{
  "id": "1701.01074",
  "version": "v1",
  "published": "2017-01-04T16:57:37.000Z",
  "updated": "2017-01-04T16:57:37.000Z",
  "title": "The role of defect and splitting in finite generation of extensions of associated graded rings along a valuation",
  "authors": [
    "Steven Dale Cutkosky"
  ],
  "comment": "22 pages",
  "categories": [
    "math.AG",
    "math.AC"
  ],
  "abstract": "Suppose that $R$ is a 2 dimensional excellent local domain with quotient field $K$, $K^*$ is a finite separable extension of $K$ and $S$ is a 2 dimensional local domain with quotient field $K^*$ such that $S$ dominates $R$. Suppose that $\\nu^*$ is a valuation of $K^*$ such that $\\nu^*$ dominates $S$. Let $\\nu$ be the restriction of $\\nu^*$ to $K$. The associated graded ring ${\\rm gr}_{\\nu}(R)$ was introduced by Bernard Teissier. It plays an important role in local uniformization. We show that the extension $(K,\\nu)\\rightarrow (K^*,\\nu^*)$ 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 blow up of $R$, $S_1$ is a local ring of a blowup of $S$, $\\nu^*$ dominates $S_1$, $S_1$ dominates $R_1$ and the associated graded ring ${\\rm gr}_{\\nu^*}(S_1)$ is a finitely generated ${\\rm gr}_{\\nu}(R_1)$-algebra. We also investigate the role of splitting of the valuation $\\nu$ in $K^*$ in finite generation of the extensions of associated graded rings along the valuation. We will say that $\\nu$ does not split in $S$ if $\\nu^*$ is the unique extension of $\\nu$ to $K^*$ which dominates $S$. We show that if $R$ and $S$ are regular local rings, $\\nu^*$ has rational rank 1 and is not discrete and ${\\rm gr}_{\\nu^*}(S)$ is a finitely generated ${\\rm gr}_{\\nu}(R)$-algebra, then $\\nu$ does not split in $S$. We give examples showing that such a strong statement is not true when $\\nu$ does not satisfy these assumptions. We deduce that if $\\nu$ has rational rank 1 and is not discrete and if $R\\rightarrow R'$ is a nontrivial sequence of quadratic transforms along $\\nu$, then ${\\rm gr}_{\\nu}(R')$ is not a finitely generated ${\\rm gr}_{\\nu}(R)$-algebra.",
  "revisions": [
    {
      "version": "v1",
      "updated": "2017-01-04T16:57:37.000Z"
    }
  ],
  "analyses": {
    "subjects": [
      "14B05",
      "14E22",
      "13B10",
      "11S15"
    ],
    "keywords": [
      "associated graded ring",
      "finite generation",
      "regular local rings",
      "dimensional excellent local domain",
      "rational rank"
    ],
    "note": {
      "typesetting": "TeX",
      "pages": 22,
      "language": "en",
      "license": "arXiv",
      "status": "editable"
    }
  }
}