Complete valuation, norm of finite extension. Proof of Propositon.

Question

I would like to ask for tips how to manage with proof of this proposition. How to show that $v$ can be uniquely extended to $v'$ ? Should I assume that $v$ can be also extended to another valuation? Also I don't know how to get equation.

Answer 1

This is a consequence of Hensel’s Lemma. Not the version that talks about roots of a polynomial, but the version that says that if the polynomial factors into two relatively prime factors over the residue field, then it factors over the original ring, and you can guarantee that one of the factors upstairs has the same degree as the factor downstairs.

I think that if you prove the following Lemma, you should be able to get the whole thing:

Lemma. Let $k$ be a field complete under the (additive) valuation $v$, and let $f(X)=X^n+c_{n-1}X^{n-1}+\cdots+c_1X+c_0$ be $k$-irreducible. Then:
(1) If $v(c_0)\ge0$, then for all $i$, $v(c_i\ge0$;
(2) If $v(c_0)>0$, then for all $i$, $v(c_i)>0$;
(3) If $v(c_0)<0$, then $v(c_0)<\min_{0<i}v(c_i)$.

An important consequence is:

Proposition. Let $k$ be a field complete under the additive valuation $v$, and let $K$ be an algebraic extension of $k$, with a valuation $V$ extending $v$. Then $\mathcal O=\{z\in K:V(z)\ge0\}$ is the integral closure in $K$ of $\mathfrak o=\{z\in k:v(z\ge0\}$.
In other words, for $z\in K$, you get $V(z)\ge0\Longleftrightarrow z\text{ is integral over }\mathfrak o$.

I think that now it should be easy to show that if $V$ and $V'$ are extensions of $v$ from $k$ to $K$, then $V=V'$. That does it for uniqueness.

For existence, use the formula given. Of course the formula guarantees multiplicativity immediately; what you need besides is the additivity: $V(a+b)\ge\min\bigl(V(a),V(b)\bigr)$. You get this by showing that if $V(z)\ge0$, then $V(1+z)\ge0$ as well. Use the fact that the Norm $N^{k(z)}_k(z)$ is, up to sign, the constant coefficient of the minimal $k$-polynomial for $z$.