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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 $k = \langle \pi \rangle \times U, \quad \langle \pi \rangle \simeq \Bbb{Z}$ ?

Hikicianka
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  • My definition of complete field requires a metric (so $v(x)$ makes little sense if $v$ is a metric). What is your definition? – Arthur Jun 19 '19 at 10:53
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    $(k,v)$ is a complete field, if $v$ is a complete, normalized valuation ok k. The valuation $v$ is called complete if every Cauchy sequence in the $v$ - topology converges to a point in $k$. $v$ is normalized if $v(k^{\times}) = \Bbb{Z}$. It's definition from Iwasawa Local Fields. – Hikicianka Jun 19 '19 at 11:04

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