I have the functor $~~~\Bbb{Q}\otimes_\Bbb{Z}-~~~$ from $\Bbb{Z}$-modules to $\Bbb{Q}$-vectorspaces. If I consider a left exact sequence of $\Bbb{Z}$-modules $$0\rightarrow A\stackrel{f}{\rightarrow} B\rightarrow C$$and then apply the functor I get $$0\rightarrow \Bbb{Q}\otimes_\Bbb{Z}A\stackrel{id_\Bbb{Q}\otimes f}{\rightarrow} \Bbb{Q}\otimes_\Bbb{Z}B\rightarrow \Bbb{Q}\otimes_\Bbb{Z}C$$ First I only want to check that $\ker(id_\Bbb{Q}\otimes f)=im(0\rightarrow \Bbb{Q}\otimes_\Bbb{Z}A)$. So more precisly I want to check that $\ker(id_\Bbb{Q}\otimes f)=0$. But my problem is that I don't see how an element looks like in $\Bbb{Q}\otimes_\Bbb{Z}A$. I mean I know that in $u\otimes v\in U\otimes V$ where $(u,v)\in U\times V$. But is it here the same so does an element looks like $q\otimes a$ for $(q,a)\in \Bbb{Q}\times A$? Because in the lecture our prof somehow wrote something like $\sum_i q_i\otimes a_i$. Why do we need a sum? Is this more useful to prove injectivity? And is my idea with $q\otimes a$ wrong or does it also works with this representation?
It would be nice if someone could explain this to me a bit in detail since I'm really confused.
Thanks for your help.
What @EricWofsey says in the comments is true, and yet in the special case of $\mathbb Q \otimes_{\mathbb Z} A$ we can rewrite a general element in the form $q \otimes a$ by "adding fractions".
Consider a general element $\sum_i q_i \otimes a_i$ where $q_i = \frac{n_i}{d_i}$ with $n_i,d_i \in \mathbb Z$ and $d_i > 0$. Let $d$ be the "least common denominator", that is, the least common multiple of the $d_i$'s. Let $p_i = d / d_i$ so $q_i = \frac{n_i p_i}{d}$. Then $$\sum_i q_i \otimes a_i = \sum_i \frac{n_i p_i}{d} \otimes a_i = \sum_i \frac{1}{d} n_i p_i \otimes a_i = \frac{1}{d} \otimes \sum_i n_i p_i a_i $$ and clearly $\sum_i n_i p_i a_i \in A$.