# If $A$ is a Dedekind domain and $I \subset A$ a non-zero ideal, then every ideal of $A/I$ is principal.

In this question I will use the following definition of a Dedekind domain:

An integral domain $A$ is a Dedekind Domain if:

1) $A$ is a Noetherian Ring.

2) $A$ is integrally closed.

3) Every non-zero prime ideal of $A$ is maximal.

I also know that every non-zero ideal $I \subset A$ can be expressed uniquely as a product of powers of prime ideals (a proof that does not use results from localization or primary decomposition can be found in Pierre Samuel's Algebraic Theory of Numbers).

With this information, and without having been taught localization (this is for an undergrad class in ANT) is it possible to prove the following statement:

If $A$ is a Dedekind Domain and $I \subset A$ a non-zero ideal, then every ideal of $A/I$ is principal.

--I should point out that I can come up with a proof when I know that in a Dedekind domain $A$ if $p \subset A$ is a non-zero prime (hence maximal) ideal, then the localization $A_p$ is a P.I.D. But, as we have not been taught localization in class I cannot use it to solve the question. I also looked at Atiyah- Macdonald, and they basically prove this as Theorem 9.3, but they use results from localization and primary decomposition.

Yes. Factor $I = \displaystyle\prod_{i =1}^n \mathfrak{p}_i^{e^i}.$ Then by the Chinese Remainder Theorem $A/I \cong \displaystyle\bigoplus_{i =1}^n A/\mathfrak{p}_i^{e^i}.$ So it is enough to show each factor $A/\mathfrak{p}_i^{e^i}$ is principal. The ideals of $A/\mathfrak{p}_i^{e^i}$ are exactly the images of the ideals of $A$ containing $\mathfrak{p}_i^{e^i},$ i.e., $\mathfrak{p}_i^{n}$ for $1\le n \leq e_i,$ under the projection map $\pi:A \rightarrow A/\mathfrak{p}_i^{e^i}.$ If $\pi(\mathfrak{p}_i) = \pi(\mathfrak{p}_i^2)$ then $\pi( \mathfrak{p}_i) = 0$ and $A/\mathfrak{p}_i^{e_i}$ is a field. Otherwise, let $\alpha \in\pi( \mathfrak{p}_i) \setminus \pi(\mathfrak{p}_i^2).$ Then $(\alpha)$ is a proper ideal such that $(\alpha) \not\subset \pi(\mathfrak{p}_i^n)$ for any $n\ge 2.$ It follows $(\alpha) = \pi(\mathfrak{p}_i).$ We conclude $\pi(\mathfrak{p}_i^n) = (\alpha^n)$ and hence $A/\mathfrak{p}_i^{e_i}$ is principal.
• Well, strictly speaking if $1 < e_1$, then $A/{p_i}^{e_i}$ is not a PID (because it is not a domain), as ${p_i}^{e_i}$ is not prime in $A$. But, I get what you mean. Thanks for the answer. Oct 14, 2011 at 5:18
• @Rankeya How do we know that the only ideals containing $p_{i}^{e_{i}}$ are powers of $p_{i}$? Oct 26, 2015 at 18:54
• @Jackson: Let $Q$ be a prime ideal of Dedekind domain which contains $P_i^{e_i}$. So, $P_i^{e_i}\subset Q$ implies (taking radicals) $P_i\subseteq Q$; since prime ideals are maximal, $P_i=Q$. [Without going into radicals, we can argue: $P_i^{e_i}\subset Q$ means $Q$ divides $P_i^{e_i}=O_iP_i\cdots P_i$; from uniqueness of factorization into primes ideals, we get $Q=P_i$. Nov 19, 2019 at 4:03
HINT $\:$ By the approximation theorem, Dedekind domain ideals $\rm\:J\!\ne\! 0\:$ are strongly two generated, i.e. for each $\rm\:0\ne i\in J\:$ there exists $\rm\:j\in J\:$ such that $\rm\:J = (i,j)\:.\:$ Therefore every ideal $\rm\: J\supset I\:$ can be presented in the form $\rm\:J = (i,j)\:$ for $\rm\:0\ne i\in I,\ j\in J\:.\:$ Hence $\rm\ J = (j)\pmod I\$ is principal.
NOTE $\$ This property characterizes Dedekind domains, i.e. a domain $\rm\:D\:$ is Dedekind iff every nonzero ideal of $\rm\:D\:$ is strongly two generated.