# $\mathbb{Z}_p\otimes_{\mathbb{Z}}\mathbb{Q}\cong\mathbb{Q}_p$ 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?

• Hint: $1/2\mathbb Z$ is just a sloppy way to write $\mathbb Z$. The coefficient is actually on the maps of the colimit, instead of the objects. Jun 1, 2022 at 3:52
• @Trebor I don't think "sloppy" is the correct word here. Jun 4, 2022 at 16:16
• @QiZhu Note that "sloppy" is not criticizing. We're all just using the notation as a tradition, and it is normal to use sloppy notation. As long as it's not confusing or ambiguous it's fine. Jun 4, 2022 at 17:54
• @Trebor Sure, I guess I just don't get what you want to convey with the word "sloppy" here because I don't see what's supposed to be "sloppy" here. (But of course in the end it doesn't matter for the arguments!) Jun 4, 2022 at 18:24

1. 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.
2. 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).
3. It might also help to remember that if $$p\nmid n,$$ then $$\frac 1n\Bbb{Z}_p = \Bbb{Z}_p.$$
4. 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.
5. 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.)