To prove $O/aO$ is finitely generated Let $R$ be one dimensional Noetherian integral domain and $K$ be it's fraction field. Let $L/K$ be a finite extension, and $O$ be the integral closure of $R$ in $L$.
Then, why $O/aO$ ($a$ is an element of ideal of $O$ and integral over $R$) is finitely generated?
I showed previously $R/aR$ is finitely generated. Can I use this result ?
Thank you in advance.
 A: For the question to make sense, I'll assume that you mean $a \neq 0$ and that we want to show that $O/a O$ is finitely generated as an $R$-module. I actually think it's easier to show a stronger statement: we will show that $O/a O$ has finite length as an $R$-module. I learned this proof from Bourbaki.
First, we show that it suffices to treat the case where $a \in R$ as follows: $a$ is integral over $R$, so that we have an equation $a^n+r_{n-1}a^{n-1}+ \dots + r_0=0$, where $r_i \in R$. Because $O$ is an integral domain, we may assume that $r_0 \neq 0$. But then the equation shows that $r_0 \in aO$ so that $r_0O \subset a O$. This implies that $O/aO$ is a quotient of $O/r_0O$ by the third isomorphism theorem. So if we can show that $O/r_0O$ has finite length as an $R$-module, so does $O/aO$. Thus we can assume that $a \in R$.
So suppose that $a \in R$ (and still $a \neq 0$.) We first show that $R/aR$ has finite length as an $R$-module. Because $aR$ is an ideal, $R/aR$ is a ring, so we can consider it as a module over itself. All $R$-submodules of $R/aR$ are in fact $R/aR$ submodules and vice versa, so it suffices to show that $R/aR$ has finite length as a module over itself. Prime ideals in $R/aR$ correspond to prime ideals in $R$ containing $a$. Because $a \neq 0$ and $R$ is an integral domain, if we have any prime ideal $\mathfrak{p}$ in $R$ that contains $a$, then $\mathfrak{p}$ properly contains the prime ideal $(0)$. Because $\operatorname{dim}R = 1$, this implies that $\mathfrak{p}$ is maximal. Thus $\operatorname{dim} R = 0$. As $R$ is Noetherian, so is $R/aR$. Thus $R/aR$ is a Noetherian zero-dimensional ring and hence it is Artinian. Thus $R/aR$ is a Noetherian and Artinian module over itself and hence $R/aR$ has finite length as a module over itself (and as a module over $R$).
Let $n=[L:K]$. Let $M$ be any finitely generated $R$-submodule of $O$. We want to show that $$\ell(M/aM) \leq n \ell(R/aR)$$ where $\ell$ denotes the length.
Because $K \otimes_R M \subset K \otimes_R O = L$, the rank of $M$ over $R$ is finite. Let $r$ be the rank of $M$ over $R$. Choose $r$ linearly independent elements $m_1, \dots, m_r \in M$. Let $M'=\sum m_i R$. Because the $m_i$ are linaerly indepedent, the sum is actually direct so that $M' \cong R^r$. For any $m \in M$, we have that $m,m_1, \dots, m_r$ is linearly dependent (or else the rank would be larger than $r$). So we have a non-trivial equation $0=rm+r_1m_1+\dots +r_rm_r$ with $r,r_i \in R$. Then $r\neq 0$ by linear independence of $m_1, \dots, m_r$. This shows that $rm \in M'$. So we have shown that for each $m \in M$, there is some non-zero $r \in R$ such that $rm \in M'$. This implies that $M/M'$ is a torsion $R$-module. Because $M$ is assumed to be finitely generated, so is $M/M'$. As $M/M'$ is a finitely generated torsion module over an integral domain, we can find some non-zero $r \in R$ such that $r \overline{m}=0$ for all $\overline{m} \in M/M'$. Thus we may consider $M'':=M/M'$ as a module over $R/rR$. As we have already seen, $R/r$ is an Artinian ring. Any finitely generated module over an Artinian ring has finite length, so $M''$ has finite length as an $R/rR$-module and hence also as an $R$-module.
