Do we have $a\otimes 1=0$ if and only if $a=0$?

Question

Let $S$ be a commutative ring with $1\neq 0$ and $M$ any abelian group. For any $a\otimes 1\in M\otimes_{\Bbb Z}S$, do we have $a\otimes 1=0$ if and only if $a=0$?

Answer 1

Let $S=\bigoplus\limits_{d=0}^{\infty}S_d$ be a graded ring and $M=\bigoplus\limits_{k=-\infty}^\infty M_k$ a graded $S$-module, then $\forall k\in\Bbb Z$, $M_k$ is a $S_0$-module, for any $a\otimes 1\in M_k\otimes_{\Bbb Z}S_0$, we have $a\otimes 1=0$ if and only if $a=0$.

$\textbf{Proof}$:

$M_k\xrightarrow{f_k} M_k\otimes_{\Bbb Z}S_0\xrightarrow{g_k} M_k$

$f_k(a)=a\otimes 1, g_k(a\otimes s)=sa$,

then $(g_kf_k)(a)=a$,

hence $f_k$ is injective.

Answer 2

Computing $\mathbb{Z}/n\otimes \mathbb{Q}$, we have $$a\otimes_\mathbb{Z} q = a\otimes_\mathbb{Z} \left(n\cdot \frac{q}{n}\right) = (a\cdot n)\otimes_\mathbb{Z} \frac{q}{n} = 0,$$ so that the tensor product is the zero module.