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    Please use this identifier to cite or link to this item: https://tkuir.lib.tku.edu.tw/dspace/handle/987654321/76936

    Title: 諾德問題與非分歧布勞爾群
    Other Titles: Noether$S Problem and Unramified Brauer Group
    Authors: 胡守仁
    Contributors: 淡江大學數學學系
    Date: 2011-08
    Issue Date: 2012-05-22 22:15:13 (UTC+8)
    Abstract: 令K 為一體, G 為一有限群。有理性問題就是討論對於G的忠實表現V , V=G是否有理。換 成代數的語言, 有理性問題就是要決定固定體K(V )G 是否有理(=純超越)。諾德問題為當V 為正則表現時的特例。我們以K(G) 表示其固定體。 如果G為p-群, 而體上有足夠的單位根, 我們已知對於秩為pn, n · 4及p = 2而n = 5的群 G ,K(G) 是有理的。Bogomolov 利用Bogomolov multiplier B0(G) 證明當秩為p6 時, 反 例存在。最近,Moravec 利用另一個同調不變量e B0(G)證明了某些秩為35的群,K(G)不有理。由 於e B0(G)的同調性, 使它的計算較為可行, 而且有已知的演算法可用。我們將運用這些演算法來 計算特定群的e B0(G)並將決定哪些秩為p5的群其Bogomolov multiplier 不為零。
    Let K be a field, G be a finite group and V be a faithful representation of G over K, rationality problem asks whether the quotient space V=G is rational. Algebraically, rationality problem asks whether K(V )G is rational, i.e. purely transcendental over K. Noether’s problem is the special case when the representation is the regular representation. In this case we denote the fixed field by K(G). If G is a p-group of order pn and the field contains enough roots of unity, it is known that K(G) is rational for n · 4 and for n = 5 when p = 2. Using Bogomolov multiplier B0(G) as an obstruction, Bogomolov constructed examples of groups of order p6 such that K(G) is not rational. Moravec used another obstruction e B0(G) to show that K(G) is not rational for some p-groups of order p5 when p = 3. In this project, we shall study the invariant e B0(G), use it to study whether K(G) is rational for other p-groups of order p5. The invariant e B0(G) is homological in nature and there is an algorithm to compute it. We shall use the algorithm to determine e B0(G) for certain groups G and classify groups of order p5 with nonzero Bogomolov multiplier.
    Appears in Collections:[數學學系暨研究所] 研究報告

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