Timeline for Complemented subalgebra in a free Lie ring
Current License: CC BY-SA 4.0
17 events
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| Sep 29, 2022 at 7:58 | comment | added | YCor | Since you're inside an abelian Lie algebra, you just mean a linear complement (i.e., as $\mathbf{Z}$-module). Now just use that a submodule of a free $\mathbf{Z}$-module is free. So you can lift the quotient. | |
| Sep 29, 2022 at 6:10 | comment | added | MANI | Let us continue this discussion in chat. | |
| Sep 28, 2022 at 13:47 | comment | added | MANI | @YCor I have edited the question sir, please have a look on this. | |
| Sep 28, 2022 at 13:47 | history | edited | MANI | CC BY-SA 4.0 |
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| Sep 28, 2022 at 12:54 | comment | added | YCor | Could you clarify in your post what exactly the question is. | |
| Sep 28, 2022 at 12:13 | comment | added | MANI | @YCor I am sorry, but my question is about the existence of complement subring in $Z(\tilde{F})$? How this can be concluded? | |
| Sep 28, 2022 at 11:46 | comment | added | YCor | How to conclude? I gave the argument in the last 3 lines of my previous comment. (I didn't justify why $F/[F,F]$ is torsion-free, but the latter point is clear: this is the "free abelian Lie algebra" on the given generating subset and is thus just the free abelian group on the basis.) | |
| Sep 28, 2022 at 11:23 | comment | added | MANI | @YCor, Yes I understand this that how $F/[F,F]$ is torsion free. But now how can I conclude the rest of things? | |
| Sep 28, 2022 at 10:50 | comment | added | YCor | I think that the whole point is that $F/[F,F]$ is a torsion-free group. It is more practical to work inside $F$ itself. If $x\in F$, the condition $nx\in [F,F]+[F,R]$ just means $nx\in [F,F]$. And $[F,F]+[F,R]=[F,F]$. Since $F/[F,F]$ is torsion-free, this implies $x\in [F,F]$. | |
| Sep 28, 2022 at 10:47 | comment | added | YCor | This is helpful to understand the question at least. I'll think about it now. | |
| Sep 28, 2022 at 10:46 | comment | added | MANI | How can this be helpful to deduce the result? | |
| Sep 28, 2022 at 10:43 | comment | added | YCor | Thanks, so it means $n\tilde{x}$, to be consistent with the additive notation. | |
| Sep 28, 2022 at 10:40 | comment | added | MANI | @YCor $\tilde{x}^n==\tilde{x}+\tilde{x}+\ldots +\tilde{x}$ ($n$-times) for the reference (Theorem 3.2 of \tilde{x}.) | |
| Sep 28, 2022 at 10:34 | comment | added | YCor | What do you mean by $\tilde{x}^n$? I don't what the $n$-power could mean in a Lie ring. | |
| Sep 28, 2022 at 10:30 | history | edited | YCor | CC BY-SA 4.0 |
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| S Sep 28, 2022 at 10:19 | review | First questions | |||
| Sep 28, 2022 at 12:24 | |||||
| S Sep 28, 2022 at 10:19 | history | asked | MANI | CC BY-SA 4.0 |