7
$\begingroup$

It is known that if $R$ is a DVR with fraction field $K,$ then the $R$-submodules of $K$ are $0,K,x^nR,$ with $n$ any integer and $x$ a generator of the maximal ideal of $R.$ I was wondering if there is a simple structure theorem for the $R$-submodules of $K^n,$ the $n$-dimensional vector space over $K$ (similar to that for finitely generated torsion-free modules over a Dedekind domain). I know that the subroups of $\mathbb{Q}$ are somewhat complicated to describe, but was wondering, in particular, if the situation gets any nicer with, say, $\mathbb{Z}_P$-submodules of $\mathbb{Q}^n,$ where $P$ is a nonzero prime ideal of $\mathbb{Z}.$

$\endgroup$
7
$\begingroup$

Indeed, it seems that the situation gets nicer, but certainly not as nice as what I depicted in my first and very flawed answer. (See the remains below and the enlightening counter-example of Wilberd van der Kallen.)

Still, the situation is as tamed as it can be in the case of a complete discrete valuation ring, see YCor's answer and his very concise proof of Joseph Rotman's theorem [1, Theorem 3].

The meaningful keywords are torsion-free modules of finite rank over a discrete valuation ring. Note that if $R$ is any domain and $K$ its fraction field, then any given torsion-free $R$-module $M$ of finite rank $n$ embeds into $K^n = M \otimes_R K$. Conversely, any $R$-submodule of $K^n$ is torsion-free of finite rank.

If $(R, p)$ is a discrete valuation ring (DVR) with maximal ideal $p$, then a torsion-free $R$-module of finite rank $n$ can be represented by an $n \times n$ unipotent matrix with coefficients in $R^{\ast}$, the $p$-adic completion of $R$. The $R$-isomorphism and quasi-isomorphism classes of the torsion-free $R$-modules of finite rank can be described via three different conditions on such matrices [2, Corollary 1.7]. We can also read into a representative matrix whether a module has a direct factor which is divisible or free [2, Corollary 1.8]. This should help build many examples of indecomposable modules such as Wilberd van der Kallen's. Other useful results can be found in [4].

If $M$ and $N$ are two isomorphic torsion-free $R$-modules of finite rank, then such modules have the same rank and the same $p\text{-rank}$, where $p\text{-rank}(M) = \dim_k(M/pM)$ and $k = R/p$. But these two invariants are not complete and most torsion-free $R$-modules of finite rank are not of the form $K^n \oplus R^m$. If $M$ embeds into $N$, $\text{rank}(M) = \text{rank}(N)$ and $p\text{-rank}(M) = p\text{-rank}(N)$, them $M$ is quasi-isomorphic to $N$ [2, Proposition 1.3][4, Lemma 1].($M$ and $N$ are quasi-isomorphic if $M$ embeds into $N$ and $M/N$ is a torsion module bounded by a power of $p$).

The focus of [4] is the class of purely indecomposable modules (pi-modules), i.e., torsion-free indecomposable modules $M$ of finite rank with $p\text{-rank}(M) = 1$, or equivalently indecomposable pure $R$-submodules of $R^{\ast}$ of finite rank. For instance, it is shown in [4, Proposition 1], that the set of isomorphism classes of pi-modules is a partially ordered set with the ascending chain condition, but not the descending one, and $R$ is its smallest element. Theorem 1 of [4] shows that the class of pi-modules is closed under direct summands and exhibit numerical invariants that determine an isomorphism class.


Edit: Here is a record of my first very wrong answer and some related developments, aggregating essentially YCor and tf_'s comments.

Luc's Claim (wrong!). Let $R$ be DVR and let $K$ be its fraction field. An $R$-submodule of $K^n$ is isomorphic to an $R$-module the form $K^m \oplus R^k$ for some non-negative integers $m$ and $k$ such that $m + k \le n$.

But the following is proved in YCor's answer, very concisely and with the exclusive use of Matlis duality.

YCor's Claim (true!). Let $R$ be complete DVR and let $K$ be its fraction field. An $R$-submodule of $K^n$ is isomorphic to an $R$-module the form $K^m \oplus R^k$ for some non-negative integers $m$ and $k$ such that $m + k \le n$.

