Prove if $F\left(\sqrt[n]{a}\right)$ is unramified or totally ramified in certain conditions

hadaasyj
2022-04-28
Answered

Prove if $F\left(\sqrt[n]{a}\right)$ is unramified or totally ramified in certain conditions

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Tatairfzk

Answered 2022-04-29
Author has **12** answers

Step 1

For (1)${x}^{n}-a$ is separable in the residue field $\frac{{O}_{F}}{\left({\pi}_{F}\right)}$ so $\frac{F\left({a}^{\frac{1}{n}}\right)}{F}$ is automatically unramified.

Note that Hensel lemma gives that${\zeta}_{q-1}\in F\left({a}^{\frac{1}{n}}\right)$ where q is the cardinality of the residue field, and ${x}^{n}-a$ is separable in the residue field so Hensel lemma again gives that ${a}^{\frac{1}{n}}\in F\left({\zeta}_{q-1}\right)$ and $F\left({a}^{\frac{1}{n}}\right)=F\left({\zeta}_{q-1}\right)$ .

For (2) take$nl+mv\left(a\right)=1$ , let $b={a}^{m}{\pi}_{F}^{nl},\text{}v\left(b\right)=1,\text{}F\left({a}^{\frac{1}{n}}\right)=F\left({b}^{\frac{1}{n}}\right)$ and ${x}^{n}-b$ is Eisenstein over $O}_{F$ so $\frac{F\left({b}^{\frac{1}{n}}\right)}{F}$ has degree n and $v\left({b}^{\frac{1}{n}}\right)=\frac{1}{n}$ , it is totally ramified.

For (1)

Note that Hensel lemma gives that

For (2) take

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