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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!

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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! –  InvisiblePanda Dec 16 '12 at 22:01
    
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. –  user18119 Dec 17 '12 at 21:43
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2 Answers

up vote 2 down vote accepted

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 $\mathbb{R}$.

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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? –  InvisiblePanda Dec 16 '12 at 22:06
    
@Randal'Thor There are several definitions of valuations. Your definition is one of them. However, usually a discrete valuation takes its values in $\mathbb{R}$. –  Makoto Kato Dec 16 '12 at 22:17
    
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! –  InvisiblePanda Dec 16 '12 at 22:27
    
@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, $v_p(1+i) = 1$) –  Sanchez Dec 16 '12 at 22:31
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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.

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