Tensor product and extension of scalars We have that $\mathbb{Z}^n\otimes_{\mathbb{Z}}\mathbb{Q}\cong \mathbb{Q}^n$. Now, consider an abelian group of the form
$$G = \mathbb{Z}^n\oplus\mathbb{Z}/p_1\mathbb{Z}\oplus\dots\oplus \mathbb{Z}/p_n\mathbb{Z}$$
Is the tensor product $G\otimes_{\mathbb{Z}}\mathbb{Q}$ still isomorphic to $\mathbb{Q}^n$? If so are the torsion elements in $G$ completely undetectable after tensoring by $\mathbb{Q}$?
 A: Yes and yes. This is a straightforward calculation using the fact that tensor products preserve colimits; we get that
$$A \otimes \mathbb{Q} \cong \mathbb{Q}^n \oplus \bigoplus_{i=1}^n \mathbb{Q}/p_i \mathbb{Q}$$
(using that tensor products preserve first direct sums and then cokernels). But $p_i \mathbb{Q} = \mathbb{Q}$ so $\mathbb{Q}/p_i\mathbb{Q} = 0$. (I've renamed the abelian group to $A$.)
In general, the kernel of $A \to A \otimes \mathbb{Q}$ is exactly the torsion subgroup of $A$. This has the following fun interpretation in homological terms. Tensoring any abelian group $A$ with the short exact sequence $0 \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \to 0$ produces a Tor exact sequence
$$0 \to \text{Tor}_1(A, \mathbb{Q}/\mathbb{Z}) \to A \to A \otimes \mathbb{Q} \to A \otimes \mathbb{Q}/\mathbb{Z} \to 0$$
(since $\mathbb{Q}$ is flat) which shows that the kernel of $A \to A \otimes \mathbb{Q}$ is exactly $\text{Tor}_1(A, \mathbb{Q}/\mathbb{Z})$, which is therefore the torsion subgroup of $A$ (this is where "Tor" gets its name from). This calculation can also be done using the slightly easier fact that $\text{Tor}_1(A, \mathbb{Z}/n\mathbb{Z})$ is the $n$-torsion subgroup of $A$ (which follows from using the Tor exact sequence given by tensoring with $0 \to \mathbb{Z} \xrightarrow{n} \mathbb{Z} \to \mathbb{Z}/n\mathbb{Z} \to 0$), then taking a filtered colimit over all $n$ and using the fact that Tor preserves filtered colimits in each variable.
