#### Vol. 11, No. 6, 2017

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

### Steven Dale Cutkosky

Vol. 11 (2017), No. 6, 1461–1488
##### 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.

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or by using our contact form. ##### Keywords
valuation, local uniformization
##### Mathematical Subject Classification 2010
Primary: 14B05
Secondary: 11S15, 13B10, 14E22
##### Milestones 