This result turns out to be known. Indeed, it follows immediately from:

Rotman's Theorem [1, Theorem 3]. A reduced torsion-free module of finite rank over a complete DVR is free.

A module over a principal ideal ring is said to be reduced if its divisible submodule is $\{0\}$.

Rotman's proof is also short but relies on the Kulikov's existence of basic submodules.

Eventually, let us note that YCor's Claim also settles:

Claim 1. Let $R$ be a complete DVR and let $K$ be its fraction field. Then $\text{Ext}_R^1(K, R) = 0$.

Claim 1, when proved by other means, can be used to prove Rotman's theorem by induction on the rank. As mentioned by tj_ in the comment, even more is true.

Claim 2. Let $R$ be a DVR with maximal ideal $p$ and let $K$ be its fraction field. Then $\text{Ext}^1(K, R) = R^{\ast}/R$ where $R^{\ast}$ is the $p$-adic completion of $R$.

Claim 2 can be used to prove the converse of Rotmans' theorem.

Converse of Rotman's Theorem. Let $R$ be a DVR with maximal ideal $p$. If every reduced torsion-free module of finite rank over $R$ is free, then $R$ is $p$-adically complete.

The proof of Claim 2 relies essentially on the computations $$\text{Hom}_R(K, K/R) \simeq K^{\ast}, \quad \text{Hom}_R(K/R, K/R) \simeq R^{\ast}$$ where $K^{\ast}$ is the $p$-adic completion of $K$. The latter can be proved for any Noetherian local ring $R$ if $K/R$ is replaced by the injective hull of the residual field of $R$; it is a building block of Matlis duality theory [3, Theorem 18.6.iv]. The former relies on the latter and holds for any DVR. Its computation is very similar to this MO computation by YCor.

Proof of Claim 2. If $R$ is any domain, then $K$ is its injective hull. If $R$ is any Dedekind domain, then $K/R$ is an injective $R$ module. (If $(R, p, k)$ is a DVR with maximal ideal $p$ and residue field $k$, then $K/R$ is moreover the injective hull of $k$.) Thus for any Dedekind domain $R$, the exact sequence $Q$ $$ 0 \rightarrow R \rightarrow K \rightarrow K/R \rightarrow 0$$ is an injective resolution of $R$. Therefore $$\text{Ext}_R^1(K, R) = H_{-1}(\text{Hom}_R(K, Q)) = \text{Hom}_R(K, K/R)/\text{Hom}_R(K, K)$$ where we have identified $\text{Hom}_R(K, K) \simeq K$ with its image in $\text{Hom}_R(K, K/R)$. From now on, the ring $R$ is assumed to be a DVR. Let $G = \text{Hom}_R(K, K/R)$ and let $G_n \subseteq G$ the $R$-submodule of $G$ consisting of the homomorphisms which vanishes on $p^nR$. Then $G_n \simeq \text{Hom}_R(K/p^nR, K/R) \simeq \text{Hom}_R(K/R, K/R) \simeq R^{\ast}$, the latter isomorphism being given by [3, Theorem 18.6.iv]. Since we have $G = \bigcup_{n \ge 0} G_n$ and $pG_{n + 1} = G_n$, we deduce that $G\simeq K^{\ast}$. It is not difficult to show that $K^{\ast}/K \simeq R^{\ast}/R$. Indeed, the kernel of the map induced by the inclusion $R^{\ast} \rightarrow K^{\ast}/K$ is $R^{\ast} \cap K = R$ while $K^{\ast} = R^{\ast} + K$ follows from the density of $R$ in $R^{\ast}$.


[1] J. Rotman, "A note on completion of modules", 1960.

[2] D. Arnold, "A duality Torsion-free modules of finite rank over a discrete valuation ring", 1969.

[3] H. Matsumura, "Commutative Ring Theory", 1986.

