I have two questions about tensoring a $\mathbf{Z}$-module with the additive group $\mathbf{Q}_p/\mathbf{Z}_p$.
Is it true that if $T$ is a finite abelian group, then we have $$T \otimes_{\mathbf{Z}} \mathbf{Q}_p/\mathbf{Z}_p = 0?$$ I've heard this fact being thrown around a few times, but I can't see why it's true. Here's why I find it confusing: if we write $\mathbf{Q}_p/\mathbf{Z}_p = \lim_{\rightarrow n }\mathbf{Z}/p^n \mathbf{Z}$ and use the fact that tensor products commute with direct limits, we should get that $$T \otimes_{\mathbf{Z}} \mathbf{Q}_p/\mathbf{Z}_p \cong p\text{-Sylow subgroup of }T,$$ where above we used the fact that $T \otimes \mathbf{Z}/p^n\mathbf{Z}$ is isomorphic to the Sylow $p$-subgroup of $T$ for all sufficiently large values of $n$. But of course, the Sylow $p$-subgroup of $T$ is certainly not always zero. Why is it that the above tensor product should always be zero? Is there a missing step here?
How do you calculate the tensor product $$\mathbf{Q}^{\times} \otimes_{\mathbf{Z}} \mathbf{Q}_p/\mathbf{Z}_p,$$ where $\mathbf{Q}^{\times}$ denotes the multiplicative group of $\mathbf{Q}$? In terms of what I've tried, I found the tensor product $\mathbf{Q}^{\times} \otimes \mathbf{Z}/p^n\mathbf{Z}$ and then tried taking the direct limit over all $n$, but the above tensor product evaluates to the rather unhelpful expression: $$\mathbf{Q}^{\times} \otimes \mathbf{Z}/p^n\mathbf{Z} \cong \mathbf{Q}^{\times} / (\mathbf{Q}^{\times})^{p^n}.$$ While the group $\mathbf{Q}^{\times} / (\mathbf{Q}^{\times})^{p^n}$ is pretty understandable, I'm not sure how to express the direct limit of these groups in a helpful way.
Thank you for the help everyone! I greatly appreciate it.
It's not true that the direct limit here is the $p$-Sylow subgroup. Keep in mind that the maps $\mathbb{Z}/p^n\mathbb{Z}\to\mathbb{Z}/p^{n+1}\mathbb{Z}$ in the direct system are multiplication by $p$. So after tensoring with $T$, you get a system with maps $T/p^nT\to T/p^{n+1}T$ that are given by multiplication by $p$. For sufficiently large $n$, $T/p^nT$ will be just the $p$-Sylow subgroup. However, the maps in the system are not the identity map but rather multiplication by $p$ from the $p$-Sylow subgroup to itself. Every element is killed by some power of $p$ so ends up becoming $0$ when it maps into the direct limit, and so the direct limit ends up being trivial.
More generally, if $T$ is a torsion abelian group and $D$ is a divisible abelian group, then $T\otimes D=0$. You can prove this quite directly: for any $t\in T$ and $d\in D$, let $n$ be the order of $t$. Since $D$ is divisible, there is an element $x\in D$ such that $nx=d$. Then $t\otimes d=t\otimes nx=nt\otimes x=0$. So every simple tensor in $T\otimes D$ is $0$ and thus $T\otimes D=0$.
As for computing $\mathbb{Q}^\times\otimes\mathbb{Q}_p/\mathbb{Z}_p$, this is easy by just understanding the structure of $\mathbb{Q}^\times$. There is an isomorphism $\mathbb{Q}^\times\cong \{1,-1\}\oplus\mathbb{Z}^{\oplus\mathbb{N}}$ where the summands in the second factor correspond to the primes (this is just unique factorization). When you tensor with $\mathbb{Q}_p/\mathbb{Z}_p$, that kills the finite group $\{1,-1\}$ and so you end up just getting a direct sum of infinitely many copies of $\mathbb{Q}_p/\mathbb{Z}_p$, one for each prime. Explicitly, the universal bilinear map $\mu:\mathbb{Q}^\times\times \mathbb{Q}_p/\mathbb{Z}_p\to(\mathbb{Q}_p/\mathbb{Z}_p)^{\oplus\mathbb{N}}$ can be described as follows. List the primes as $(p_n)_{n\in\mathbb{N}}$. Then $\mu(\pm\prod_n p_n^{d_n},x)$ is the sequence $(d_nx)_{n\in\mathbb{N}}\in(\mathbb{Q}_p/\mathbb{Z}_p)^{\oplus\mathbb{N}}$.