If $\{e_1,...,e_n\}$ is a $\Bbb Z$-basis of $\Bbb Z^n$ then $\{e_1\otimes 1,...,e_n\otimes 1\}$ is a $\Bbb Q$-basis of $\Bbb Z^n\otimes \Bbb Q$

Question

Suppose $\{e_1,...,e_n\}$ is a basis of $\Bbb Z^n$ over $\Bbb Z$ (not necessarily the standard basis). It is well-known that $\Bbb Z^n\otimes \Bbb Q$ is an $n$-dimensional $\Bbb Q$-vector space. I want to prove that $\{e_1\otimes 1,...,e_n\otimes 1\}$ is a basis of $\Bbb Z^n\otimes \Bbb Q$. My attempt is as follows:

It suffices to show that $\{e_1\otimes 1,...,e_n\otimes 1\}$ spans $\Bbb Z^n\otimes \Bbb Q$ over $\Bbb Q$, but since $\Bbb Z^n\otimes \Bbb Q$ is generated by elements of the form $a\otimes b$ with $a\in \Bbb Z^n$ and $b\in \Bbb Q$, it suffices to show that $\{e_1\otimes 1,...,e_n\otimes 1\}$ generates any element of the form $a\otimes b$. Given such an element, we may write $a=c_1e_1+...+c_ne_n$ for some $c_i\in \Bbb Z$. Then $a\otimes b=\sum_i bc_i (e_1\otimes 1)$ and the proof is complete.

Is my proof correct? Or is there another way to prove this?

Answer 1

Your proof is entirely correct, though you might want to be more explicit at every step:

1. Every element of $\Bbb{Z}^n\otimes\Bbb{Q}$ is of the form $$z=\sum_{k=1}^mx_i(a_i\otimes b_i),$$ for some positive integer $m$, some $x_1,\ldots,x_m\in\Bbb{Z}$, some $a_1\ldots,a_m\in\Bbb{Z}^n$ and some $b_1,\ldots,b_m\in\Bbb{Q}$.
2. Expressing the rational number $b_i$ with one common denominator shows that $$z=\sum_{k=1}^mx_i(c_i\otimes\tfrac{p_i}q),$$ for some positive integer $m$, some $x_1,\ldots,x_m\in\Bbb{Z}$, some $c_1\ldots,c_m\in\Bbb{Z}^n$, some $p_1,\ldots,p_m\in\Bbb{Z}$. and some nonzero integer $q$.
3. Because the tensor product is $\Bbb{Z}$-bilinear you can rewrite the above as $$z=\sum_{k=1}^m((x_ip_ic_i)\otimes\tfrac{1}q)=\left(\sum_{k=1}^m(x_ip_ic_i)\right)\otimes\frac1q,$$ so every element of $\Bbb{Z}^n\otimes\Bbb{Q}$ is of the form $a\otimes\tfrac1q$ for some $a\in\Bbb{Z}^n$ and some nonzero integer $q$.
4. A subset of $\Bbb{Z}^n\otimes\Bbb{Q}$ generates $a\otimes\tfrac1q$ over $\Bbb{Q}$ if and only if it generates $a\otimes1$.
5. Because the $e_i$ generate $\Bbb{Z}^n$ over $\Bbb{Z}$, the $e_i\otimes1$ generate all elements of the form $a\otimes1$ over $\Bbb{Z}$, and hence also over $\Bbb{Q}$.
6. This shows that $\{e_i\otimes1:\ 1\leq i\leq n\}$ is a generating set of $n$ elements over $\Bbb{Q}$ for the $n$-dimensional $\Bbb{Q}$-vector space $\Bbb{Z}^n\otimes\Bbb{Q}$, and therefore it is a basis.