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    Please use this identifier to cite or link to this item: https://tkuir.lib.tku.edu.tw/dspace/handle/987654321/46958


    Title: 諾德問題之研究
    Other Titles: A Study of Noether$S Problem
    Authors: 胡守仁
    Contributors: 淡江大學數學學系
    Date: 2009
    Issue Date: 2010-04-15 15:40:49 (UTC+8)
    Abstract: 令K 為一體, G 為一有限群。G 作用於有理函數體K(xg : g 2 G):h ¢ xg = xhg, 此 處g; h 2 G. K(G) = K(xg : g 2 G)G 表示其固定體. 諾德問題就是要決定K(G) 在K 之上是否為有理的(=純超越。) 這是代數幾何中有理性問題的特殊情形。代數幾何中的有理性 問題研究的是對於任何K上的代數簇X, K(X)是否為純超越, 也就是X 99K Pn 是否為雙有 理等價。 如果體上有足夠的單位根, 我們已知對於秩為pn, n · 4的群K(G) 是有理的。Bogomolov 証明當秩為p6 時, 反例存在。因此就漏掉了當秩為p5 的情形。當p = 2 時, 我們證明 了秩為32的群, K(G) 是有理的[CHKP]。我們將研究當p 為奇質數秩為p5 的群時,K(G) 的有理性我們也將研究某些雙有理不變量, 這對於有理性問題的研究十分有用。 Let K be any field and G be a finite group. G acts on the rational function field K(xg : g 2 G) by h ¢ xg = xhg for any g; h 2 G. Denote by K(G) = K(xg : g 2 G)G the fixed field. Noether’s asks whether K(G) is rational (= purely transcendental) over K. This is a special case of the rationality problem in algebraic geometry which asks to determine for a given algebraic variety X over K whether K(X) is purely transcendental over K or equivalently, whether X 99K Pn is birationally equivalent. For fields containing enough roots of unity, it is known that K(G) is rational for groups of order pn for n · 4. Bogomolov proved that counterexample exists for groups of order p6. Thus there is a gap for the case of p5. For p = 2, it is shown that for groups of order 32, K(G) is rational [CHKP]. We shall now look into the case of groups of order p5 for odd prime p. We will also study some birational invariants such as Bogomolov multiplier which will aid us in the study of rationality problems
    Appears in Collections:[數學學系暨研究所] 研究報告

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