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 $a=0$?

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


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

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

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

hence $f_k$ is injective.

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    $\begingroup$ Are you kind to explain the notation? $M_k$, $S_0$??? $\endgroup$ – user26857 Jun 13 at 15:06
  • $\begingroup$ @user26857 graded ring and graded module $\endgroup$ – Born to be proud Jun 13 at 15:52
  • $\begingroup$ 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? $\endgroup$ – user26857 Jun 14 at 5:53
  • $\begingroup$ @user26857 $S=\bigoplus_{d=0}^{\infty}S_d,M=\bigoplus_{-\infty}^\infty M_i$. $\endgroup$ – Born to be proud Jun 14 at 8:38
  • $\begingroup$ 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. $\endgroup$ – user26857 Jun 14 at 11:31

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