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

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 $$a=0$$?

• Try computing $\mathbb{Z}/n \otimes_{\mathbb{Z}} \mathbb{Q}$. – rogerl Jun 13 at 12:11
• See for example math.stackexchange.com/questions/1902109/… – Arnaud D. Jun 13 at 12:16
• @ArnaudD. If $M$ is a $S$-module, do we have $a\otimes 1=0$ if and only if $a=0$? – Born to be proud Jun 13 at 12:30
• @rogerl If $M$ is a $S$-module, do we have $a\otimes 1=0$ if and only if $a=0$? – Born to be proud Jun 13 at 12:31

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.

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.

• Are you kind to explain the notation? $M_k$, $S_0$??? – user26857 Jun 13 at 15:06
• @user26857 graded ring and graded module – Born to be proud Jun 13 at 15:52
• And what has this to do with your question? I can understand that you could be interested in the case when M is an S-module, but from where came up with the grading? – user26857 Jun 14 at 5:53
• @user26857 $S=\bigoplus_{d=0}^{\infty}S_d,M=\bigoplus_{-\infty}^\infty M_i$. – Born to be proud Jun 14 at 8:38
• I know what's a graded ring and a graded module. Btw, in this case every $M_k$ is an $S_0$-module. Why say 'If"? And I still don't understand what you proved. – user26857 Jun 14 at 11:31