Quotient of noetherian domain has finite length 
Let $A$ be a noetherian one-dimensional domain, and let $\mathfrak a\ne 0$ be an ideal of $A$. Is it true that $A/\mathfrak a$ has finite length as an $A$-module?

Apparently Neukirch's Algebraic Number Theory uses this fact to define a function $\operatorname{ord}\colon K^*\to \mathbb Z$ given by $\operatorname{ord}(x/y)=\operatorname{length}_A A/xA-\operatorname{length}_A A/yA$, where $K$ is the field of fractions of $A$.
My attempt was as follows: since $A$ is noetherian, $\mathfrak a$ contains the product of a finite number of nonzero prime ideals $\mathfrak p_1\cdots\mathfrak p_m$. If $\mathfrak p\supseteq \mathfrak a$, then $\mathfrak p\supseteq \mathfrak p_1\cdots\mathfrak p_m$, so we conclude that some $\mathfrak p_j\subseteq \mathfrak p$. Since $\mathfrak p_j$ is maximal, we have $\mathfrak p=\mathfrak p_j$. This shows that $A/\mathfrak a$ has only a finite number of prime ideals. But I couldn't go farther then that.
 A: First, note that $A/\mathfrak{a}$ is zero dimensional. Indeed, $\mathfrak{a}$ is a nonzero ideal of $A$, and the prime ideals of $A/\mathfrak{a}$ are in bijective correspondence with the prime ideals of $A$ which contain $\mathfrak{a}$. Since $A$ is one dimensional, and $\{0\}$ is a prime ideal of $A$, it follows that every prime ideal of $A/\mathfrak{a}$ is maximal.
Next, recall that any ring which is zero dimensional and Noetherian must also be Artinian. A good reference for this fact is Theorem 8.5 of Atiyah and MacDonald's "Introduction to Commutative Algebra".
Hence, $A/\mathfrak{a}$ is Artinian as a module over itself, and therefore Artinian as an $A$-module, since the $A$-module structure on $A/\mathfrak{a}$ factors through $A/\mathfrak{a}$. Likewise, an identical observation shows that $A/\mathfrak{a}$ is Noetherian as an $A$-module. By (e.g.) Proposition 6.8 of Atiyah-MacDonald, any module over a commutative ring which satisfies both chain conditions has finite length, so this proves the claim.
A: Question: "But I couldn't go farther then that."
Answer: There is a result called the "Krull-Akizuki theorem" (Matsumura, Thm 11.7) saying the following: If $A$ is a one dimensional noetherian integral domain with field of fractions $K$ and $K \subseteq L$ is a finite extension of $K$ with $A \subseteq B \subseteq L$ a sequence of rings, then $B$ is a noetherian (at most) one dimensional integral domain with the property that  if $0 \neq J \subseteq B$ a non trivial ideal it follows $length_A(B/J) < \infty$.
From this your claim follows with $K=L, A=B$.
The book Mat gives a complete proof.
