Timeline for Units in the group ring over fours group after Gardam
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Sep 10, 2023 at 22:18 | history | edited | YCor | CC BY-SA 4.0 |
fixed typo
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| Mar 22, 2021 at 15:25 | comment | added | ARG | ah yes... I overlooked that. So "stability by extension" remains unclear. Interesting! | |
| Mar 22, 2021 at 13:46 | comment | added | YCor | @ARG but the unit conjecture is not true for finite groups, so this doesn't say much about stability by extensions. | |
| Mar 22, 2021 at 13:41 | comment | added | ARG | That's a very nice insight indeed! And of course now we know this does not work for [non-split] extensions (since the group $P$ is a finite extension of $\mathbb{Z}^3$). [But there is no reason to believe it would have held for extensions.] | |
| Mar 22, 2021 at 11:01 | comment | added | YCor | @ARG actually the even more concise way (and intuitive, to me) is that the set of rings for which such a conjecture holds is stable under taking subrings, directed limits, and ultraproducts. This is enough to pass from finite fields to arbitrary domains. [By the way this is also known at the level of the group (passes to subgroups, directed limits, and ultraproducts)]. Of course this reflects here the principle that every system of polynomial equations and inequalities that has a solution in a domain has a solution in a finite field and this can be written in detail, quantitatively etc. | |
| Mar 22, 2021 at 10:07 | comment | added | ARG | thanks for this concise presentation. Still it's sometimes interesting to get the full meanders that can lead someone to a given conclusion. | |
| Mar 22, 2021 at 10:01 | comment | added | YCor | @ARG here are the details in a more concise way. Suppose the result known for all finite fields. Let $K$ be a domain and suppose that $KG$ fails one of the conjectures (has a idempotent, zero divisor, unit, nilpotent...). Passing to a finitely generated subdomain we can suppose $K$ finitely generated. Since every f.g. domain is residually a finite field, taking a suitable quotient we get a finite field $F$ with a counterexample of the same conjecture for $FG$. | |
| Mar 22, 2021 at 9:57 | comment | added | ARG | For those who are interested in the details of "finite fields $\implies$ all fields" see this other post | |
| Mar 22, 2021 at 7:53 | history | edited | YCor | CC BY-SA 4.0 |
added mention to other answer
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| Mar 21, 2021 at 22:51 | history | edited | YCor | CC BY-SA 4.0 |
simplified proof of lemma 1
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| Mar 21, 2021 at 22:45 | history | edited | YCor | CC BY-SA 4.0 |
simplified proof of lemma 1
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| Mar 21, 2021 at 21:20 | comment | added | YCor | @ARG As regards your question 1: of course the zero divisor conjecture is stronger when the algebra is larger. So I'm just assuming something a bit stronger than it for $KG$. In the known instances, the proof might be the same for all fields. Also, for given $G$, the zero divisor (resp. idempotent) conjecture for every field is equivalent to the zero divisor (resp. idempotent) conjecture for every finite field, by a simple classical approximation argument. | |
| Mar 21, 2021 at 21:15 | comment | added | YCor | @ARG For 2: I need algebraically closed in Lemma 1, it disappeared by some cut-paste accident (it's clearly false otherwise, just considering a nontrivial finite extension). I corrected | |
| Mar 21, 2021 at 21:15 | history | edited | YCor | CC BY-SA 4.0 |
added missing assumption
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| Mar 21, 2021 at 21:04 | comment | added | ARG | nice! I also wrote up a "clean" version of my comments here; the arguments work for any commutative ring (but without the algebraic closure). Which raises two questions: 1- is it easier to get the zero divisor conjecture on the algebraic closure than on the base field? 2- did you mean "Since the degree is positive, $P$ has a root $s$ in $\overline{K}$"? | |
| Mar 21, 2021 at 20:39 | history | answered | YCor | CC BY-SA 4.0 |