[4] D. Arnold, M. Dugas and K. Rangaswamy, "Torsion-free modules of finite rank over a discrete valuation ring", 2004.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Luc, I'll keep watching for you to post what you discover. I expected the structure to be somewhat like what you proposed, but was unable to get started. I was hoping for submodules to look like the direct sum of some $K$'s, some $R$'s, and some fractional ideals (which, of course, are isomorphic to $R$). I will try to work on what you have provided. $\endgroup$ – Chris Leary Jun 5 '18 at 2:11
  • 1
    $\begingroup$ @LucGuyot is your initial claim true for a complete DVR? this might be a simple consequence of Matlis duality. $\endgroup$ – YCor Jun 7 '18 at 8:25
  • 1
    $\begingroup$ Concerning the last claim: More is true: $Ext(K,R)=\hat{R}/R$ for $R$ a DVR and $\hat{R}$ its completion (I leave it as an exercise in homological algebra). BTW: The structure of finite-rank modules over complete DVR can also be proved from $Ext(K,R)=0$ by induction on the rank. $\endgroup$ – tj_ Jun 8 '18 at 21:50
  • 1
    $\begingroup$ The formula $Ext(K,R)=\hat{R}/R$ shows that the converse of what you call YCor's claim (a result which is shortly mentioned in [2] and was probably known for some time before) is also true, i.e. a DVR is complete iff each module of finite rank is isomorphic to $R^i \oplus K^j$. $\endgroup$ – tj_ Jun 8 '18 at 22:01
  • 1
    $\begingroup$ @LucGuyot you get $Ext^1(K,R)=\hat{R}/R$ directly from an injective resolution of $R$. Possibly Matlis duality is a sledgehammer in the case of a PID, let alone a DVR. At some point one gets $Hom(K/R,K/R)$, and this is equal to $\hat{R}$: this is where the completion shows up (of course Matlis' duality makes this "obvious" but it's not needed to check this). $\endgroup$ – YCor Jun 9 '18 at 0:09
6
$\begingroup$

Passing to the spanned subspace, it is enough to consider those submodule that span $K^n$ as a $K$-module. Then up to composition by some element of $\mathrm{GL}_n(K)$, we can suppose that the submodule $V$ contains $R^n$.

So this reduces to classifying submodules of $(K/R)^n$. This is an artinian module, actually a module over $\hat{R}$, the completion of $R$. Write $S=K/R$. Matlis duality $\mathrm{Hom}(-,S)$ yields a natural correspondence between $\hat{R}$-submodules of $S^n$ and the quotient $\hat{R}$-modules of $\hat{R}^n$. In particular, modulo composition by an element of $\mathrm{GL}_n(\hat{R})$, every submodule of $S^n$ can be written as $S^k\times F$, where $F=\prod_{i=k+1}^nF_i\subset S^{n-k}$ has finite length. Actually, we can suppose $F=0$, composing beforehand by another element of $\mathrm{GL}_n(K)$.

In particular, Luc's initial claim is correct when $R$ is complete (but fails as soon as it's not complete and $n\ge 2$).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This is an interesting observation, and I will see if this helps with what I am trying to do. Thank you. $\endgroup$ – Chris Leary Jun 7 '18 at 14:43
  • 1
    $\begingroup$ Luc, I wrote on purpose "submodules of $\hat{R}^n$: it's a kind of "orthogonal" bijection, mapping a submodule to the set of homomorphisms into $S$ that vanish on the submodule. Indeed if it maps $N$ to $N'$, then the quotient by $N'$ is the Matlis dual of $N$. Actually to write properly, I should have written "correspondence between the set of submodules... and the set of submodules...". $\endgroup$ – YCor Jun 8 '18 at 17:32
  • $\begingroup$ @YCor Thanks for explaining this, and sorry for the trouble. The naive way (that is mine) to read your line about the correspondance is to assume that the correspondance is the Matlis duality $N \mapsto N' = \text{Hom}_{\hat{R}}(N, S)$. But what you mean is different, your are referring to another correspondance, namely $N \mapsto N^{\ast} = \ker(\hat{R}^n \twoheadrightarrow N')$. This is now clear for me. $\endgroup$ – Luc Guyot Jun 8 '18 at 20:11
4
$\begingroup$

Fix a prime $p$ and let $x$ be a $p$-adic integer that is not a rational number. Now let $M$ be the group of pairs of rational numbers $(r,s)$ so that $rx-s$ is a $p$-adic integer. Then $M$ maps onto $\mathbb Q$ but it does not contain a copy of $\mathbb Q$, because that would mean that some $s/r$ would not just be a good approximation of $x$, but a perfect one.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.