Timeline for Freeness of a Z[x]-module
Current License: CC BY-SA 3.0
22 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Mar 12, 2014 at 10:21 | history | edited | Stefan Kohl♦ | CC BY-SA 3.0 |
Made the matrix at the end display correctly (wasn't LaTeX'ed before).
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| Jun 13, 2013 at 16:53 | comment | added | François Brunault | Thank you for computing the matrix further! You're right, I did a mistake when computing $f(2)$, I forgot to report the expression of $f(-1)$, now it should be ok. Actually, there seems to be some freedom in the choice of the matrix entries, just like several Bézout identities are possible. | |
| Jun 13, 2013 at 16:51 | history | edited | François Brunault | CC BY-SA 3.0 |
Corrected entries of the matrix
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| Jun 13, 2013 at 13:55 | comment | added | Stefan Kohl♦ | While David's answer certainly answers the question itself with minimal effort, your answer completely describes the structure of $R$. I like your description, so I have now accepted your answer. | |
| Jun 13, 2013 at 13:54 | vote | accept | Stefan Kohl♦ | ||
| Jun 13, 2013 at 13:45 | comment | added | Stefan Kohl♦ | I have computed some more entries of the matrix given at the end of the answer -- see gap-system.org/DevelopersPages/StefanKohl/problems/mat.txt. (For the GAP code to compute the matrix, see gap-system.org/DevelopersPages/StefanKohl/problems/zxmatrix.txt). | |
| Jun 13, 2013 at 13:30 | comment | added | Stefan Kohl♦ | Thank you very much! -- Now the construction seems clear to me. Though I get for the fourth line of the matrix (3 -2 2 6) rather than (3 0 -2 6). -- The latter does not fit (mod 3) with (0 1 2 0) -- or have I made some mistake? | |
| Jun 13, 2013 at 8:44 | comment | added | François Brunault | The entries of the matrix were incorrect due to the change of numbering, now corrected. | |
| Jun 13, 2013 at 8:43 | history | edited | François Brunault | CC BY-SA 3.0 |
corrected the matrix entries
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| Jun 12, 2013 at 11:47 | comment | added | François Brunault | Now edited the answer in order to incorporate the various comments. | |
| Jun 12, 2013 at 11:45 | history | edited | François Brunault | CC BY-SA 3.0 |
Incorporating various comments into the answer for convenience of the reader
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| Jun 12, 2013 at 10:20 | comment | added | François Brunault |
Sorry for not making the argument very precise. To define $f$ on the whole of $\mathbf{Z}$, we define $f$ succesively at $0$, $1$, $-1$, $2$, $-2$, $3$, $-3$ and so on. Each time $f(n)$ is defined as $M \cdot a_n$ plus some linear combination of $a_m$ where $m$ ranges over the previously used integers. The point is that in this way we ensure that all possible pairs $\{m,n\}$ are visited, so all necessary congruences hold. I agree that with an arbitrary well-ordering of $\mathbf{Z}$ there might be some problems (I have not checked this).
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| Jun 11, 2013 at 21:28 | comment | added | Stefan Kohl♦ | I don't think it is immediate "to adapt the idea to define $f$ on the whole of $\mathbb{Z}$". Maybe it is possible, but I don't see it. As you write it down, the construction works only in one direction. Also taking a well-ordering on $\mathbb{Z}$ does not seem to do the job, as it would not respect the congruences. The matrix you give at the end of your answer is only for the positive integers. -- Maybe you can make your argumentation precise? | |
| Jun 11, 2013 at 17:14 | comment | added | François Brunault | And a reference for the last fact is : mathoverflow.net/questions/10239/… | |
| Jun 11, 2013 at 15:41 | comment | added | François Brunault | Also needed is the fact that $\mathbf{Z}^{\mathbf{Z}}$ is not free, the reason I know for this goes as follows : if $\mathbf{Z}^{\mathbf{Z}} \cong \bigoplus_{x \in X} \mathbf{Z}$ then $X$ is uncountable and $\operatorname{Hom}(\mathbf{Z}^{\mathbf{Z}},\mathbf{Z}) \cong \mathbf{Z}^X$, but it is known that the left hand side is isomorphic to the free abelian group with basis $\mathbf{Z}$, thus is countable. | |
| Jun 11, 2013 at 14:01 | comment | added | François Brunault | @David : This is something I learnt thanks to MO :) If $R$ is a PID, then every submodule of a free $R$-module is also free. For a reference see e.g. the comments to this answer mathoverflow.net/questions/16953/… | |
| Jun 11, 2013 at 12:52 | comment | added | David E Speyer | For the benefit of the rest of us, why can't a free $\mathbb{Z}$-module (of some uncounteable rank) contain a copy of $\mathbb{Z}^{\mathbb{Z}}$? | |
| Jun 11, 2013 at 12:37 | history | edited | François Brunault | CC BY-SA 3.0 |
Added details to give the structure of R as Z-module.
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| Jun 11, 2013 at 9:21 | comment | added | Stefan Kohl♦ | Very nice -- thank you very much!! -- Though I don't see yet the isomorphism to $\mathbb{Z}^\mathbb{Z}$. If you could explain that a little more, I'll accept your answer, as it reveals much more of the structure of $R$. | |
| Jun 11, 2013 at 7:58 | comment | added | François Brunault | On second thought, this argument seems to show that $R$ is actually isomorphic to $\mathbf{Z}^{\mathbf{Z}}$ as a $\mathbf{Z}$-module. | |
| Jun 11, 2013 at 7:46 | history | answered | François Brunault | CC BY-SA 3.0 |