I have a question about an argument in the proof of Proposition 7.1.38 from Liu's "Algebraic Geometry and Arithmetic Curves" (page 263):
We have two Noetherian rings $A, B=O_{X}(X)$ with fraction fields $K(Y)=\operatorname{Frac}(A)$, $K(X)=\operatorname{Frac}(B)$.
Since $[K(X):K(Y)]=n $ finite $K(X)$ is a finite dimensional vector space over $K(Y)$ generated by $b_1,..., b_n \in B$.
Therefore $K(X)= \langle b_1,..., b_n \rangle_{K(Y)}$.
We set $M:= \sum_{1 \le i \le n} Ab_i$
By construction $B/M$ is a torsion $A$-module.
My QUESTION is why does this already imply that $B/M$ has finite length as $A$-module?
The point $y$ is a point of codimension 1. Hence $\mathrm{dim}(\mathcal{O}_{Y,y}) = 1$. So $\mathrm{dim}(A) = 1$ and recall that $A$ is local. It has only one non-zero prime ideal $\mathfrak{m}$.
Any finitely generated $A$-module $M$ has a sequence of submodules $ 0 = M_0 \subset M_1 \subset \cdots \subset M_n = M$ such that $M_i/M_{i-1}$ is isomorphic to $A/\mathfrak{p}_i$ for some prime ideal $\mathfrak{p}_i$ of $A$ (for reference of this fact, see Serre, Local Algebra, I, corrolary 2).
Let now $M$ be a torsion $A$-module. Given the structure of $A$, the $\mathfrak{p}_i$ are either $0$ or $\mathfrak{m}$. If any of the $\mathfrak{p}_i$ is $0$, then, $M$ would have a torsion-free element. So all of the $\mathfrak{p}_i$s are $\mathfrak{m}$ and the sequence is a composition series for $M$. By definition, the existence of such a series means that $M$ is of finite length.