# Incomplete Proof from Neukirch about modules and algebraic integers

There is a theorem in Neukirch which says

If $L | K$ is separable and $A$ is a principle ideal domain, then every finitely generated $B$-submodule $M \neq 0$ of $L$ is a free $A$-module of rank $[L:K]$

Where $K$ are the fractions of $A$ and $L$ is a finite extension of $K$ and $B$ are the integers of $A$ in $L$.

Recalling that $L = \{\frac{b}{a} \}_{b \in B, a \in A}$, the proof is given

Let $M \neq 0$ be a finitely generated $B$-submodule of $L$ and $(\alpha_i)_{1..n}$ a basis of $L|K$. Multiplying by an element of $A$, we can arrange for $a_i \in B$. Let $d =\text{disc}(\alpha_1,...,\alpha_n).$ By (2.9), $dB \subset A\alpha_1 + ... + A\alpha_n$, and in particular, $\text{rank}(B) \leq [L:K]$, and since a system of generators of the $A$-module $B$ is also a system of generators of the $K$-module $L$, $\text{rank}(B) = [L:K]$. Let $(\mu_i)_{1..r} \in M$ be a system of generators of the $B$-module $M$. There is an $a \neq 0 \in A$ such that $a \mu_i \in B$ so that $aM \subset B$. Then $$adM \subset dB \subset A\alpha_1 + ... + A\alpha_n = M_0$$ Since $M_0$ is a free $A$-module, so is $adM$, and hence also $M$.

Now there is a missing step here...

$[L:K] = \text{rank}(B) \le \text{rank}(M) = \text{rank}(adM) \le \text{rank}(M_0) = [L:K]$

How did we get $\text{rank}(B) \le \text{rank}(M)$?

Take an element $m\in M\setminus 0$. Then the elements $\alpha _im\in M$, $i=1,\ldots ,n$, are linearly independent over $A$, that is the submodule $Bm$ of $M$ has rank $n$. The rank does not decrease when extending a module - this is true for modules over domains. Hence you get what you want.
• Can you clarify please why do the elements $\alpha_i m$ generate $Bm$ as an $A$-module? I think that you use $dB \subseteq A\alpha_1 + \cdots + A\alpha_n$, but how do you get rid of $d$? Feb 14 at 11:09
Another possible way to phrase this. Note $$M$$ is a free $$A$$-module so $$M \otimes_A K$$ is a nonzero $$K$$-vector space. But $$M$$ is a $$B$$-module as well, so $$M \otimes_A K$$ is also a nonzero $$(B \otimes_A K)$$-vector space, or in other words a nonzero $$L$$-vector space. So its $$K$$-dimension must be at least $$[L:K]$$.