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Let $n$ a positive integer and $k \in \mathbb{Z}^n$ such that for all integer $a \geq 2$ and $h \in \mathbb{Z}^n$ we have $k \neq ah$. Here $\cdot$ is the scalar product.

Is it true that $\{x \in \mathbb{Z}^n : k \cdot x = 0\} \cong \mathbb{Z}^{n-1}$ (group isomorphism)?

The answer is yes for $n=1,2$ but the general case seems difficult to me.

Thanks for any suggestions.

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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 $j\neq i$. – darij grinberg Mar 8 '12 at 18:32
One of your $\mathbb Z^n$'s should be $\mathbb Z^{n-1}$. – darij grinberg Mar 8 '12 at 18:39
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 $\{x \in \mathbb{Z}^n : k \cdot x = 0\} \cong \mathbb{Z}^{n-1}$ :-( – user21706 Mar 8 '12 at 20:14
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). – darij grinberg Mar 8 '12 at 20:29
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 $S_k \cong S_{k^\prime}$. – user21706 Mar 9 '12 at 9:53

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