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Let $f(x)$ be a polynomial of degree $m$ over $\mathbb{Q}_{p}$ with all roots $r_{i}$ such that $\operatorname{ord}_{p} r_{i} = \frac{1}{p}$. Why does $p$ have to divide the ramification index of $\mathbb{Q}_{p}(r_{i})/\mathbb{Q}_{p}$?

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1 Answer 1

If you extend the $p$-adic valuation $\operatorname{ord}_p$ to a finite degree extension field $K/\mathbb{Q}_p$, then its image -- the value group -- is a subgroup of $\mathbb{Q}$ of the form $\frac{1}{e} \mathbb{Z}$ for a unique positive integer $e$, which is called the ramification index of the extension.

With this definition the answer to your question is almost immediate: you are assuming that you have an element $r_i$ of $K$ with $\operatorname{ord}_p(r_i) = \frac{1}{p}$. Therefore the value group of $K$ contains $\frac{1}{p}$, and $\frac{1}{e} \mathbb{Z}$ contains $\frac{1}{p}$ if and only if $p$ divides $e$.

(If you are working from some other definitions, let us know...)

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