-free. (For intuition, think about avoiding -submodules like or . One could also phrase this in terms of suitable tensor products, e.g. by in the torsion-free case, or a “semi-localization” of at a finite set of primes in the torsion case.)Then is a pure submodule of , hence a direct summand of by the pure submodule criterion (Theorem 2)—indeed, is f.g. over the noetherian ring , hence finitely presented.

*Proof:* Suppose for some (finitely many) in we have for some matrix . The key idea is localization: if we temporarily “add denominators to ”, then the system of equations will be easier to solve. Then in the spirit of Bezout’s identity, we’ll take a suitable linear combination to get a solution with denominator .

Let be a prime of , and the corresponding multiplicative subset. Then is a -saturated -submodule of , i.e. (via exactness of localization) is torsion-free. But is a DVR, hence a PID, so by the structure theorem over PIDs (specifically Exercise 3), is a free -module. Consequently (following the reasoning from Exercises 3 and 4), is a direct summand of .

Thus (by Example 1) is a pure -submodule of . But , so by Definition 1 (of purity) we have for some .

It follows that for each prime of , there exists with . Finally, it suffices to find an -linear combination of equal to . But this is easy: the -ideal generated by the is not contained in for any prime ideal (because it contains the element ), so , i.e. there exists a finite -linear combination of equal to .

Remark 11We can avoid localizing at sets containing zerodivisors by instead carefully semi-localizing, i.e. working with of the form (note that ). This actually makes the final paragraph easier; we just need to check is still a PID, which follows from a “pretty strong approximation theorem” (essentially the Chinese remainder theorem (CRT) for Dedekind domains) argument.

Corollary 4 (Torsion part (of f.g. module over Dedekind domain) breaks off, “right splitting approach” via pure submodule criterion)If is f.g. over a Dedekind domain , then is a direct summand of .

*Proof:* is torsion-free, so we may apply Theorem 3.

Remark 12Note that the proof here doesn’t (at least explicitly) use on the classification/characterization of torsion-free modules over . This contrasts with the standard development where one uses Exercise 6 (specifically, that torsion-free modules are projective) to prove the right splitting of , which is in fact also the approach for PIDs given in Exercises 3 and 4.

By the previous corollary, , where is torsion-free, so it remains to (separately) classify decompositions of torsion and torsion-free modules .

Corollary 5 (Decomposition of f.g. torsion-free modules over Dedekind domains, “right splitting approach” via pure submodule criterion)If is f.g. torsion-free over a Dedekind domain , then for (therankof ) fractional ideals , and fractional ideals of are indeed (of rank and hence) indecomposable.

*Proof:* In view of Theorem 3, our strategy is to find -saturated submodules of . We may think of as living inside a f.d. -vector space (equivalently, localize w.r.t. the set ). Then we’ll show that the -coordinates correspond to indecomposable direct summands of , which will be isomorphic to fractional ideals of (which are generally not principal—this is why the Smith normal form approach seems difficult—see Remark 4).

Let in . Saturate the submodule to get (in general it will no longer be principal), where . Then , so for some fractional ideal of , which (since is saturated) breaks off as a direct summand of . The -dimension of the localization/tensored-up -vector space decreases by each time we break a fractional ideal off, so the eventual decomposition is for some fractional ideals , with each indeed indecomposable (as it has corresponding -dimension ).

The hard work is done, but for completeness we include the proof of the torsion classification:

Theorem 6 (Classification of f.g. torsion modules over Dedekind domains)If is torsion, then is an -module (here we’ve used CRT), which induces a decomposition (there are certainly more concrete ways to think about all of this). (One could probably phrase this in terms of primary decompositions, but it doesn’t seem particularly worthwhile.) Each localized is easy to classify, since is a PID.(Actually, without the CRT prime power decomposition, we could’ve directly noted that is a PID (cf. Remark 11)—though the proof happens to use “pretty strong approximation”/CRT (but in a different way).)