Show $\mathbb{Q}\otimes_{\mathbb{Z}}\mathbb{Q}\cong\mathbb{Q}\otimes_{\mathbb{Q}}\mathbb{Q}$ as left $\mathbb{Q}$-modules. I'm trying to work this problem out of Dummit and Foote, and I'm having a bit of confusion with the notation. I thought the fact that there is a $\mathbb{Z}$ with the tensor symbol in the first module means we're considering $\mathbb{Q}\otimes\mathbb{Q}$ as a module over $\mathbb{Z}$, and the $\mathbb{Q}$ with the tensor on the right means we are considering it as a module over $\mathbb{Q}$? What am I misunderstanding?
 A: As Tobias Kildetoft points out, $\mathbb{Q}$ is both a $(\mathbb{Q},\mathbb{Z})$-bimodule and a $(\mathbb{Z},\mathbb{Q})$-bimodule making $\mathbb{Q}\otimes_{\mathbb{Z}}\mathbb{Q}$ a $(\mathbb{Q},\mathbb{Q})$-bimodule. Also, $\mathbb{Q}$ is a $(\mathbb{Q},\mathbb{Q})$-bimodule, making $\mathbb{Q}\otimes_{\mathbb{Q}}\mathbb{Q}$ a $(\mathbb{Q},\mathbb{Q})$-bimodule.
Now, I claim that $\mathbb{Q}\otimes_{\mathbb{Z}}\mathbb{Q}$ and $\mathbb{Q}\otimes_{\mathbb{Q}}\mathbb{Q}$ are both isomorphic to $\mathbb{Q}$ as $(\mathbb{Q},\mathbb{Q})$-bimodules. Indeed, let
$$\phi:\mathbb{Q}\to \mathbb{Q}\otimes_{\mathbb{Z}}\mathbb{Q}$$ be the map
$\phi(r)=1\otimes r$. Then, 
\begin{align*}
\phi(\frac{m}{n}r)&=1\otimes\frac{m}{n}r\\
&=m\otimes\frac{1}{n}r\\
&=\frac{mn}{n}\otimes\frac{1}{n}r\\
&=\frac{m}{n}\otimes\frac{n}{n}r\\
&=\frac{m}{n}(1\otimes r)\\
&=\frac{m}{n}\phi(r)
\end{align*}
Also, $\phi(r\frac{m}{n})=\phi(r)\frac{m}{n}$, so $\phi$ is a bimodule map. It is clearly injective and a computation analogous to the one above shows that it is surjective since 
$$\sum_i\frac{m_i}{n_i}\otimes r_i=\sum_i1\otimes\frac{m_i}{n_i}r_i=1\otimes\sum_i\left(\frac{m_i}{n_i}r_i\right).$$
The analogous map $\psi:\mathbb{Q}\to \mathbb{Q}\otimes_{\mathbb{Q}}\mathbb{Q}$ is also clearly a bimodule isomorphism.
