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Let $K \subset L$ a finite extension of number fields. Let $O_K$ and $O_L$ be the rings of integers of $K$ and $L$. The goal of this question was to elaborate that $O_L$ is finite $O_K$-module. Although the question is answered, Jeroen remarked that this claim could alternatively proved using following Lemma from Neukirch's Algebraic Number Theory:

Let be a basis of finite extension $L/K$. Multiplying them by suitable elements from $O_K$ ('clearing the denominators') we can assume that all $\alpha_i$ lie in $O_L$. Then

Lemma (cf. Neukirch, I$.2.9$). Let $d \in K$ be the discriminant of $\alpha_1,\ldots,\alpha_n$. Then $d\mathcal{O}_L \subseteq \mathcal{O}_K \alpha_1 + \cdots + \mathcal{O}_K \alpha_n$.

Then we have a chain of inclusions $$d\mathcal{O}_L \subseteq \mathcal{O}_K \alpha_1 + \cdots + \mathcal{O}_K \alpha_n \subseteq \mathcal{O}_L.$$

But I not know how from this we can conclude $O_L = \mathcal{O}_K \alpha_1 + \cdots + \mathcal{O}_K \alpha_n$. Equivalently how can I show that for any $a \in O_L$ which by the Lemma can be written as $a=\frac{ad}{d}= \sum \frac{c_i}{d} \alpha_i$ we can conclude that $c_i/d \in O_K$.

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1 Answer 1

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Note that in this case

$$\mathcal O_L\subseteq d^{-1}\alpha_1\mathcal O_K+\cdots+d^{-1}\alpha_n\mathcal O_K$$

and the latter is a free $\mathcal O_K$-module and in particular finitely generated. We may consider the special case $K=\Bbb Q$ and $\mathcal O_K=\Bbb Z$. Then, $\mathcal O_L$ is as a submodule of a finitely generated module over a Noetherian ring is itself finitely generated. This by passes the usual trace argument in so far as this argument is hidden in the lemma. Using this on $\mathcal O_K\subseteq\mathcal O_L$ for a finite extension of number fields $L/K$ then gives your result.

However, I am not completely sure if the $\alpha_i$ do form an integral basis necessarily. The argument I presented only proves that $\mathcal O_L$ is itself finitely generated and hence finite over $\mathcal O_K$ containing the latter (which is the question asked in the title and what the linked answer is concerned with; please correct me if your questions was something else).

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  • $\begingroup$ clearly since we deal with number fields the $O_K$ has $\mathbb{Z}$-module structure, and therefore $A:=d^{-1}\alpha_1\mathcal O_K+\cdots+d^{-1}\alpha_n\mathcal O_K$ as well. But why is $A$ also a free $\mathbb{Z}$-module? $\endgroup$
    – user267839
    Commented Feb 3, 2021 at 2:35
  • $\begingroup$ by the way what you precisely mean by 'usual trace argument'? $\endgroup$
    – user267839
    Commented Feb 3, 2021 at 2:35
  • $\begingroup$ @IsaktheXI The trace argument in the post you linked is to the best of my knowledge the standard approach for showing that the ring of integers of a number field is finitely generated. $\endgroup$
    – mrtaurho
    Commented Feb 3, 2021 at 2:38
  • $\begingroup$ @IsaktheXI I meant to write free $\mathcal O_K$-module. Using this asserstion then once for the trivial case $K=\Bbb Q$ generalizes it to arbitrary number fields, if I'm not mistaken. But I might've messed something up here. But this part is not really relevant for the question so I'll just remove it for now. $\endgroup$
    – mrtaurho
    Commented Feb 3, 2021 at 2:42
  • $\begingroup$ about this 'usual trace trick':In the linked post leoli1 as well Jeroen van der Meer used trace in their proofs. So as far as I understood you correctly when one talks about 'usual trace argument' in this context, conventially one means the argument used by leoli1, right? That is this diagonal embedding of a $O_K$-module in $O_K^n$ for $n$ big enough via $(Tr_{L/K}(\cdot y_1),...Tr_{L/K}(\cdot y_n)$, right? $\endgroup$
    – user267839
    Commented Feb 3, 2021 at 3:18

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