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This is probably well known in algebraic number theory, in particular Minkowski lattice theory, but I am given an integer vector of dimension $n$ whose components are relatively prime, meaning there is an integer linear combination of them that equals 1. Is it true that I can always find $n-1$ other integer vectors such that they form an element of $SL(n, \mathbb{Z})$? I tried to think in terms of cofactor expansion of determinant, but that was only useful for $n=2$. Thinking geometrically also bore no fruit. Thanks!

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2 Answers 2

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HINT: This is a particular case of Smith normal form theorem.

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  • $\begingroup$ Thanks for the hint. With further help of TedShifrin I got it finally. So we apply the Smith normal form to a n x 1 matrix, using product of elementary matrices P and Q, which both have determinant +/-1. Then $P'Q'$ would give the desired extension. $\endgroup$
    – John Jiang
    Dec 26, 2016 at 6:23
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You can find the construction of such a basis in M. Newman's book "Integral Matrices". It's Theorem II.$1$ on Page 13.

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