map $\iota: A \rightarrow \mathbb{Q}\otimes_{\mathbb{Z}} A$, where $A$ is a finite abelian group of order n

Question

Hi everyone I'm trying to teach myself some basic module theory using Dummit & Foote's Abstract Algebra. The second example on page 363 confuses me a bit. It says:

"Let $R = \mathbb{Z}$, $S = \mathbb{Q}$ and let $A$ be a finite abelian group of order n. In this case the $\mathbb{Q}$-module $\mathbb{Q}\otimes_{\mathbb{Z}}A $ obtained by extension of scalars from the $\mathbb{Z}$-module $A$ is 0. ... In particular, the map $\iota:A \rightarrow \mathbb{Q}\otimes_{\mathbb{Z}} A$" is the zero map. "

I was a little bit confused about why $\iota$ is a zero map. To me, it's enough to show that $\forall a \in A$, the image $1\otimes a$ is modded out. So I wrote $1\otimes a=1\otimes(2n+1)a-1\otimes na-1\otimes na$, which is technically in the form of the element that is modded out. But I'm skeptical because ultimately it is just a trivial equation as $na=0, \forall a\in A$....

Any help will be appreciated!

Answer 1

Every element in $A$ has finite order. Suppose $x\in A$ has order $m$. Then $$\frac pq\otimes x=\frac{mp}{mq}\otimes x=\cdots?$$

Answer 2

$$1\otimes a =\frac{1}{n}\otimes na =0$$ In general tensoring with $\mathbb{Q}$ will kill all torsion.