Timeline for Automorphisms of a matrix in Smith normal form?
Current License: CC BY-SA 3.0
8 events
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| Sep 16, 2011 at 10:10 | comment | added | Amritanshu Prasad | The moves work with matrices where the $i$th row is taken modulo $p^{\lambda_i}$. So, you only need to invert modulo some power of $p$.The Birkhoff moves are not moves on the matrices per se (so maybe my answer does not answer you question), but rather on matrices where the entries are taken modulo some congruences. | |
| Sep 12, 2011 at 21:22 | comment | added | Will Orrick | @Amritanshu Prasad : Thank you (and thanks to the other commenters) for your answer. While I continue to think about what you have said, I will ask what I'm sure is a naive question. A $p$-free integer is always a unit in $\mathbf{Z}/p^λ\mathbf{Z}$, but not in $\mathbf{Z}$. Therefore it would seem that move 1 is not generally invertible in the context of the original question. I confess that I have not yet fully understood certain aspects of your answer, so perhaps this is taken care of somehow? | |
| Sep 12, 2011 at 6:21 | history | edited | Amritanshu Prasad | CC BY-SA 3.0 |
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| Sep 11, 2011 at 23:23 | history | edited | Amritanshu Prasad | CC BY-SA 3.0 |
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| Sep 11, 2011 at 23:20 | comment | added | Amritanshu Prasad | You are absolutely right. | |
| Sep 11, 2011 at 13:29 | comment | added | Wilberd van der Kallen | One has a map from the group we seek to the automorphism group of $A$ with as kernel the group consisting of the maps that can be written as identity plus a homomorphism from ${\mathbb Z}^n $ to $D{\mathbb Z}^n $. | |
| Sep 11, 2011 at 12:09 | comment | added | Ralph | Which group is meant by "This is acually" Aut(A) ? If $D=\lambda \cdot I$ then $GL_n(\mathbb{Z}) \cap DGL_n(\mathbb{Z})D^{-1} = GL_n(\mathbb{Z})$ surely isn't Aut(A). | |
| Sep 11, 2011 at 11:32 | history | answered | Amritanshu Prasad | CC BY-SA 3.0 |