English

Unramified Brauer groups for groups of order p^5

Commutative Algebra 2011-09-15 v1 Algebraic Geometry Number Theory

Abstract

Let kk be any field, GG be a finite group acting on the rational function field k(xg:gG)k(x_g : g\in G) by hxg=xhgh\cdot x_g=x_{hg} for any h,gGh,g\in G. Define k(G)=k(xg:gG)Gk(G)=k(x_g : g\in G)^G. Noether's problem asks whether k(G)k(G) is rational (= purely transcendental) over kk. It is known that, if \bC(G)\bC(G) is rational over \bC\bC, then B0(G)=0B_0(G)=0 where B0(G)B_0(G) is the unramified Brauer group of \bC(G)\bC(G) over \bC\bC. Bogomolov showed that, if GG is a pp-group of order p5p^5, then B0(G)=0B_0(G)=0. This result was disproved by Moravec for p=3,5,7p=3,5,7 by computer computing. We will give a theoretic proof of the following theorem (i.e. by the traditional bare-hand proof without using computers). Theorem. Let pp be any odd prime number. Then there is a group GG of order p5p^5 satisfying B0(G)0B_0(G)\neq 0 and G/[G,G]Cp×CpG/[G,G] \simeq C_p \times C_p. In particular, \bC(G)\bC(G) is not rational over \bC\bC.

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Cite

@article{arxiv.1109.2966,
  title  = {Unramified Brauer groups for groups of order p^5},
  author = {Akinari Hoshi and Ming-chang Kang},
  journal= {arXiv preprint arXiv:1109.2966},
  year   = {2011}
}