# Show that $1\otimes (1,1,\ldots)\neq 0$ in $\mathbb{Q} \otimes_{\mathbb{Z}} \prod_{n=2}^{\infty} (\mathbb{Z}/n \mathbb{Z})$.

Show that $1\otimes (1,1,\ldots)\neq 0$ in $\mathbb{Q} \otimes_{\mathbb{Z}} \prod_{n=2}^{\infty} (\mathbb{Z}/n \mathbb{Z})$.

Here’s what I tried:

If $1\otimes (1,1,\ldots)= 0$, then $1\otimes (1,1,\ldots)= (-1)\otimes (1,1,\ldots)=1\otimes (-1,-1,\ldots)$, but $1\neq -1$ in $\mathbb{Z}/n\mathbb{Z}$ for all $n$.

I am not sure that this is valid…

#### Solutions Collecting From Web of "Show that $1\otimes (1,1,\ldots)\neq 0$ in $\mathbb{Q} \otimes_{\mathbb{Z}} \prod_{n=2}^{\infty} (\mathbb{Z}/n \mathbb{Z})$."

No, your argument isn’t correct: if $1\otimes x=1\otimes y$ then not necessarily $x=y$.

I’d do this as follows: if $M$ is a $\mathbb Z$-module, then $\mathbb Q\otimes_{\mathbb Z}M\simeq S^{-1}M$, where $S=\mathbb Z\setminus\{0\}$. The isomorphism is given by $\dfrac ab\otimes x\mapsto\dfrac{ax}{b}$. This way $1\otimes(1,1,\dots)$ corresponds to the fraction $\dfrac{(1,1,\dots)}{1}$, and if this equals $\dfrac01$ then there is $m\in\mathbb Z$, $m\ne0$ such that $m(1,1,\dots)=(0,0,\dots)$, that is, $m=0$ in $\mathbb Z/n\mathbb Z$ for all $n\ge2$. This shows that $m=0$, a contradiction.