Let $L|K$ be a finite field extension, $v$ a non-archimedean valuation and $w$ an extension to $L.$ If $\mathcal{O}_L$ is the integral closure of the valuation ring $\mathcal{O}_v$ of $v$ in $L,$ show that the localization $\mathcal{O}_{\mathfrak{B}}$ of $\mathcal{O}_L$ at the prime ideal $\mathfrak{B} = \{\alpha \in \mathcal{O}_L | w(\alpha) >0\}$ is the valuation ring $\mathcal{O}_w$ of $w.$

This is an exercise from Neukirch's book "Algebraic Number Theory", pg. 166 which I know how to solve. However, I feel that a more conceptual proof should be available but I seem unavailable to find one. Is there a "good" reason to see why the exercise is true? If $v$ is discrete, the exercise is obvious, and I would want some reason / solution which would be as convincing to me as I am about the veracity of the claim in that situation. I am sorry if this question is vague. I am OK with high-powered machinery to prove this theorem, in fact, I would be happy if it was used. I think (?) that we would be done if we could prove that the extension $\mathcal{O}_\mathfrak{B} \subset \mathcal{O}_w$ is integral. If we could prove directly that $\mathcal{O}_\mathfrak{B}$ is a valuation ring, we would also be done. But neither of these facts are obvious to me.

For sake of convenience, I include the solution I have at the moment:
It is easy to see that $\mathcal{O}_{\mathfrak{B}} \subset \mathcal{O}_w.$ Conversely, if $x \in \mathcal{O}_w$ let $$a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 =0$$ be an algebraic equation for $x$ with coefficients in $K.$ Let $a_i$ be the coefficient farthest to the left with the lowest valuation. Let $b_j=a_j/a_i,$ and divide our equation by $a_i$ to get a new algebraic equation $$b_nx^n + b_{n-1}x^{n-1} + \cdots + b_1x+b_0=0.$$ Note that $v(b_j) >0$ if $j >i$ and that $v(b_j) \geq 0$ for all $i.$ Divide the equation with $x^i$ to get $$(b_nx^{n-i}+ \cdots + b_{n-i+1}x+1)+x^{-1}(b_{n-2}+ \cdots + b_0x^{-i+1})=0.$$ Let $w= (b_nx^{n-i}+ \cdots + b_{n-i+1}x+1)$ and $y= b_{n-2}+ \cdots + b_0x^{-i+1}.$ We then have that $x = -y/w.$ We now show that $y \in \mathcal{O}_L,$ and that $w \in \mathcal{O}_L, w \not \in \mathfrak{B},$ thus proving our inclusion. We note that $\mathcal{O}_L$ is the intersection in $L$ of all valuation rings containing $\mathcal{O}_v.$ Using this fact, one shows without too much trouble that our claim is true.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.