Timeline for When is $GL_m(R)$ generated by elementary and diagonal matrices?
Current License: CC BY-SA 3.0
29 events
| when toggle format | what | by | license | comment | |
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| S Jan 6, 2017 at 18:46 | history | bounty ended | CommunityBot | ||
| S Jan 6, 2017 at 18:46 | history | notice removed | CommunityBot | ||
| Dec 29, 2016 at 22:12 | comment | added | Ilya Bogdanov | The question would look much better if the embedding of $U(R)$ was there, and it is not too late to fix this. | |
| S Dec 29, 2016 at 16:48 | history | bounty started | Sam Williams | ||
| S Dec 29, 2016 at 16:48 | history | notice added | Sam Williams | Canonical answer required | |
| Dec 28, 2016 at 16:49 | comment | added | Sam Williams | @YCor Yeah, pitfalls of staring at something for too long. Thanks for the input. Hopefully now the question is clear to all. | |
| Dec 28, 2016 at 16:11 | comment | added | YCor | @Sam I agree it seems it was obvious to you, which is a good start. It wasn't to me, as you might have noticed. As regards your second question, the answer is clear from my previous comment :) | |
| Dec 28, 2016 at 15:59 | comment | added | Sam Williams | @YCor It seemed obvious that that is what I was asking, particularly when I said (up to a unit). But hopefully now we can agree on what I am asking? | |
| Dec 28, 2016 at 15:11 | comment | added | YCor | OK; this lengthy discussion would have been avoided if you had asked whether $GL_m(R)$ is generated by elementary and diagonal (invertible) matrices. | |
| Dec 28, 2016 at 15:10 | history | edited | YCor | CC BY-SA 3.0 |
changed the title to make it fit with the question
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| Dec 28, 2016 at 13:47 | comment | added | Sam Williams | @YCor I also edited the question to hopefully make it a bit clearer what I mean. | |
| Dec 28, 2016 at 13:45 | history | edited | Sam Williams | CC BY-SA 3.0 |
added 146 characters in body
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| Dec 28, 2016 at 13:42 | comment | added | Sam Williams | @YCor As far as I know, it is quite natural to embed $U(R)$ into $GL_m(R)$ by considering $U(R)$ as $GL_1(R)$ and mapping $GL_m(R)$ into $GL_{m+1}(R)$ via the stabilization $X\mapsto\begin{pmatrix}X&0\\0&1\end{pmatrix}$ | |
| Dec 28, 2016 at 12:01 | comment | added | ACL | @YCor (Unless $R=\mathbf Z/2\mathbf Z$.) | |
| Dec 28, 2016 at 11:51 | history | edited | YCor | CC BY-SA 3.0 |
n changed to m when applicable
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| Dec 28, 2016 at 11:30 | answer | added | Ram Krishna Verma | timeline score: -3 | |
| Dec 28, 2016 at 3:12 | comment | added | YCor | @SamWilliams it's still a bit vague. $U(R)$ is contained in $R^\times$. How do you embed it into $GL_m$? The element $r\in R^\times$ could correspond to the scalar matrix $(r,r,\dots,r)$, or to $(r,1,\dots,1)$, which is not equivalent. | |
| Dec 28, 2016 at 0:43 | comment | added | Sam Williams | @R. van Dobben de Bruyn Yep, I mean commuting variables | |
| Dec 28, 2016 at 0:35 | comment | added | R. van Dobben de Bruyn | Just to clarify: when you write $D[t_1,\ldots,t_n]$, do you mean that the $t_i$ commute with each other (and with all scalars)? | |
| Dec 28, 2016 at 0:27 | comment | added | Sam Williams | @YCor Thanks, I fixed the typesetting. Well it isn't false for $D$ commutative. In this case $GL_m(R)=U(R)SL_m(R)$. Now, if $m\geq 3$ then Suslin's stability theorem tells us $SL_m(R)=E_m(R)$. So, any $X\in GL_m(R)$ can be written as a product of elementary matrices (up to a unit). By the way, I am writing the units of $R$ as $U(R)$. | |
| Dec 28, 2016 at 0:25 | history | edited | Sam Williams | CC BY-SA 3.0 |
edited body
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| Dec 27, 2016 at 20:15 | comment | added | YCor | By the way you should use two different letters for the size of matrices and the number of indeterminates, unless you really want them to match for any precise reason. | |
| Dec 27, 2016 at 20:15 | comment | added | YCor | Yes: I just said that this is trivially false in the case $D$ is commutative, because of the determinant condition. | |
| Dec 27, 2016 at 18:58 | history | edited | Sam Williams | CC BY-SA 3.0 |
added 15 characters in body
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| Dec 27, 2016 at 18:54 | comment | added | Sam Williams | @YCor What I mean is, any invertible matrix $X$ can be written as a product of elementary matrixes $X=E_1\ldots E_t$ (give or take some unit) | |
| Dec 27, 2016 at 18:48 | comment | added | YCor | If $D$ is commutative you don't need any theorem: computing the determinant trivially implies $E_n(R)\neq GL_n(R)$. Or do you mean something else? Suslin's theorem indeed says that $E_n(R)=SL_n(R)$. | |
| Dec 27, 2016 at 18:45 | history | edited | YCor |
edited tags
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| Dec 27, 2016 at 16:18 | review | First posts | |||
| Dec 27, 2016 at 16:58 | |||||
| Dec 27, 2016 at 16:15 | history | asked | Sam Williams | CC BY-SA 3.0 |