Why does unit group of valuation ring contains $K^{\times}$ Let $K$ be a number field, $K_v$ be a completion of $K$ by non-archimedean valuation $v$ of $K$ and $U_v$ be a unit group of valuation ring $R_v$.
Then why $K^{\times}\subset U_v$?(for almost all $v$)
This implies principal idele is idele.
Edit this is wrong

Let $K = \Bbb{Q}$ then $$\Bbb{A_Q} = \{ a_\infty\prod_p a_p \in \Bbb{R} \times \prod_p \Bbb{Q}_p, \text{ for all but finitely many } p, a_p \in \Bbb{Z}_p\}$$ 
it is a commutative unital ring whose $1$ is $1_\infty \prod_p 1_p$ (with $1_p$ the $1$ of $\Bbb{Q}_p$) and
$$\Bbb{A_Q}^\times = \{ a \in \Bbb{A_Q}, \exists b \in \Bbb{A_Q}, ab = 1\}$$
$$=\{ a_\infty\prod_p a_p \in \Bbb{R}^\times \times \prod_p \Bbb{Q}_p^\times, \text{ for all but finitely many } p, a_p \in \Bbb{Z}_p^\times\}$$
For $x \in \Bbb{Q}$ let $x_p$ its image in $\Bbb{Q}_p$ then the diagional embedding $x \mapsto x_\infty \prod_p x_p$ is a ring homomorphism $\Bbb{Q \to A_Q}$ and it sends $\Bbb{Q^\times \to A_Q^\times}$.
With $x = \frac{n}{m}$ we have $x \in \Bbb{Z}_p^\times$ whenever $p \nmid nm$.
The other thing to know is that $\prod_p \Bbb{Z}_p = \varprojlim \Bbb{Z} /(n)$ is the set of limits of sequences of integers that converge $\bmod n$ for every $n$.
