Timeline for Discrete valuations for which Abhyankar inequality is strict
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 5, 2013 at 5:39 | comment | added | Lars | Dear Manish, this seems to work, thanks! | |
| Jul 4, 2013 at 8:27 | comment | added | Manish Kumar | Actually if you define valuation on $k[x,y,z]$ by putting $z=xy+x^2y^2+x^3y^3+\ldots$ and taking y-ord then it seems you get a counter example. As the residue field of the valuation will be k(x) and the center of this valuation on all monidal transformation is always codimension 2. | |
| Jul 3, 2013 at 19:58 | comment | added | Lars | Hi Manish, it is not true that every discrete valuation on k(X)/k has a divisor as a center on some model. This property is equivalent to the Abhyankar inequality being an equality. In particular, a discrete valuation $v$ has a divisor as center on some model if and only if $trdeg_k k(v)=trdeg_k k(X)-1$. There is an example for a discrete valuation which does not have this property in the paper by Vaquié which I cite. | |
| Jul 3, 2013 at 17:28 | comment | added | Manish Kumar | Note that resolution of singularities (in fact uniformization itself) implies that $X$ has a model so that the center of $v$ is regular. | |
| Jul 3, 2013 at 17:25 | comment | added | Manish Kumar | Dear Lars, Thanks for the clarification. I think for every valuation $v$ of $k(X)/k$ one should be able find a model of $X$ on which the center of $v$ is a divisor. If the center of $v$ on $X$ is regular then if one takes a sequence of monoidal transformation of $X$ along $v$ then eventually the center will become a divisor. Basically look at the valuations of a system of parameters of the ideal defining the center of $X$. Their gcd is 1. After the monidal transformation these numbers will go down for the center of $v$ on this new model. Eventually one would get that the center is a divisor. | |
| Jul 3, 2013 at 13:36 | comment | added | Lars | Dear Manish, thanks for your comment! It is true that a discrete valuation for which equality holds in Abhyankar's inequality does not necessarily have a divisor as center on a given model ($\mathbb{A}^2$ in your example). But as soon as equality holds, there exists some model on which the valuation has a divisor as center; if I am not mistaken, then in your example one such model would be the spectrum of $k[X,Y,X/Y]$, which is an open subset of the blow-up of $\mathbb{A}^2$ in the origin. | |
| Jul 3, 2013 at 12:59 | comment | added | Manish Kumar | This is not an answer to your precise question but on the comment "$v$ comes from a divisor" by which I suppose you mean "the center of $v$ is a divisor". Let $A=k[X,Y]_{(X,Y)}$ and $v(f(X,Y))=ord_t(f(t,t))$. Then $v$ is a discrete valuation dominating $A$. Clearly, $k_A=k$ and check that $k(v)$ contains $k(X/Y)$. I think it is also true that $k(X/Y)$ is algebraically closed in $k(X,Y)$. So by Abhyankar's inequality $k(v)=k(X/Y)$. But of course the center of $v$ is of co-dimension 2 but equality holds in Abhyankar's inequality. | |
| Jul 2, 2013 at 17:07 | history | asked | Lars | CC BY-SA 3.0 |