$\mathbb{Z}_p\otimes_{\mathbb{Z}}\mathbb{Q}\cong\mathbb{Q}_p$ through colimit I've been working to compute some tensor products and am wondering if there's a clean way to see that $\mathbb{Z}_p\otimes_{\mathbb{Z}}\mathbb{Q}\cong\mathbb{Q}_p$ using the fact that tensor products (being left adjoint) commute with colimits. My first thought is to expand $\mathbb{Q}$ as a direct limit to get
$$\mathbb{Z}_p\otimes_{\mathbb{Z}}\mathbb{Q}=\mathbb{Z}_p\otimes_{\mathbb{Z}}(\text{colim}(\mathbb{Z}\rightarrow\frac{1}{2}\mathbb{Z}\rightarrow\frac{1}{6}\mathbb{Z}\rightarrow\ldots))=\text{colim}(\mathbb{Z}_p\otimes\mathbb{Z}\rightarrow\mathbb{Z}_p\otimes\frac{1}{2}\mathbb{Z}\rightarrow\mathbb{Z}_p\otimes\frac{1}{6}\mathbb{Z}\rightarrow\ldots)$$
$$=\text{colim}(\mathbb{Z}_p\rightarrow\mathbb{Z}_p\otimes\frac{1}{2}\mathbb{Z}\rightarrow\mathbb{Z}_p\otimes\frac{1}{6}\mathbb{Z}\rightarrow\ldots)$$
It seems to me like one couldn't cancel the other $\mathbb{Z}$'s because you'd eventually divide by $0$ in the colimit depending on what $p$ is. I suspect I'm certainly going about this wrong. Is there a slick way to see that this tensor product is $\mathbb{Q}_p$ using colimits? If not, what's another way to understand this tensor product?
 A: It looks like there are a few things that might be confusing you here. Here are some suggestions/hints to get you started.

*

*Remember that $\Bbb{Z}_p$ is not $\Bbb{F}_p$! In particular, $p\neq 0$ in $\Bbb{Z}_p,$ but it appears you're thinking it is based on your division by $0$ comment.

*Building off the first point, it may be helpful to remember that in fact, $\Bbb{Q}_p = \Bbb{Z}_p\left[\frac1p\right].$ For your final calculation, it might also be useful to remember how to express the localization $R\left[\frac 1f\right]$ as a colimit for $f\in R$ (this may also help you decide how you want to express $\Bbb{Q}$ as a colimit in the first place, given the answer you want to obtain).

*It might also help to remember that if $p\nmid n,$ then $\frac 1n\Bbb{Z}_p = \Bbb{Z}_p.$

*Don't forget that you can't get $\Bbb{Q}$ just by taking the colimit of
$$
\Bbb{Z}\to\frac12\Bbb{Z}\to\frac{1}{2\cdot 3}\Bbb{Z}\to\dots\to\frac{1}{2\cdot3\cdot5\cdot\dots\cdot p_n}\Bbb{Z}\to\cdots.
$$
In particular, you will never get denominators that have $2^2$ in them. Perhaps you didn't intend to take the colimit of the above, but the pattern in your sequence of maps is not entirely clear from only two terms.

*As Trebor said in a comment above, you can think of $\frac1n\Bbb{Z}$ as simply $\Bbb{Z}$ via the isomorphism of $\Bbb{Z}$-modules
\begin{align*}
\Bbb{Z}&\to\frac1n\Bbb{Z}\\
1&\mapsto\frac1n
\end{align*}
(I will let you fill in the details). In particular, this lets you express $\Bbb{Q}$ as a colimit of a diagram of the form
$$
\Bbb{Z}\xrightarrow{f_1}\Bbb{Z}\xrightarrow{f_2}\cdots
$$
for some choice of maps $f_1,f_2,\dots.$ (As in point 4 above, be careful with what each $f_i$ really is.)

See if you can take it from here!
