Is $\mathbb{Q}\otimes_\mathbb{Z}\mathbb{Z}/n=0$?

Is $\mathbb{Q}\otimes_\mathbb{Z}\mathbb{Z}/n=0$?

Because \begin{equation}\frac{a}{b}\otimes_\mathbb{Z}1=\frac{na}{nb}\otimes_\mathbb{Z}1=\frac{a}{nb}\otimes_{\mathbb{Z}}n=\frac{a}{nb}\otimes_\mathbb{Z}0=0?\end{equation}

Solution 1:

Yes. In general, $\mathbb Q\otimes_{\mathbb Z} A$ will be a vector space over $\mathbb Q$.

In general, every element $b\in B\otimes_{\mathbb Z}\mathbb Z/\left<n\right>$ with have order a divisor of $n$ - that is, $nb=0$.

So you have a vector space over $\mathbb Q$ in which every element has finite order. This can only be the trivial group.

This exact example is used in the Wikipedia page for tensor products.