is the idele class group a flat Z-module? If $K$ is a local or global field and I define $C_K$ to be the idele class group if $K$ is global and I let $C_K=K^{\times}$ is $K$ is local, then as a $\mathbb{Z}$-module will it be flat? i.e. If $$ 0 \longrightarrow A \longrightarrow B \longrightarrow X \longrightarrow 0$$ is exact will $$ 0 \longrightarrow A \otimes C_K \longrightarrow B \otimes C_K \longrightarrow X \otimes C_K \longrightarrow 0$$  always be exact?
Thank you
 A: For a $\mathbf{Z}$-module $M$, flat is the same as torsion-free. So if $K$ is a local field of characteristic different from $2$, then $1\neq -1$, so $-1\in K^\times$ is a non-zero $\mathbf{Z}$-torsion element and thus $K^\times$ is not $\mathbf{Z}$-flat. If $K$ is global, then for each prime $v$ of $K$, $K_v^\times$ injects into the idele class group of $K$, so again, you'll have non-zero torsion as long as the characteristic of $K$ is not $2$. 
If the characteristic of $K$ is $2$, then it suffices (by the injectivity of local multiplicative groups into the idele class group) to treat the case of $K$ local. Then I think the answer is: $K^\times$ is $\mathbf{Z}$-torsion-free, i.e., has no non-trivial roots of unity, if and only if the residue field of $K$ is of order $2$. Indeed, the choice of a uniformizer for the ring of integers of $K$ gives a $k$-algebra isomorphism $K\cong k((T))$ with $k$ the residue field (this follows from the structure theory of complete discrete valuation rings of equal positive characteristic). If $k\neq\mathbf{F}_2$, then $k$ has non-trivial roots of unity, and hence so does $K$. If $k=\mathbf{F}_2$, then there can be no non-trivial roots of unity because reduction $K^\times\rightarrow k^\times$ is bijective on roots of unity of order prime to $2$ in $K$.
EDIT: My answer above does not rule out the possibility when $K$ has characteristic $2$ and residue field $\mathbf{F}_2$ that the idele class group has torsion which does not come from the multiplicative groups of a single completion.  What this would mean (unless I'm mistaken) is that there is a number $\alpha\in K^\times$ ($K$ is global now) which is an $n$-th power in every completion for some $n\geq 1$ but which is not an $n$-th power in $K$...so I guess the answer is furnished by the Grunwald-Wang theorem. Perhaps somebody who knows this theorem (mainly its exceptional cases) better than I do can comment. My understanding is that it says that ``usually" such an $\alpha$ cannot exist, but that there are cases (probably involving the prime $2$) where it can. 
SECOND EDIT: Actually I'm not sure Grunwald-Wang will solve this, because even in cases where we know that $\alpha\in K^\times$ is an $n$-th power in $K$, there might still be a non-trivial element of $C_K$ whose $n$-th power is $\alpha$ (other than the obvious trivial element which exists by our assumption that $\alpha\in (K^\times)^n$). This is actually the case in my answers above involving roots of unity. The element $1\in K^\times$ certainly has a square root in $K$, but it also has a non-trivial square root in $C_K$ as long as the characteristic is not $2$. 
