Unramified Brauer groups for groups of order p⁵

2011/09/14 by Hoshi, Akinari, Kang, Ming-chang
#12F12 #13A50 #14E08 #14M20 #Algebraic Geometry (math.AG) #Commutative Algebra (math.AC) #FOS: Mathematics #Number Theory (math.NT)

paper · doi:10.48550/arxiv.1109.2966

Abstract

Let k be any field, G be a finite group acting on the rational function field k(x_g : g∈ G) by h⋅ x_g=x_hg for any h,g∈ G. Define k(G)=k(x_g : g∈ G)^G. Noether's problem asks whether k(G) is rational (= purely transcendental) over k. It is known that, if \bC(G) is rational over \bC, then B₀(G)=0 where B₀(G) is the unramified Brauer group of \bC(G) over \bC. Bogomolov showed that, if G is a p-group of order p⁵, then B₀(G)=0. This result was disproved by Moravec for p=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 p be any odd prime number. Then there is a group G of order p⁵ satisfying B₀(G)≠ 0 and G/[G,G] ≃ C_p × C_p. In particular, \bC(G) is not rational over \bC.

Related

• Noether's problem and unramified Brauer groups
• Noether's problems for groups of order 243
• Noether's problem for p-groups of order p⁵
• Degree three unramified cohomology groups and Noether's problem for groups of order 243
• Groups of order p⁵ and their unramified Brauer groups