Rings of integers are noetherian (question about a specific proof)

I'm reading Neukirch's book about Algebraic number theory and I totally fail to understand something. At one point he proves that the ring of integers $\mathcal{O}_K$ of a number field $K$ is noetherian in the following manner:

"every ideal is a finitely generated $\mathbb{Z}$-module by (2.10) and therefore a fortiori a finitely generated $\mathcal{O}_K$-module."

Now 2.10 is the claim that for an integrally closed, principle ideal domain A, F its field of fractions, L a finite separable extension of F and B the integral closure of A in L, we have that "every finitely generated B-submodule $M\ne 0$ of L is a free A-module of rank $[L:F]$".

I fail to see how 2.10 is used here. I guess in our case $F=\mathbb{Q}, L=K, A=\mathbb{Z}, B=\mathcal{O}_K$. So I know that every finitely generated $\mathcal{O}_K$ submodule (in other words, ideal) is a free $\mathbb{Z}$-module, which is additional information; but how can I assume that the ideal is finitely generated in the first place? That's what I need to prove!

Solution 1:

Prop 2.10 is that

"every finitely generated $B$-submodule $M≠0$ of $L$ is a free $A$-module of rank $[L:F]$".

The $B$-submodule of $L$ generated by $1\in L$, namely $B$, is a finitely generated $B$-submodule of $L$. Thus $B$ is a free $\mathbb{Z}$-module of rank $[L:F]$ by the proposition, so $B$ is a finitely generated $\mathbb{Z}$-module.

Because $\mathbb{Z}$ is noetherian and $B$ is a finitely generated $\mathbb{Z}$-module, any $\mathbb{Z}$-submodule of $B$ (in particular, any ideal of $B$) will also be a finitely generated $\mathbb{Z}$-module.

This is then where the quote