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?

  • $\begingroup$ 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. $\endgroup$
    – reuns
    Commented Jun 27, 2022 at 21:21
  • $\begingroup$ 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 $\endgroup$
    – Babu
    Commented Jun 27, 2022 at 22:10
  • $\begingroup$ 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. $\endgroup$
    – reuns
    Commented Jun 27, 2022 at 22:17
  • $\begingroup$ Hmm I think my doubt has roots here $\endgroup$
    – Babu
    Commented Jun 27, 2022 at 22:27
  • $\begingroup$ 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 $\endgroup$
    – Babu
    Commented Jun 27, 2022 at 22:33

1 Answer 1


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 $3\not | b.$

  • $\begingroup$ I had understood what you are saying from reading the definition itself. I feel the question I put is un addressed $\endgroup$
    – Babu
    Commented Jun 27, 2022 at 22:00
  • $\begingroup$ @EthakkaappamwithChai I revised my answer to match your question a little better. As for when the valuation ring coming from a given field is unique, I'm not sure. $\endgroup$
    – subrosar
    Commented Jun 28, 2022 at 6:15

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