We say a ring $R$ is a DVR if is a local integral domain and the maximal ideal is principal. I want to show that this ring admits a discrete valuation $\nu: K(R) \rightarrow \mathbb{Z} \cup\{0\}$ that satisfies:
$1)$ $ \nu(a+b) \geq \min\{\nu(a),\nu(b)\}$
$2)$ $\nu(ab) = \nu(a)+ \nu(b)$
$3)$ $\nu(a)=\infty \Leftrightarrow a=0$
$4)$ $ R = \{x \in K(R) : \nu(x)\geq 0\}$
My idea is that: I need to define $\nu(x)=0$ for every unit which are the elements that are not in the maximal ideal. And $a$ is the generator of the maximal ideal we should define $\nu(a)=1$. I would like to define $\nu(x)$ as the exponent of $a$ in the factorization of $x$. However, this works only if $R$ is a UFD. I would appreciate any help to generalize this idea to the case without assuming UFD.
What you try to show is false without Noetherian condition. See e.g. https://mathoverflow.net/questions/155621/condition-for-a-local-ring-whose-maximal-ideal-is-principal-to-be-noetherian
Define $\mu(a) = \sup \{n\in\mathbb N: x^n \mid a\}$ where $(x)$ is the maximal ideal of $R$. The hard part is to show (3), i.e. if $\mu(a)=\infty$ then $a=0$, which is the same as to say $I = \cap_{n=0}^\infty\mathfrak m^n=\{0\}$. Note that $\mathfrak m I=I$, if $R$ is assumed to be Noetherian then by Nakayama's lemma, $I=\{0\}$.
(1) is straightforward, (2) can be deduced from (3).
From here, it's easy to show that any nonzero element in $R$ can be uniquely decomposed as $x^nu$ where $u$ is a unit. So a DVR is a UFD, but it's not so easy to show this directly.