I encountered the phrase "normalized valuation" similar to the following:

Let $A_i$ be the valuation ring $k[x_1,...,x_n]_{\langle x_i\rangle}$ and $v_i$ be the normalized valuation defined by $A_i$.

I didn't know this term before, and a short internet search did not help me.

What I know: we can define a map $k[x_1,...,x_n]\smallsetminus\{0\}\to\mathbb{Z}$ by sending $f=gx_i^{n_f}$ with $x_i\nmid g$ to $n_f\in\mathbb{Z}$. Then extend this to $v:Q(k[x_1,...,x_n])^*=k(x_1,...,x_n)^*\to\mathbb{Z}$ via $\frac{f}{g}\mapsto n_f-n_g$, and this is a discrete valuation on $k(x_1,...,x_n)$ with $k[x_1,...,x_n]_v=k[x_1,...,x_n]_{\langle x_i\rangle}$ as its discrete valuation ring. Is the map $v$ already the $v_i$ mentioned above? Is it "normalized", and what does this mean?

Also, the definition of the map $v$ relies on $k[x_1,...,x_n]$ being a UFD. So by the same argument, $R_{\langle p\rangle}$ is a DVR if $R$ is a UFD and $p\in R$ prime. So I guess this does no longer hold for rings of the form $R_\mathfrak{p}$ where $\mathfrak{p}\subset R$ is prime in general? What about $k[x_1,...,x_n]_{\langle x_1,x_2\rangle}$?

Thank you!

  • $\begingroup$ I just found a comment by QiL on "normed valuations", where he says that's the case if the value of a uniformizing element is 1. In my case, if I'm not mistaken, a uniformizing element would be $\frac{x_i}{1}$, with value $1-0=1$. Is normed = normalized maybe? If yes, I'll also gladly accept any answer elaborating on the question(s) in my last paragraph above! $\endgroup$ Commented Dec 16, 2012 at 22:01
  • $\begingroup$ In that message, the OP wrote wrongly normed instead of normalized. I don't know what is a normed valuation, but a normalized discrete valuation is as in Makoto's answer. $\endgroup$
    – user18119
    Commented Dec 17, 2012 at 21:43

2 Answers 2


A normalized discrete valuation $v$ of a field $K$ means that $v$ is a discrete valuation of $K$ such that $v(K^*) = \mathbb{Z}$. In general, $v(K^*)$ can be any discrete subgroup of $\mathbb{R}$.

  • $\begingroup$ Hello @Makoto! Are you sure about that? I'm a bit confused now, since we defined a valuation of a field $K$ in a totally ordered group $G$ as a group homomorphism $v:K^*\to G$ with the known properties. A discrete valuation was then if $G=\mathbb{Z}$ and if $v$ was surjective. By my definition, any discrete valuation would then be normalized? $\endgroup$ Commented Dec 16, 2012 at 22:06
  • $\begingroup$ @Randal'Thor There are several definitions of valuations. Your definition is one of them. However, usually a discrete valuation takes its values in $\mathbb{R}$. $\endgroup$ Commented Dec 16, 2012 at 22:17
  • $\begingroup$ Okay, so the normalized part has nothing to do with what in @QiL's comment has been written as "normed"? Just wondering, I don't really see through, since the original poster of this seems to have taken "normed" = "normalized", or I don't get him right ;) I just want to clear out any misunderstandings on my side here! $\endgroup$ Commented Dec 16, 2012 at 22:27
  • 1
    $\begingroup$ @Randal'Thor, there are natural valuations that you want to consider discrete but the group is not $\mathbb{Z}$. Consider for example the 2-adic valuation $v_2$ of $\mathbb{Z}$. $2 = (1+i)^2$ (up to unit) in $\mathbb{Z}[i]$. If you want to extend $v_2$ to $v_p$ where $p = (1+i)$, a possible way is to define $v_p(2) = 1$ (so that $v_p$ restricted back to $\mathbb{Z}$ is $v_2$), but then $v_p(1+i) = 1/2$. For cases like this, one may want to scale the valuation back (in this case, $v_p(1+i) = 1$) $\endgroup$
    – user27126
    Commented Dec 16, 2012 at 22:31
  • $\begingroup$ In others words: the valuation map is surjective. $\endgroup$ Commented Jul 21, 2019 at 0:07

For general prime ideals $\mathfrak{p}\subseteq k[x_1,\ldots, x_n]$, the local ring $k[x_1,\ldots, x_n]_\mathfrak{p}$ will not be a valuation ring, and hence will not define a valuation. But that does not mean you cannot consider valuations on such a ring; in fact, it is a very interesting thing to do!

Take, for instance, the local ring $R = \mathbb{C}[x,y]_{(x,y)}$. This is not a valuation ring, but it makes sense to consider valuations $\nu\colon R\to \mathbb{R}\cup\{+\infty\}$. In fact, a rather interesting book (The Valuative Tree by Favre and Jonsson) has been written about valuations $\nu\colon R\to \mathbb{R}\cup\{+\infty\}$ satisfying the properties:

  1. $\nu$ only takes nonnegative values on $R$.
  2. $\nu(z) = 0$ for all $z\in \mathbb{C}^\times$.
  3. $\nu(f)>0$ for all $f$ in the maximal ideal $\mathfrak{m}$ of $R$.

Such $\nu$ are called centered valuations. Such valuations also have a notion of normalized. We say a centered valuation $\nu$ is normalized if $\min_{f\in \mathfrak{m}} \nu(f) = 1$. It turns out that the set of normalized centered valuations on $R$ has an interesting combinatorial structure: it is an $\mathbb{R}$-tree.

The study of spaces of valuations on rings is becoming a more popular subject these days, sometimes falling under the heading of nonarchimedean analytic geometry or Berkovich geometry.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .