Let R be an integral domain, let Frac(R) denote the field of fractions of R. Then as Frac(R) contains R it is an R-module in an usual way. Now suppose every proper submodule of the R-module Frac(R) is cyclic. Then what can we say about R ? Can such integral domains be characterized in some way ?
If I wanted the field of fractions Frac(R) itself to be cyclic (or even finitely generated) as an R-module then I know that R would become a field. But my condition is only on proper submodules.
Answer
Any proper ideal of R is a proper submodule of Frac(R), so R must be a PID. If p is any prime element of R, then M={r/pn:r∈R,n∈N} is a submodule of Frac(R) which is not cyclic. Thus M must be all of Frac(R), which means every element of R is divisible by pn for some n. This means p is the only prime of R, so R has at most one prime and hence is either a DVR or a field.
Conversely, if R is a field, your statement clearly holds. If R is a DVR with uniformizer p, let M be a proper submodule of Frac(R). If M has elements of arbitrarily small valuation, then M is all of Frac(R). Otherwise, let d∈Z be the minimal valuation of an element of M; then M is generated by pd and hence is cyclic.
So every proper submodule of Frac(R) is cyclic iff R is either a field or a DVR.
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