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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jun 14, 2016 at 20:58 vote accept Qayum Khan
Jun 14, 2016 at 5:57 comment added David Benjamin Lim What you want is the Artin-Tate Lemma.
Jun 14, 2016 at 4:29 answer added abx timeline score: 1
Jun 14, 2016 at 0:37 comment added Qayum Khan @abx: That is an absolutely perfect answer. It has taken me some time to get the book and verify the statement and the proof. Please shift your comment to an official answer, so that it is credited as answer and the question can be closed. For the record, the second part of that Theorem 2 (with ground ring $K=\mathbb{Z}$) states that the fixed subring $R^G$ is also finitely generated as a ring.
Jun 12, 2016 at 7:41 comment added abx The only hypothesis you need is $R$ finitely generated (no characteristic 0, or japanese, or...). See Bourbaki's Commutative algebra V, §1, no. 9, Theorem 2.
Jun 12, 2016 at 5:16 comment added Qayum Khan @Qiaochu Yuan: Thank you --- I edited the setup to add "of characteristic 0." Some sort of characteristic-zero counter-example was produced in 1977 by C L Chuang and P H Lee. Sadly, I'm not able to get access to this article or the two mentioned in the remark, so I can't study them.
Jun 12, 2016 at 5:13 history edited Qayum Khan CC BY-SA 3.0
@Qiaochu Yuan: added "of characteristic 0" to the setup. Some sort of characteristic-zero counter-example was produced in 1977 by C L Chuang and P H Lee. Sadly, I'm not able to get access to this article or the two in the remark.
Jun 12, 2016 at 4:48 comment added Qiaochu Yuan I don't understand your remark. Where are you assuming or showing that $R$ has characteristic $0$?
Jun 12, 2016 at 2:01 history asked Qayum Khan CC BY-SA 3.0