# Non trivial valuation over a finite extension $F/\mathbb{Q}_p$ is equivalent to the one induced by $v_p$

I'm stuck on a question in an exercise on p-adic numbers. Actually, let $$p$$ a prime number, $$F/\mathbb{Q}_p$$ a finite extension of fields, $$\mathcal{O}$$ the integral closure of $$\mathbb{Z}_p$$ on $$F$$, $$\mathfrak{p}$$ the maximal ideal of $$\mathcal{O}$$.

We have already shown that : $$\forall l \in \mathbb{N}_{\geq 2}$$, $$\gcd(l,p)=1 \;\forall x \in \mathfrak{p} \; \exists y \in \mathcal{O} \quad y^l=1+x$$, using Hensel's lemma.

Now, the next question is :

Deduce from this that for any non trivial valuation $$w$$ on $$F$$, if we are noting $$\mathcal{O}_w$$ its valuation ring, we have : $$\mathcal{O} \subset O_w$$.

(We know as well that if we are noting : $$v(x) = v_p(N_{F/\mathbb{Q}_p}(x))$$, then $$\mathcal{O} = \{ x \in F \; | \; v(x) \geq 0\}$$)

To do this, I thought to show that $$\mathbb{Z}_p \subset O_w$$, as $$O_w$$ is integrally closed on $$F$$, we would have $$\mathcal{O} \subset O_w$$. Now, if $$x \in \mathbb{Z}_p$$, $$x = \frac{a}{b}$$, where $$\gcd(p,b)=1, \gcd(a,b)=1$$. We also know that the restriction to $$\mathbb{Q}$$ of $$|.|_w$$ (which is the n.a absolute value corresponding to the valuation) is either trivial or $$|.|_q$$ for a prime $$q$$, by Ostrowski's theorem. So : $$w(x) = w(\frac{a}{b}) = w(a)-w(b)$$. If $$w$$ is trivial over $$\mathbb{Q}$$, $$w(x) = 0$$ and we have the result. Otherwise, $$w(x) = v_q(a) - v_q(b)$$. But then I'm stuck, cause actually the result I have to prove indicates that $$q=p$$, but there is no way (I found) from here to conclude this (and it's actually quite normal, cause this is what implies the next question).

So, anyone could help me, please ?

Thank you !

• "Deduce from this that for any non trivial discrete valuation w on F" otherwise the statement is untrue. And you can make the Hensel lemma part very concrete from the binomial series $(1+x)^{1/l}= \sum_{k\ge 0} {l\choose k} x^l$ since it has integer coefficients it converges for $v(x)>0$ Commented Oct 12, 2019 at 6:26
• @reuns: Is your reasoning why this fails without assuming $w$ is discrete similar to the one at the end of my answer, or do you have a different argument or even explicit example which does not rely on the axiom of choice? Commented Oct 12, 2019 at 17:08

This is true for a non-trivial discrete (nonarchimedean) valuation $$w$$. Namely, we want to show $$\mathcal{O} \subseteq \mathcal{O}_w$$ or equivalently $$w(x) \ge 0$$ for all $$x\in \mathcal{O}$$, which I will write as $$w(\mathcal O) \ge 0$$. Now from the fact which you have, one deduces that for all $$x \in \mathfrak{p}$$ and all $$l \in \mathbb{N}_{\geq 2}$$ with $$\gcd(l,p)=1$$, $$l$$ divides $$w(1+x)$$. By discreteness of $$w$$ this enforces

$$w(1+\mathfrak{p}) =0.$$

Now the set $$1+\mathfrak{p}$$ makes up the most interesting part of $$\mathcal{O}$$. Actually, I assume you know, or can infer easily from what you know, that the multiplicative group of units $$\mathcal{O}^\times$$ has a direct decomposition

$$\mathcal{O}^\times \simeq \mu_F \times (1+\mathfrak{p}),$$

where $$\mu_F$$ is the group of roots of unity contained in $$F$$; and further the multiplicative monoid $$\mathcal{O} \setminus\lbrace 0 \rbrace$$ has a direct decomposition

$$\mathcal{O} \setminus\lbrace 0 \rbrace \simeq \pi^{\mathbb N_0} \times \mathcal{O}^\times$$

where $$\pi$$ is an element with $$v(\pi)=1$$. Now from the definition of a root of unity it is immediate that

$$w(\mu_F) = 0$$

as well, hence $$w(\mathcal{O}^\times) =0$$. So if, finally, we can show that

$$w(\pi) \ge 0,$$

then we are done. To do that, I would think it suffices e.g. that up to some element $$u \in \mathcal{O}^\times$$ (for which we already know that $$w(u)=0$$), some power of $$\pi$$ is $$p \in \mathbb Z$$, and $$w(p) \ge 0$$ because $$w$$ is a non-archimedean valuation. [Update: A much easier argument I saw in this answer by user reuns is: $$w(\pi) < 0$$ would, via the ultrametric maximum principle, contradict the already established $$w(1+\pi)=0$$.]

As for why I think this is not true without the discreteness assumption: The axiom of choice notoriously allows for field isomorphisms $$\mathbb C_p \simeq \mathbb C_q$$ for all primes $$p,q$$. Since your $$F$$ is contained in $$\mathbb C_p$$, via such an isomorphism we can pull back the canonical ("$$q$$-adic") non-archimedean valuation of $$\mathbb C_q$$ (which no longer has a discrete value group) to $$\mathbb C_p$$, and then restrict it to $$F$$, call this $$w$$. Now if the above statement still held true, it would follow that $$\frac{1}{q} \in \mathcal{O} \subseteq \mathcal{O}_w$$; however, that isomorphism, like any field isomorphism, fixes $$\mathbb Q$$, and of course in the $$q$$-adic valuation $$v_q(\frac{1}{q})=w(\frac{1}{q}) <0$$. So what the proof above shows is that such a restriction of a $$q$$-adic value to some local field inside $$\mathbb C_p$$, besides not being explicit, can never be a discrete valuation -- interesting!

• Thank you very much for this very interesting and complete answer ! That's more that what I could have hoped ! Commented Oct 12, 2019 at 18:26