Trace map argument for number fields Let $K \subset L$ a finite (and therefore inevitably algebraic)
extension of number fields. Let $O_K$ and $O_L$ be the ring of integers
of $K$ and $L$. The goal is to show that that $O_L$ is finite
$O_K$-module. The later means that there exist
$l_1, l_2,..., l_m \in O_L $ such that $O_L = l_1 \cdot O_K 
+ l_2 \cdot O_K + ... + l_m \cdot O_K$.
Now in my lecture notes is noted that this result follows from
an 'easy application of the so called trace map argument'.
I have to confess that I nowhere heard about such
'trace map argument' in that context and I would very thankful
if somebody has an idea what it is and elaborate how this
argument is used here.
My ideas: There exist a $K$-bilinear 'trace map' $Tr: L \times L \to K, (x,y) 
\mapsto tr(x \cdot y)$ where $x \cdot y$ is regarded as $K$-linear
endomorphism $L \to L$. Since we are dealing with number fields
it is known that $Tr$ is not degenerated (the extension is separated).
The field extension $K \subset L$ is assumed to be finite, so there exist
a $K$-basis $x_1, ..., x_m \in L$ with $L \cong K^m \cong
x_1K \oplus x_2K \oplus ... \oplus x_mK$.
Now seemingly the trick is to find other basis
$y_1, y_2, ..., y_m \in O_L$ (probably by 'clearing denominators
of the $x_i$) and to show that these $y_i$ are still linear independent $(1)$
over $K$ and for every $l \in L$ holds:
$l \in O_L $ iff $l = \sum_{i=1}^m a_i y_i $ with $a_i \in O_K$ $(2)$
I conjecture that $(2)$ can be derived by the mentioned 'trace map
argument' while $(1)$ can be proved independently. But it is also not
clear how. The $x_i$ form a $K$-basis and not a $L$-basis, therefore
multiplying the basis elements by $x_i$ by elements from $L^*$
could destroy their linear independence, or not?
 A: Yes, this sounds reasonable. Starting with a basis $\alpha_1,\ldots,\alpha_n$ of an extension $L / K$, you can multiply them with elements of $\mathcal{O}_K$ to assume that they in fact lie in $\mathcal{O}_L$. After that, one has the following result, whose proof I'll more or less quote.
Lemma (cf. Neukirch, I.2.9). Write $d$ for the discriminant of $\alpha_1,\ldots,\alpha_n$. Then $d\mathcal{O}_L \subseteq \mathcal{O}_K \alpha_1 + \cdots + \mathcal{O}_K \alpha_n$.
Proof: If $\alpha = c_1 \alpha_1 + \cdots + c_n \alpha_n \in \mathcal{O}_L$, with coefficients $c_i$ in $K$, then the $c_i$ are a solution of the system of linear equations
$$\operatorname{Tr}_{L/K}(\alpha_i\alpha) = \sum_j \operatorname{Tr}_{L/K}(\alpha_i\alpha_j)c_j,$$
and as $\operatorname{Tr}_{L/K}(\alpha_i \alpha) \in \mathcal{O}_K$, they are given as the quotient of an element of $\mathcal{O}_K$ by the determinant $\det \operatorname{Tr}_{L/K}(\alpha_i \alpha_j) = d$. Therefore the $dc_j$ live in $\mathcal{O}_K$, and the desired result follows.
I presume this is enough to infer finite generation of $\mathcal{O}_L$ over $\mathcal{O}_K$.
A: In order to get the basis $y_1,\dots,y_m$ we need to show that $K\mathcal{O}_L=L$, so we can multiply the $x_i$ with some non-zero elements $r_i\in K$ to get elements $y_i=r_ix_i$ in $\mathcal{O}_L$ which will then of course still form a $K$-basis of $L$. To see $K\mathcal{O}_L=L$, let $x\in L$. As $x$ is algebraic over $K$ there are some $a_i$ with $$x^n+a_1x^{n-1}+\dots+a_n=0$$
Let $d\in \mathcal{O}_K$ be a common denominator of the $a_i$. By multiplying the above equation with $d^nx$ we see that $dx$ satisfies a monic polynomial equation over $\mathcal{O}_K$, hence $dx\in \mathcal{O}_L$ and it follows that $x\in K\mathcal{O}_L$.
It follows as argued at the beginning that we may choose a basis $y_1,\dots,y_m$ of $L/K$ in $\mathcal{O}_L$. Let $B$ be the $\mathcal{O}_K$-submodule of $\mathcal{O}_L$ spanned by the $y_i$. Define $B^\#$ to be the dual module of $B$, i.e. $$B^\#=\{x\in L\mid Tr_{L/K}(xy)\in\mathcal{O}_K \text{ for all } y\in B\}$$ By definition of $B^\#$ we have a well defined map $B^\#\to \mathcal{O}_K^n$ given by $x\mapsto (Tr_{L/K}(xy_1),\dots,Tr_{L/K}(xy_m))$. Since $L/K$ is separable the trace form is non-degenerate, hence $B^\#\to\mathcal{O}_K^n$ is injective. As $\mathcal{O}_K$ is noetherian it follows that $B^\#$ is a finite $\mathcal{O}_K$-module. Since $\mathcal{O}_L\subset B^\#$ using again that $\mathcal{O}_K$ is noetherian we see that $\mathcal{O}_L$ is finite over $\mathcal{O}_K$.
(Actually the fact that $\mathcal{O}_K$ is noetherian is usually proved by the same argument, but over $\Bbb Z$)
