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Let $(x_1,\dots,x_r)$ be a non-zero element of $\mathbb{Z}^r$, and let $h$ be the highest common factor of $x_1, \dots, x_r$. Show that there is an isomorphism $\mathbb{Z}^r \to \mathbb{Z}^r$ taking $(x_1,\dots,x_r)$ to $(1, 0, 0,\dots,0)$ if and only if $h=1$.


One direction isn't too bad; if $\phi$ is such an isomorphism then $ \phi\big((x_1,\dots,x_r)\big) = (1,0,\dots,0) $ $ \phi\big(h(\tfrac{x_1}{h},\dots,\tfrac{x_r}{h})\big) = (1,0,\dots,0) $ $ h \phi\big((\tfrac{x_1}{h},\dots,\tfrac{x_r}{h})\big) = (1,0,\dots,0). $ But then $h|1$, so we must have $h=1$. I really have no idea how to go about proving the other direction... so any help would be appreciated.

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    Homomorphisms can only increase the (absolute value of the) gcd, so isomorphisms must preserve it.2011-05-29

3 Answers 3

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Let $H$ be the cyclic subgroup of $\mathbb{Z}^r$ generated by $(x_1,\ldots,x_r)$. If $h=1$, then $t(y_1,\ldots,y_r) = (x_1,\ldots,x_r)$ implies $t = \pm{1}$, and so $\mathbb{Z}^r/H$ is torsion-free and hence free abelian. So $H$ has a complement in $\mathbb{Z}^r$, which must be isomorphic to $\mathbb{Z}^{r-1}$, and the result follows easily.

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If there is an isomorphism $h$ mapping $(x_1,\dots,x_n)$ to $(1,0,\dots,0)$, its inverse $h^{-1}$ maps $(1,0,\dots,0)$ to $(x_1,\dots,x_n)$, and the matrix $A$ of $h^{-1}$ with respect to the canonical basis of $\mathbb Z^n$ then has $(x_1,\dots,x_n)$ as its first column. Since $A$ must have determinant $1$ or $-1$, the $\gcd$ of each of its columns is $1$.

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Just go through the extended Euclidean algorithm for expressing $h$ as an integer linear combinantion of $x_1,\ldots,x_r$. Try the case $r=2$ and see how $(x_1,x_2)$ is reduced to $(h,0)$.

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    @Fadah, $\phi(1,0)=(a,-x_2)$, $\phi(0,1)=(b,x_1)$, for instance. Note that the determinant is 1, which means $\phi$ is invertible.2011-05-30