Basis of $G\otimes_{\mathbb{Z}}\mathbb{R}$ for a finitely generated abelian group $G$.

Question

Suppose that $G$ is a finitely generated abelian group, i.e. there exists a finite generating set $\{x_{1},...,x_{n}\}$ in $G$ such that every element of $G$ can be written as $\sum_{i=1}^{n}\lambda_{i}x_{i}$ for some $\lambda_{i}\in\mathbb{Z}$. Then I want to find a basis for $G\otimes_{\mathbb{Z}}\mathbb{R}$. My guess would be that the elements $x_{i}\otimes 1$ form a basis for this real vector space. It is easy to see that they span the whole space, since every element of $G\otimes_{\mathbb{Z}}\mathbb{R}$ can be written as $\sum_{i=1}^{n}x_{i}\otimes a_{i}$. My struggle lies in proving the linearly independency, i.e. suppose that $$\sum_{i=1}^{n}r_{i}(x_{i}\otimes 1) = 0,$$ then $$\sum_{i=1}^{n}x_{i}\otimes r_{i} = 0.$$ I want to conclude that $r_{i}=0$, but I don't see how. Any help is appreciated.

Answer 1

You can use the classification of finitely generated abelian groups so that $G = \mathbb{Z}^n \oplus A$, where $A$ is a direct sum of finitely many cyclic abelian groups. Note that tensoring with $\mathbb{R}$ will kill all torsion, and clearly $\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{R} \cong \mathbb{R}$, so you don't need to go through the trouble of using a basis.

Note that what you were saying was not technically correct, since your generating set might contain torsion elements, so the elements you wrote down might not be exactly a basis. The argument above does show that the basis of the real vector space will be the elementary tensors induced from the basis of the free part of the abelian group.