On page 267 of Commutative Algebra with a View Toward Algebraic Geometry by David Eisenbud is
Theorem 11.3 (Krull-Akizuki Theorem). If $R$ is a one-dimensional Noetherian domain with quotient field $K$, and $L$ is a finite extension field of $K$, then any subring $S$ of $L$ that contains $R$ is Noetherian and of dimension $\leq$ 1, and has only finitely many ideals containing a given nonzero ideal of R. In particular, the integral closure of $R$ in $L$ is Noetherian.
But I don't understand why $S$ has only finitely many ideals containing a given nonzero ideal of $R$.
Here is what I got.
Let $I$ be a nonzero ideal of $R$. Then $IS$ is a nonzero ideal of $S$. Then from the proof of the theorem $S/IS$ is an $R$-module of finite length. Does the finiteness of $S/IS$ imply it has only finitely many ideals?
Thank you very much for reading. I just edited the question.