$G$ be a free abelian group of rank $k$ , let $S$ be a subset of $G$ linearly independent over $\mathbb Z$ , then is it true that $|S| \le k$?

Question

Let $G$ be a free abelian group of rank $k$ , let $S$ be a subset of $G$ linearly independent over $\mathbb Z$ , then is it true that $|S| \le k$ ?

score 1 · Answer 1 · answered Sep 02 '16 at 16:26

Consider $G\otimes_ZQ$ it is a $Q$-vector space of dimension $k$. Let $ZS$ the subgroup of $G$ generated by $S$. The dimension of $ZS\otimes_ZQ$ is the cardinality of $S$, which is inferior to $k$ as a subvector space of a $k$-dimensional vector space.

score 0 · Accepted Answer · answered Sep 02 '16 at 16:26

0

Yes, this is true. Introducing an independent generating set, we can just as well assume we are talking about $\mathbb{Z}^k$ and we can consider embedded in $\mathbb{Q}^k$.

Thus, a set $S$ with $|S| \gt k$ will be dependent over the rationals (I assume you know the result for vector spaces). Take this relation and multiply by an integer to clear denominators. You then have a dependence over the integers.

answered Sep 02 '16 at 16:26

quid

42,135

thanks , it is really simple ... do you think there's an elementary argument for this http://math.stackexchange.com/questions/1912076/g-be-a-free-abelian-group-of-rank-k-let-s-be-a-subset-of-g-that-genera?rq=1 too ? ( the argument here has no immediate analogue in the generating case as far as I can see ... but still ) – Sep 02 '16 at 16:34

$G$ be a free abelian group of rank $k$ , let $S$ be a subset of $G$ linearly independent over $\mathbb Z$ , then is it true that $|S| \le k$?

2 Answers2