3
$\begingroup$

For an integral domain $R$ let $\mathrm{Frac}(R)$ denote its field of fractions. Then $R$ is embedded in $\mathrm{Frac}(R)$ and we can consider $\mathrm{Frac}(R)$ as an $R$-module.

Can we characterize all non-field integral domains $R$ such that every proper non-zero submodule of the $R$-module $\mathrm{Frac}(R)$ is projective ?

If $R$ satisfies my condition, then since every $R$-submodule of $K$ is projective hence so are in particular the fractional ideals of $R$, and hence every non-zero fractional ideal of $R$ is invertible, thus $R$ is a Dedekind domain. But $R=\mathbb Z$ is a Dedekind domain which does not satisfy my condition . So the family of integral domains, that satisfies the condition I stated , should be Dedekind domain + something more ; I can't figure out what this something more should be .

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Local pid's satisfy this property. It doesn't appear that semilocal ones (say with two maximal ideals) do, since if $x \in M_1 \setminus M_2$, then $R[x^{-1}]$ doesn't look like it is finitely generated as an $R$-module ... $\endgroup$
    – David Handelman
    Sep 23, 2017 at 18:30
  • 1
    $\begingroup$ @DavidHandelman :Yes , local PID's i.e. fields and DVR's do satisfy the property , but I don't know whether this is necessary or not. In fact if it could just be shown that domains satisfying my condition are local then we would be done as local Dedekind domains are either fields or DVR en.wikipedia.org/wiki/Discrete_valuation_ring $\endgroup$
    – user111524
    Sep 23, 2017 at 20:10

1 Answer 1

Reset to default
5
$\begingroup$

An integral domain $R$ such that every proper non-zero $R$-submodule of $\text{Frac}(R)$ is projective is a local principal ideal ring. (The converse is David Handelman's comment above).

Indeed, we have

Claim. Let $R$ be a bounded factorization domain, e.g., a Noetherian domain, with at least two maximal ideals. Then $\text{Frac}(R)$ contains a non-zero proper $R$-submodule which is not projective.

Proof. Let $\mathfrak{p}$ and $\mathfrak{q}$ be to distinct maximal ideals of $R$. Let $p \in \mathfrak{p} \setminus \mathfrak{q}$ and let $q \in \mathfrak{q} \setminus \{ 0 \}$. Then $1/q \notin R[1/p]$, since otherwise we would get $p \in \mathfrak{q}$, a contradiction. Thus $R[1/p] \subsetneq \text{Frac}(R)$. But $R[1/p]$ is not projective over $R$ because it doesn't embed into a free $R$-module. Indeed, $1 \in R[1/p]$ is a non-zero element which is divisible by any power of $p$ and no such element exists in a free $R$-module.

Improve this answer
$\endgroup$
10
  • $\begingroup$ could you please elaborate on why $1/q \in R[1/p]$ would imply $\mathfrak{p} = \mathfrak{q}$ ? $\endgroup$
    – user111524
    Sep 29, 2017 at 14:34
  • $\begingroup$ And do you mean $\mathfrak{p} , \mathfrak{q}$ are two distinct maximal ideals ? Otherwise you can't probably get an element from the relative complement of each , because it might happen that one of the prime ideals is just strictly contained in the other $\endgroup$
    – user111524
    Sep 29, 2017 at 14:41
  • 1
    $\begingroup$ May be we don't need $R$ to be a Dedekind domain for this claim ? If $1/q \in R[1/p]$ , then $rq=p^n$ for some $n \ge 1$ . Then $p^n=rq \in \mathfrak q$ , and since $\mathfrak q$ is prime ideal , so $p \in \mathfrak q$ , contradicting how we chose $p$ $\endgroup$
    – user111524
    Sep 29, 2017 at 14:45
  • $\begingroup$ @users Agreed, I edited my answer accordingly. $\endgroup$
    – Luc Guyot
    Sep 29, 2017 at 15:06
  • $\begingroup$ Why a free $R$-module cannot have an element divisible by all powers of a prime ? If $R$ is a field then a free $R$-module can contain such element , but say I am assuming $R$ is not a field . Can you explain why a free $R$-module cannot have such an element ? I can conclude if $R$ is a UFD , but I am not assuming that ... $\endgroup$
    – user111524
    Sep 29, 2017 at 15:29

Your Answer

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