Timeline for A kind of orthogonal subgroup
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 9, 2012 at 9:55 | comment | added | user21706 | Now, as you said we can work like the euclidean algorithm and write $k = h_{i_1, j_1}(k_1)$, $k = h_{i_2, j_2}(k_2)$...$k = h_{i_u, j_u}(k_u)$ where $k_u$ is one of $e_1, e_2, \ldots, e_n$ and so $S_k \cong S_{k_u} \cong \mathbb{Z}^{n-1}$. It is OK? | |
| Mar 9, 2012 at 9:53 | comment | added | user21706 |
I try to follow your advice: let $e_1, e_2, \ldots, e_n$ be the standard base of the $\mathbb{Z}-$module $\mathbb{Z}^n$, for any $i \neq j$ we have that $h_{i,j} : \mathbb{Z}^n \to \mathbb{Z}^n : x \mapsto x + (x \cdot e_i) e_j$ is an isomorphism. For any $k \in \mathbb{Z}$ we definite $S_k = \{x \in \mathbb{Z}^n : k \cdot x = 0\}$. If $k = h_{i,j}(k^\prime)$ then $k \cdot x = 0$ iff $k^\prime \cdot x = -(k^\prime \cdot e_i)(x \cdot e_j)$ iff $k^\prime \cdot x^\prime = 0$ where $x^\prime = h_{j,i}(x)$, in conclusion $S_k \cong S_{k^\prime}$.
|
|
| Mar 8, 2012 at 20:29 | comment | added | darij grinberg | Okay, actually forget about my reformulation of your problem, and just apply my proof to your original problem. When you replace $a_i$ by $a_i-a_j$, the module $\left\lbrace x\in\mathbb Z^n\mid k\cdot x=0\right\rbrace$ transforms in a rather simple way (which you should be able to figure out on your own; I am in a lecture right now). | |
| Mar 8, 2012 at 20:14 | comment | added | user21706 |
Thanks, all right, so exists $y \in \mathbb{Z}^n$ such that $k \cdot y = 1$, but I do not understand why this fact implies that $\{x \in \mathbb{Z}^n : k \cdot x = 0\} \cong \mathbb{Z}^{n-1}$ :-(
|
|
| Mar 8, 2012 at 20:10 | history | edited | user21706 | CC BY-SA 3.0 |
added 4 characters in body
|
| Mar 8, 2012 at 18:39 | comment | added | darij grinberg | One of your $\mathbb Z^n$'s should be $\mathbb Z^{n-1}$. | |
| Mar 8, 2012 at 18:32 | comment | added | darij grinberg | Yes, and the proof imitates the euclidean algorithm. In elementary terms, your question is: If $a_1$, $a_2$, ..., $a_n$ are some integers which have no common divisors except $1$ and $-1$, prove that some $\mathbb Z$-linear combination of $a_1$, $a_2$, ..., $a_n$ is $1$. Prove this by successive lowering of $\left|a_1\right|+\left|a_2\right|+...+\left|a_n\right|$ through replacing some $a_i$ by $a_i-a_j$ or $a_i+a_j$ for some $j\neq i$. | |
| Mar 8, 2012 at 18:28 | history | asked | user21706 | CC BY-SA 3.0 |