Let $(k,v)$ be any complete field and $U = \{x | x \in k^{\times}, v(x)=0 \}$. Let $\langle \pi \rangle $ denote the cyclic subgroup of $k^{\times}$, generated by a prime element $\pi$. My question is
why isomorpism $k^{\times}/U \simeq v(k^{\times}) = \Bbb{Z}$ implies that $k = \langle \pi \rangle \times U, \quad \langle \pi \rangle \simeq \Bbb{Z}$ ?