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I am reading the proof of Proposition 2.10 in Neukirch. Here is the result, given the classic $AKBL$ setup.

Prop. If $L|K$ separable and $A$ is a PID, then every finitely generated $B$-submodule $M\neq 0$ of $L$ is a free $A$-module of rank $[L:K]$. In particular, $B$ admits an integral basis over $A$.

Proof: Let $M\neq 0$ be a finitely generated $B$-submodule of $L$ and $\alpha_1,\ldots,\alpha_n$ a basis of $L|K$. Multiplying by an element of $A$, we may arrange the $\alpha_i$ to lie in $B$. We then have $dB\subseteq A\alpha_1+\cdots+A\alpha_n$ by Lem 2.9 ($d$ is discriminant), in particular, rank$(B)\leq[L:K]$, etc.

I am wondering how rank $(B)$ is even defined. Aren't we suppsoed to prove that $B$ is a free $A$-module? So how does this final assertion work and why is it true?

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  • $\begingroup$ What is $B{}{}$? $\endgroup$ Sep 27, 2020 at 17:34
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    $\begingroup$ If $A$ is a PID, every finitely generated, torsion-free $A$-module is free. $\endgroup$ Sep 27, 2020 at 18:03

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I have been struggling to understand Neukirch's proof here as well, and I shall share my understanding of his proof below. I'm also a new learner on algebraic number theory, so I'm sincerely sorry if there are mistakes and misunderstandings.

Personally, I think that one should first prove a weaker result:

Proposition: Let $A$ be an integrally closed integral domain with field of fraction $K$, and let $B$ the integral closure of $A$ in a separable extension $L$ of $K$ of degree $m$. Then there exists free $A$-submodules $M$ and $M^{\prime}$ of $L$, such that $M \subset B \subset M^{\prime}$.

And in paricular,

Corollary $(\star)$: When $A$ is noetherian, $B$ is a finitely generated $A$-module, and furthermore, when $A$ is a PID, $B$ is free of rank $[L:K]$.

The corollary gives the second result (after In particular) of Proposition (2.10) in Neukirch. So it has already been enough if we are only interested in the existence of integral basis of $B$ over $A$. I shall provide a proof of the entire Proposition (2.10) in Neukirch using the above corollary.

PROOF of the quoted proposition and corollary $(\star)$: The proof of the above proposition and corollary can be found as Proposition 2.29 in Milne's note "Algebraic Number Theory", or [Zariski-Samuel] Vol.1, Ch.V., Sect. 4.

PROOF of (2.10): So to prove (2.10), Let $(\mu_i)_{i=1, \ldots, r} \in M$ be a system of generators of the $B$-module $M$. Then there is an $a \neq 0 \in A$ such that $a \mu_i \in B$ for $i=1, \ldots, r$. (This is the standard trick introduced on Page 8). So we see that $aM \subset B$. By Corollary $(\star)$, $B$ is a free $A$-module, where $A$ is a PID, hence $aM$ is a free $A$-module with $\text{rank}_A(aM) \leq [L:K]$. So $M$ is also a free $A$-module with $\text{rank}_A(aM) \leq [L:K]$.

To show that $\text{rank}_A(M) = [L:K]$, it remains to show that $\text{rank}_A(aM) \geq [L:K]$. This is discussed in the post: Incomplete Proof from Neukirch about modules and algebraic integers.

Just to restate the discussion there (Some of the arguments below are directly copied there from Hagen Knaf's answer.), let $(\alpha_i)_{i=1, \ldots, n}$ be a basis of $L|K$. Multiplying by an element of $A$, we can arrange for $\alpha_i \in B$. (Still the trick introduced on Page 8) Take an element $m \in M \setminus 0$. Then the elements $\alpha_i m\in M$, $i=1,\ldots ,n$, are linearly independent over $A$, which means that the $A$-submodule $Bm$ of $M$ has rank $n$. Hence $$ \text{rank}_A (M) \geq \text{rank}_A (Bm) = n = [L:K]. $$

This is what we desired.

Sorry for any possible mistakes above. And I'm still wondering if there is a good way to prove (2.10) directly, or understand what Neukirch means.

THANKS: Thanks Angina Seng for the comments in this post, and Hagen Knaf for his answer in the linked post above!

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