Timeline for Units in the group ring over fours group after Gardam
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 25, 2021 at 13:53 | history | edited | Giles Gardam | CC BY-SA 4.0 |
add update
|
| Apr 12, 2021 at 19:53 | comment | added | Dima Pasechnik | just wanted to point out that as GAP was apparently used in the work, it ought to be in references. gap-system.org/Contacts/cite.html | |
| Mar 21, 2021 at 20:40 | comment | added | YCor | If I'm correct $(KP)^\times$ has no torsion beyond scalars (so torsion forms a central subgroup naturally isomorphic to $K^*$), for arbitrary field $K$; this is a bit long for a comment and I wrote the argument as an answer mathoverflow.net/a/387096/14094 | |
| Mar 12, 2021 at 22:56 | comment | added | ARG | Actually, the argument works in any ring of characteristic $p$, since $\epsilon(x)^{n-1}$ will be a multiplicative inverse of $\epsilon(x)$. So good new Giles: any torsion element in any ring of positive characteristic will contradict zero divisors! | |
| Mar 12, 2021 at 22:35 | comment | added | ARG | Assume $x^n=1$ in $K[G]$ for $K$ a field of characteristic $p$. WLOG assume $n$ is minimal. Then if $p|n$, $0=x^n-1=(x^{n/p}-1)^p$ so $x^{n/p}-1$ and some power of $x^{n/p}-1$ form a zero divisor [or contradicts minimality]. Otherwise one has $\epsilon(x)^n = 1$. So look at $x' = x/\epsilon(x)$ (we are in a field after all so $1/\epsilon(x)$ exists). Then $\epsilon(x')=1$ and $x'^n=1$.Then $0=x'^n-1=(x'-1)(x'^{n-1}+\cdots+x'+1)$. Since $\epsilon(x')=1$, one has that $\epsilon(x'^{n-1}+\cdots+x'+1)=n \neq 0$ as $(n,p)=1$. The factorisation of $x'^n -1$ contradicts zero divisor. | |
| Mar 12, 2021 at 22:35 | comment | added | ARG | @IJL many thanks, but I found the "correct" argument (which is I believer much simpler); see next comment... | |
| Mar 12, 2021 at 14:59 | comment | added | IJL | An element of a field that is a root of $x^{n-1}+\cdots+x+1-0$ is an $n$th root of 1. So for $n$ a prime that does not divide $p-1$ there won't be any such elements. | |
| Mar 12, 2021 at 13:46 | comment | added | ARG | are there $\epsilon \in \mathbb{F}_p$ with $\epsilon^n=1$, $2<n <p-1$ and $ \epsilon^{n-1} + \cdots + \epsilon + 1 =0$? It's not too hard to check that this is not the case for $p=3$ or 5. But then this fails with $p=7$: one has $n=4$ and $\epsilon(x) = -1$. However one can then exclude 4-torsion using $x^4-1 = (x^2-1)(x^2+1)$, since $\epsilon(x)^2 = -1$ has no solution with $p=7$. So is there a general argument for any $p$? | |
| Mar 12, 2021 at 13:37 | comment | added | ARG | Very nice! Just to correct my previous comment: assume $x^n=1$ in $\mathbb{F}_p[G]$. WLOG assume $n$ is the first integer to do so. Then if $p|n$, $0=x^n-1=(x^{n/p}-1)^p$ so $x^{n/p}-1$ and some power of $x^{n/p}-1$ form a zero divisor. If $n=2$ then $x^2-1=(x-1)(x+1) =0$ contradicts zero divisor again. If $n$ is $>2$ and coprime to $p$, $0=x^n-1=(x-1)(x^{n-1}+\cdots+x+1)$. Now sending $y \in \mathbb{F}_p[G]$ to the sum of its coefficients $\epsilon(y)$ (augmentation map), one has that $\epsilon(x)^n=1$. Since $\epsilon(x)^p=\epsilon(x)$, then $\epsilon(x)^{n \text{ mod}(p-1)}=1$. So ... | |
| Mar 11, 2021 at 18:25 | history | answered | Giles Gardam | CC BY-SA 4.0 |