If $D'$ is a Noetherian $D$ module, then $D'$ is Noetherian as $D'$ module?

Question

I am studying the Proposition: Let $D$ be a Dedekind domain, $F$ its field of fractions, $E$ a finite dimensional extension field of $F$ and $D'$ the subring of $E$ of $D$ integral elements. Assume that $E/F$ is a finite separable field extension. Then $D'$ is a finitely generated $D$-module.

I have to show that $D'$ is Noetherian. It is clear from the Proposition that $D'$ is a Noetherian $D$ module. Why does it follow that $D'$ is Noetherian as $D'$ module?

Would you help me, please? Thank you in advance.

Answer 1

Every ideal of $D'$ is a $D$-submodule of $D'$, so if you have an ascending chain

$$I_1 \subseteq I_2 \subseteq \cdots$$

of ideals of $D'$, then it must terminate, because $D'$ is a Noetherian $D$-module.

Answer 2

If I understand correctly your question, here is a sort of answer that you might like.

Hints and facts:

Because of the situation we have (which is really good in fact since the rings we have very good structures) the extension $D \subset D^{\prime}$ is integral, and $D$ is normal (as Dedekind domain is integrally closed). Hence we have very good correspondence between the prime ideals inside $D$ and primes inside $D^{\prime}$ respectively. Commonly in bibliography are referred as Going Down, Going up theorems (if you haven't heart them so far is a good chance to meet them because are one of the basics in commutative algebra, otherwise skip them) and you can find them in any book of commutative algebra, included this online notes http://www.mathematik.uni-kl.de/~gathmann/class/commalg-2013/main.pdf (page 89). So what this correspondence establishes, is that prime ideals in $D^{\prime}$ are giving prime ideals in $D$ and vice versa. In a Dedekind domain every ideal is a product of primes, and in commutative case only, finite product of finitely generated ideals is finitely generated ideal, so in order to prove your question it's enough to restrict our attention to prime ideals in $D^{\prime}$, since the latter is Dedekind domain too (is integrally closed by its definition, Noetherian domain and since our extension $D \subset D^{\prime}$ is integral has Krull dimension $1$ as well). Now from all the above facts your question maybe is more clear now. Instead of writing out explicitly the proof (which is kind of long-scale in fact) it's good I think to work it out alone and if you have any queries please do let me know!

The above result you can find it out as Krull-Akizuki theorem in many textbooks. (also a short proof can be found on wiki article too)