Timeline for Using Dunwoody's results on cohomological dimension to learn about a von Neumann regular group ring
Current License: CC BY-SA 3.0
15 events
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| Apr 7, 2018 at 2:01 | comment | added | Benjamin Steinberg | You are welcome. Sorry I didn't make progress on question 2. | |
| Apr 7, 2018 at 0:52 | vote | accept | rschwieb | ||
| Apr 5, 2018 at 19:37 | comment | added | Benjamin Steinberg | The proof that a finitely generated torsion group has a fixed point on the tree is explained pretty well in Section I 6.4 and 6.5 of Serre's tree book and can be read without looking at the rest of the book. The proof that the fundamental group of a finite graph of groups with finite vertex groups has a free subgroup of finite index is Section II 2.6 of Serre. You will also need that whenever a finitely generated group G acts on a tree it has a minimal invariant subtree which has finite quotient. This is on page 20 of Dunwoody-Dicks. | |
| Apr 5, 2018 at 19:17 | comment | added | rschwieb | By the way, what book would you recommend as a good introduction to the ideas behind the last paragraph? | |
| Apr 5, 2018 at 18:50 | comment | added | Benjamin Steinberg | This proof only handles the coubtable case. | |
| Apr 5, 2018 at 18:47 | history | edited | Benjamin Steinberg | CC BY-SA 3.0 |
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| Apr 5, 2018 at 17:57 | comment | added | Benjamin Steinberg | I made the answer to question 1 self-contained except I assumed Bass-Serre theory that a finitely generated torsion group acting on a tree has a fixed point or that a finitely generated group acting on a tree with finite vertex stabilizers has a free subgroup of finite index. | |
| Apr 5, 2018 at 17:55 | history | edited | Benjamin Steinberg | CC BY-SA 3.0 |
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| Apr 5, 2018 at 17:47 | history | edited | Benjamin Steinberg | CC BY-SA 3.0 |
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| Apr 5, 2018 at 17:32 | history | edited | Benjamin Steinberg | CC BY-SA 3.0 |
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| Apr 5, 2018 at 17:26 | comment | added | Benjamin Steinberg | I have to think more about question 2. I am use to finite dimensional algebras where every projective module for a self-injective algebra is also injective. | |
| Apr 5, 2018 at 17:26 | history | edited | Benjamin Steinberg | CC BY-SA 3.0 |
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| Apr 5, 2018 at 17:18 | comment | added | rschwieb | Renault doesn't argue with anything obviously cohomological. Everything he did was purely a ring theoretic or group theoretic reduction, backstopped by a lemma which says that the group ring of a Prufer group can't be self-injective. Maybe someday when I understand these things better though, I'll see a hint of the cohomology | |
| Apr 5, 2018 at 17:13 | history | edited | Benjamin Steinberg | CC BY-SA 3.0 |
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| Apr 5, 2018 at 17:07 | history | answered | Benjamin Steinberg | CC BY-SA 3.0 |