Now let $n \in \Bbb N$ and consider the exact sequence $0 \to M' \to M \to M'' \to 0$. Tensor products are right exact, so upon tensoring with $R/a^nR$, we obtain the exact sequence
$$M'/a^nM' \to M/a^nM \to M''/a^nM'' \to 0$$
This yields $$\ell(M/a^nM) \leq \ell(M'/a^nM) + \ell(M''/a^nM'') \leq \ell(M'/a^nM)+\ell(M'')$$
We now want to relate $\ell(M/a^nM)$ to $\ell(M/aM)$ and $\ell(M'/a^nM')$ to $\ell(M'/aM')$. To do this, let $X$ be any torsion-free $R$-module. We claim that for all $n$, we have $\ell(X/a^nX)=n\ell(X/aX)$. To see this, we use induction. The case $n=1$ is clear. For the inductive step, consider the exact sequence obtained from the third isomorphism theorem:
$$0 \to a^nX/a^{n+1}X \to X/a^{n+1}X \to X/a^n X \to 0$$
This implies that $$\ell(X/a^{n+1}X)=\ell(a^nX/a^{n+1}X)+\ell(X/a^n X)$$
Now because $X$ is torsion-free, multiplication by $a^n$ induces an isomorphism $X \cong a^n X$. This isomorphism sends $aX \subset X$ to $a^{n+1}X \subset a^n X$. Thus we obtain an isomorphism $X/aX \cong a^nX/a^{n+1}X$. We can now finish the induction step. We have that
$$\ell(X/a^{n+1}X)=\ell(X/aX)+\ell(X/a^nX)$$
But by the inductive hypothesis, $\ell(X/a^nX)=n\ell(X/aX)$ so that we obtain  $\ell(X/a^{n+1}X)=(n+1)\ell(X/aX)$, as desired.
We can apply this result both for $X=M$ and for $X=M'$. Thus our inequality on lenghts may be rewritten as
$$n \ell(M/aM) \leq n \ell(M'/aM')+\ell(M'')$$
Now dividing by $n$, we obtain
$$\ell(M/aM)\leq \ell(M'/aM') +\frac{\ell(M'')}{n}$$
Now let $n \to \infty$ and use that $\ell(M'')$ is finite. From this, we get that
$$\ell(M/aM) \leq \ell(M'/aM')$$
But $M' \cong R^r$, so $M'/aM' \cong (R/aR)^r$. Thus $\ell(M'/aM') = r \ell(R/aR)$. So we get $$\ell(M/aM) \leq r \ell(R/a R) \leq n \ell(R/aR)$$
Now back to $O$. We can prove that $\ell(O/aO) \leq n \ell(R/aR)$. Take any finitely generated submodule $\overline{M} \subset O/aO$. Lift each generator of $\overline{M}$ to an element of $O$ and let $M$ be the submodule of $O$ generated by those $M$. Then by construction, we have that if $\pi:O \to O/aO$ is the projection map, we get that $\pi(M) = \overline{M}$. So $\overline{M} \cong M/(M \cap aO)$. $aM \subset M \cap aO$, so $\overline{M}$ is a quotient of $M/aM$. We know that $\ell(M/aM) \leq n \ell(R/aR)$, thus $\ell(\overline{M}) \leq n \ell(R/aR)$.
Now if $X$ is any module and we have an exhaustion $X=\bigcup X_i$ by submodules $X_i$, then $\ell(X)=\sup \ell(X_i)$. As $O/aO$ is the union of all its finitely generated submodules and the length of each such submodule is bounded by $n\ell(R/aR)$, we have that $\ell(O/aO) \leq n \ell(R/aR)$, so $\ell(O/aO) \leq n \ell(R/aR)$. Thus $O/aO$ is of finite length. (And in particular finitely generated.)
