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Oct 12, 2017 at 11:44 answer added Jan-Christoph Schlage-Puchta timeline score: 0
Oct 9, 2017 at 23:11 comment added user6976 @DavidHandelman: I had the same idea (see my answer). But this does not seem much easier than the Smith normal form because the GCD is the largest diagonal element of the Smith form.
Oct 9, 2017 at 22:45 comment added David Handelman Take the gcd of all the determinants of the square matrices coming from $k$ vectors at a time. This will be 1 iff the set generates the standard copy of ${\bf Z}^k$. Worst possible case is that $n$ choose $k$ subsets will be required, but if it is a generating set, far fewer subsets should be required.
Oct 9, 2017 at 17:02 answer added user6976 timeline score: 0
Oct 9, 2017 at 0:20 answer added Richard Stanley timeline score: 9
Oct 8, 2017 at 23:57 history asked Mikhail Tikhomirov CC BY-SA 3.0