# Definitions:

DVR:There is some discrete valuation $$ν$$ on the field of fractions $$K$$ of $$R$$ such that $$R = \{0\} \cup \{x\in K : ν(x) ≥ 0\}$$ A discrete valuation is an integer valuation on a field $$K$$; that is , a function: $$v:K \to \mathbb{Z} \cup \{ \infty \}$$

Satisfying the conditions:

$$v(x \cdot y) = v(x) + v(y)$$

$$v(x+y) \geq \text{min} \{ v(x), v(y) \}$$

$$v(x) = \infty \iff x=0$$

# The problem

My concern is, there are many discrete valuation maps which one could actually put on the field of fractions, so do we get a different ring for each one we put? When it is it true that the the ring out of putting the discrete valuation function is unique?

• The discrete valuation on $K$ is just $v(a/b)=v(a)-v(b)$ where $v$ is the discrete valuation on $R$, that is $v(a)=n$ if $a\in (\pi)^n-(\pi)^{n+1}$ where $(\pi)$ is the maximal ideal. Commented Jun 27, 2022 at 21:21
• What is the point you want to say? I think you are very brilliant in this area but I am often unable to decrypt your comment to a form which is understandable. Maybe if you added more elaborate explanation. I am still a begineer at this :( (also same for other comment ) @reuns
– Babu
Commented Jun 27, 2022 at 22:10
• DVR : the ring $R$ comes with a discrete valuation $v$ satisfying a few conditions. It also extends to $Frac(R)$, and you can recover $R$ from $Frac(R)$ and $v$. This gives several equivalent definitions of DVR. There is not much more to say, only to look at examples. Commented Jun 27, 2022 at 22:17
• Hmm I think my doubt has roots here
– Babu
Commented Jun 27, 2022 at 22:27
• So you're saying you actually don't need field of fraction to define a discrete evaluation ring? But then if we have a field of fraction is the ring we get from it's discrete evaluations unique? @reuns
– Babu
Commented Jun 27, 2022 at 22:33

In general, a DVR arising from a field will not be unique. Consider the field $$\mathbb{Q}.$$ For primes $$p$$, let $$\textrm{ord}_p:\mathbb{Z}\to \mathbb{N}\cup \{\infty\}$$ be such that $$\textrm{ord}_p(k)$$ is the largest $$n$$ such that $$p^n|k$$ and $$\textrm{ord}_p(0)=\infty.$$ Next, define $$\nu_p:\mathbb{Q}\to \mathbb{Z}\cup \{\infty\}$$ so that $$\nu_p(\frac{a}{b})=\textrm{ord}_p(a)-\textrm{ord}_p(b).$$ The function $$\nu_p$$ is a valuation for all primes $$p$$, but different $$p$$ give different valuation rings. Let $$R_p$$ be the valuation ring induced by $$\nu_p$$. In this case, different $$\nu_p$$ lead to different valuation rings. For example $$R_2$$ is the ring of fractions $$\frac{a}{b}$$ such that $$2\not|b,$$ while $$R_3$$ is the ring of fractions of the form $$\frac{a}{b}$$ such that $$3\not | b.$$