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I know at least one method of constructing a convex valuation ring of rank n (but it is rather complicated).. What are the easiest methods of doing this.. given a natural number n I want to have a valuation ring (preferably convex) whose rank is n. I have heard you can do this with polynomials and power series, but I am not really sure how this is done.

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What is a convex ring? –  Qiaochu Yuan Dec 18 '09 at 2:06
    
Is books.google.com/… the same definition you are using? –  David Speyer Dec 18 '09 at 2:42
    
Let A be a valuation ring with quotient field K. Suppose furthermore that K is totally order by <. A is said to be convex if for any x,y in A and z in K with x<z<y one has z is in A. –  Jose Capco Dec 18 '09 at 2:48
    
@David : Yes it is –  Jose Capco Dec 18 '09 at 2:51
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1 Answer

up vote 7 down vote accepted

Here is a way to build a convex valuation ring with valuation group any ordered abelian group. This is inspired by Gerald Edgar's notes on transseries, which I learned about here. If we take $(G, \prec)$ to be $\mathbb{Z}^n$ ordered by lexicographic order, we will get a valuation of rank $n$.

Let $G$ be any ordered abelian group. Define $\mathbb{R}((G))$ to be the field of formal power series of the form $$\sum_{0 \leq k_1, \ldots, k_r \leq \infty} a_{k_1 \ldots k_r} \ t^{h+k_1 g_1 +\cdots k_r g_r}.$$ Here $r$ can be any positive integer, $a_{k_1 \ldots k_r}$ is any sequence of real numbers, $h$ is any element of $G$ and $g_1$, ... $g_r$ are any positive elements of $G$. The symbol $t$ is a formal parameter.

For example, $\mathbb{R}((\mathbb{Z}))$ is the field of formal Laurent series.

We add, multiply and divide such formal series by the obvious formal manipulations. We leave it to the reader to check that the sum or product of such power series is another such power series, as is the reciprocal of such a series, and that we never need to add more than finitely many coefficients together, so there are no issues of convergence.

There is a valuation $v:\mathbb{R}((G)) \to G$, sending a power series to the least $g$ for which $t^G$ appears with nonzero coefficient. The valuation ring of this valuation is clearly $\mathbb{R}[[G]]$; the ring of power series where $h$ (above) is nonnegative.

Finally, I have to tell you the ordering on $\mathbb{R}((G))$. A power series $f$ is positive if the coefficient of $t^{v(f)}$ is positive.

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Thanks :) I also read something along these lines.. so this is what is meant by construction by power series. –  Jose Capco Dec 18 '09 at 4:37
    
wouldnt an easier definition of R((G)) be the formal sums \sum a_gt^g (with all but finitely many a_g's nonzero) and R[[G]] be \sum a_g t^g that are in R((G)) with all a_g=0 for g<0 –  Jose Capco Dec 18 '09 at 5:23
    
Do you mean all but finitely many a_g's are zero? This is fine, except that it isn't a field. The valuation extends to the fraction field, so you can do that. I thought you'd rather have something more "power series" like. The construction you propose is more "polynomial" and less "power series". –  David Speyer Dec 18 '09 at 5:54
    
Oh yes.. but the new R[[G]] will still be a valuation ring of rank n over its fraction field. The fraction field can get the ordering : f/g is positive iff a_v(f)a_v(g) is positive. This ordering should make the valuation ring convex in its fraction field. –  Jose Capco Dec 18 '09 at 6:52
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