# For number fields $L/K$ the ring of integers $O_L$ is finite over $O_K$ as an immediate consequence of a Lemma from Neukirch

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 $$c_i/d \in O_K$$.

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).

• 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? Commented Feb 3, 2021 at 2:35
• by the way what you precisely mean by 'usual trace argument'? Commented Feb 3, 2021 at 2:35
• @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. Commented Feb 3, 2021 at 2:38
• @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. Commented Feb 3, 2021 at 2:42
• 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? Commented Feb 3, 2021 at 3:18