The role of defect and splitting in finite generation of extensions of associated graded rings along a valuation

Abstract

Suppose that $R$ is a 2-dimensional excellent local domain with quotient field $K$, $K^{\ast }$ is a finite separable extension of $K$ and $S$ is a 2-dimensional local domain with quotient field $K^{\ast }$ such that $S$ dominates $R$. Suppose that ${\nu }^{\ast }$ is a valuation of $K^{\ast }$ such that ${\nu }^{\ast }$ dominates $S$. Let $\nu$ be the restriction of ${\nu }^{\ast }$ to $K$. The associated graded ring ${gr}_{\nu }\left(R\right)$ was introduced by Bernard Teissier. It plays an important role in local uniformization. We show that the extension $\left(K,\nu \right)\to \left(K^{\ast },{\nu }^{\ast }\right)$ of valued fields is without defect if and only if there exist regular local rings $R_1$ and $S_1$ such that $R_1$ is a local ring of a blowup of $R$, $S_1$ is a local ring of a blowup of $S$, ${\nu }^{\ast }$ dominates $S_1$, $S_1$ dominates $R_1$ and the associated graded ring ${gr}_{{\nu }^{\ast }}\left(S_1\right)$ is a finitely generated ${gr}_{\nu }\left(R_1\right)$-algebra.

We also investigate the role of splitting of the valuation $\nu$ in $K^{\ast }$ in finite generation of the extensions of associated graded rings along the valuation. We say that $\nu$ does not split in $S$ if ${\nu }^{\ast }$ is the unique extension of $\nu$ to $K^{\ast }$ which dominates $S$. We show that if $R$ and $S$ are regular local rings, ${\nu }^{\ast }$ has rational rank 1 and is not discrete and ${gr}_{{\nu }^{\ast }}\left(S\right)$ is a finitely generated ${gr}_{\nu }\left(R\right)$-algebra, then $S$ is a localization of the integral closure of $R$ in $K^{\ast }$, the extension $\left(K,\nu \right)\to \left(K^{\ast },{\nu }^{\ast }\right)$ is without defect and $\nu$ does not split in $S$. We give examples showing that such a strong statement is not true when $\nu$ does not satisfy these assumptions. As a consequence, we deduce that if $\nu$ has rational rank 1 and is not discrete and if $R\to R^{\prime }$ is a nontrivial sequence of quadratic transforms along $\nu$, then ${gr}_{\nu }\left(R^{\prime }\right)$ is not a finitely generated ${gr}_{\nu }\left(R\right)$-algebra.