Let $R$ be a Discrete Valuation Ring with fraction field $K$. Will this imply any proper $R$-submodule of $K$ is finitely generated (hence a fractional ideal)?
I know $K$ is not finitely generated as $R$-module if $R$ is not a field, but how about proper submodules? I can't think of an obvious counter example.
On
The answer is yes. The key is to use the fact that every element of $R$ can be written as $u\pi^k$, where $k \ge 0$ and $u\in R^\times$.
As an $R$-module, $K$ is the union (direct limit) of free rank 1 submodules:
$$\bigcup_{i=1}^\infty R\pi^{-i}$$
Let $M$ be a submodule of $K$. Suppose $M$ is infinitely generated, then let $m_1,m_2,\ldots$ be (possibly uncountably many) generators. Then letting $v(\cdot)$ denote "valuation", then the set $\{v(m_i) : i\in\mathbb{N}\}$ must be an infinite subset of $\mathbb{Z}$. Furthermore, it must have infinitely many distinct negative numbers, since if it had only finitely many negative numbers, then let $-n$ be the lowest amongst them, then we would have $M\subset R\pi^{-n}$ which is clearly free of rank 1. Then, by multiplying by suitable units, this implies that $\pi^{-k}\in M$ for all $k\ge 0$, which is to say that $M = R[\pi^{-1}] = K$.
It is not hard to prove that for any valuation ring $R$ the $R$-submodules of $K$ are totally ordered by inclusion.
Now let $M$ be an $R$-submodule of $K$, $M\ne 0, K$. If $M\subseteq\pi^nR$, then there are two cases: if $n\ge 0$ then $M$ is an ideal of $R$ and it is principal, generated by a power of $\pi$, or $n<0$ and in this case $\pi^{-n}M$ is an ideal of $R$. Otherwise, $\pi^nR\subseteq M$ for all $n\in\mathbb Z$. But $\bigcup_{n\in\mathbb Z}\pi^nR=K$. (If $x\in K$, $x\ne0$ we have two cases: $x\in R$, or $x^{-1}\in R$. Suppose $x^{-1}\in R$. Then $x^{-1}=\pi^ru$ with $r\ge0$ and $u\in R$ invertible. Therefore $x=\pi^{-r}u^{-1}\in\pi^{-r}R$.) This entails $M=K$, a contradiction.