Let $K$ be a local field and $K'$ be it's finite extension. And there exists $a∈K'$ such that $v(a)$ is not integer.
Then, ramification index $e$ of $K'/K$ is at least 2, in other words, $K'/K$ is ramified extension.
I heard this holds from definition of ramification index, https://www.lmfdb.org/knowledge/show/lf.ramification_index.
But I'm having trouble. I tried to use another characterization of ramification index, but in vain.
Thank you in advance.
I assume that $v$ is normalized in such a way so that the value group of $v$ on $K$ is $\Bbb Z$. Let $\pi$ be a prime element of $K$. Note that $K'/K$ is unramified iff $\pi$ is also a prime of $K'$. If $K'/K$ were unramified $\pi$ would still be a prime of $K'^\times$, so every element $a$ of $K'$ has a decomposition $\pi^mu$ for some unit $u\in\mathcal O_{K'}^\times$, in particular $m=v(a)\in\Bbb Z$. Thus, if some $a\in K'^\times$ has non-integral valuation the extension has to be ramified.
The same argument shows that if the ramification index is $e$, then the value group of $v$ on $K'$ is $\frac{1}{e}\Bbb Z$.