finite extensions of discrete valuation fields: A method to find a basis

Suppose that $L|K$ is a finite extension of discrete valuation fields. Namely $w$ is a discrete valuation on $L$ extending a valuation $v$ on $K$. Now consider the respective rings of integers $\mathcal O_L$ and $\mathcal O_K$; I don't understand the following fact used, without any explanations, in Neukirch's book to prove Proposition II.6.8:

If $\{x_1,\ldots,x_m\}\subset\mathcal O_L$ is a basis of $\mathcal O_L$ as $\mathcal O_K$-module, then $\{x_1,\ldots,x_m\}$ is a basis of $L$ as $K$-vector space.

• I didn't really study much of modules and their bases yet, but I think that it's simply because these elements are in $\mathcal{O}_{L}$ and hence in $L$, and if you have your scalars in $\mathcal{O}_{K}$, then in particular they are in $K$ and the same proofs hold for linear independence and span. Btw, I'm also studying ANT with Neukirch's book and I'm at the same part of the book .. would be nice to have someone to chat with about it; my e-mail is pedro_souza996@yahoo.com.br, in case you're interested. Sorry if there is a chat system here that I'm not aware of – Shoutre Nov 17 '15 at 0:00
• Wait a moment: an element $y\in\mathcal O _L$ can be written as $y=\sum a_ix_i$. How would you write $y^{-1}\in L$ in terms of the $x_i$'s? It is not obvious in my opinion. – manifold Nov 17 '15 at 0:03
• Hmm, yeah, maybe I'm wrong and span doesn't hold, but certainly the proof for linear independence holds and this is enough to prove the inequality. I am reading the proof for the proposition you mentioned and didn't find where he says that. – Shoutre Nov 17 '15 at 0:16
• It's when the author wants to prove that $L|K$ separable implies that $[L:K]=ef$. He only finds a $\mathcal O_K$-basis for $\mathcal O_L$. – manifold Nov 17 '15 at 0:22
• @Shoutre, I'm interested in chatting with you, but at the moment I'm very busy. I may contact you in the next weeks. – manifold Nov 17 '15 at 0:30

Take an element $y\in L$. There is an element $a\in\mathscr{O}_K$, not zero, for which $ay\in\mathscr{O}_L$, hence $ay=\sum_i a_ix_i$ for some $a_i\in\mathscr{O}_K$. Now multiply both sides by $a^{-1}$ to get $y$ as a $K$-linear combination of the $x_i$. They are also $K$-linearly independent by a similar scaling-to-$\mathscr{O}_L$ argument.
• To find such an element $a\in\mathcal O_K$ I think you need that the valuation $v$ on $K$ should be surjetive. – manifold Nov 17 '15 at 0:23
• No, I'm wrong. The image $v(K)$ is $t\mathbb Z$ for some natural number $t$. Therefore I can find some $a\in \mathcal O_K$ with valuation $v(a)$ bigger than any chosen natural number. – manifold Nov 17 '15 at 0:28
• Dear @manifold, This only depends on $\mathscr{O}_K$ being a domain and $\mathscr{O}_L$ the integral closure of $\mathscr{O}_K$ in $L$. If $y\in L$, then because $L/K$ is algebraic, there is an equation $y^n+a_{n-1}y^{n-1}+\cdots+a_1y+a_0=0$ with $a_i\in K$. Let $b\in\mathscr{O}_K$ be a common denominator for the $a_i$ and multiply the equation through by $a=b^n$ to see that $ay$ is integral over $\mathscr{O}_K$, hence in $\mathscr{O}_L$. – Keenan Kidwell Nov 17 '15 at 13:48
$\mathcal O_L$ is a free $\mathcal O_K$-module of rank $m$, and $\{x_1,\dots,x_m\}\subset\mathcal O_L$ is a basis. Then this is a basis for $L=S^{-1}\mathcal O_L$ over $K=S^{-1}\mathcal O_K$, where $S=\mathcal O_K\setminus\{0\}$.
(This is a standard fact: if $F$ is a free $R$-module and $\{x_1,\dots,x_m\}\subset F$ is a basis over $R$, then $S^{-1}F$ is a free $S^{-1}R$-module and $\{\frac{x_1}1,\dots,\frac{x_m}1\}\subset S^{-1}F$ is a basis over $S^{-1}R$.)