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?
Best Answer
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 $\mathcal{O}_K$.