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Let $K$ be a number field and $v$ be it's one of $K$'s non-archimedian valuation. Then, I would like to prove $K_v(a^{1/m}) /K_v$ is unramified if only if $v(a)≡0 \pmod m$.

This is from Silverman's 'the arithmetic of elliptic curves', p213. I know unramified extension of local field is in bijection with extensions of the residue field. Thus, the unramified extension is generated by roots of unity of order prime to the character of residue field of local field.

But I don't have tactics to judge given extension is unramified or not.

Thank you in advance.

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If $v(a)\not\equiv 0\pmod m$, ramification is easy: just consider the valuation of the element $a^{1/m}$.

The converse is a little subtler than you make it seem, and depending on how exactly you phrase it, it need not be true: for instance, if $m=p$ coincides with the residue characteristic of $v$, $a=1$ and we interpret $a^{1/m}$ as a primitive $p$-th root of unity, the extension will be ramified.

Silverman includes a number of assumptions which exclude this case: he takes $v\not\in S$, and a couple paragraphs above it is stated that we assume this imples $v(m)=0$. Assuming this is the case and that $v(a)\equiv 0\pmod m$, note that, by multiplying $a$ by some $m$-th power, we may assume $v(a)=0$. Now, the polynomial $X^m-a$ has discriminant $\pm m^ma^{m-1}$, which under all these assumptions has valuation zero. This implies the discriminant of the extension $K_v(\sqrt[m]{a})/K_v$ is a $v$-adic unit, which implies the extension is unramified.

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  • $\begingroup$ May I ask two questions ? ①Sorry to bother, but could you tell me the first paragraph? From $v(a^{1/m})＝ v(a)/m$ is not zero mod $m$, why can I say ramification? ② Let $L/K$ be extension of local field $K$, then $L/K$ is unramified is equivalent to discriminant of $L/K$ is $v-$adic unit ? $\endgroup$
    – Duality
    Commented Sep 14, 2021 at 11:30
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    $\begingroup$ 1. If $v(a)\neq 0\pmod m$, then $v(a^{1/m})$ is not an integer. Existence of such elements is more or less the definition of ramification. 2. Yes $\endgroup$
    – Wojowu
    Commented Sep 14, 2021 at 11:53
  • $\begingroup$ I'm confused. $v$ is valuation on $K$, and $a^{1/m}$ is not necessarily in $K$, what do you mean by $v(a^{1/m})$? Are you thinking some kind of extension of valuation ? $\endgroup$
    – Duality
    Commented Sep 15, 2021 at 12:18
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    $\begingroup$ @Nekojiru A valuation on a local field uniquely extends to all its algebraic extensions. It is not uncommon to denote it by the same symbol. $\endgroup$
    – Wojowu
    Commented Sep 15, 2021 at 12:46
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    $\begingroup$ @stillconfused I'm assuming you are referring to the last part. The discriminant of the extension divides the discriminant of any of the elements of $O_L$ which generate $L/K$. Since the discriminant of $a^{1/m}$ is a unit, so must be the discriminant of the extension. $\endgroup$
    – Wojowu
    Commented Jul 26, 2023 at 18:28

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