prime ideals in extensions of dedekind domains

Question

Let $A$ be a Dedekind domain and $B$ its integral closure on a finite extension of its field of fractions. Is it true that if $\mathfrak{I}$ is an ideal of $B$, then factorizing it as $\beta_1^{e_1}\cdots\beta_r^{e_r}$, $\beta_i\cap A$ goes through all the prime ideals of $A$ that contain $\mathfrak{I}\cap A$? Can you prove it? Thank you.

Answer 1

Let $K, L$ be the quotient fields of $A, B$. Let $\mathfrak P$ be a prime ideal of $B$ which lies over the prime $\mathfrak p$ of $A$, and let $e$ be the ramification index. If you restrict the $\mathfrak P$-adic valuation of $L$ to $K$, you get that $ \nu_{\mathfrak P}(x) = e \cdot \nu_{\mathfrak p}(x)$ for all $x \in K$.

Lemma: Let $t \geq 1$. Then $\mathfrak P^t \cap A = \mathfrak p^d$, where $d$ is the smallest integer which is $\geq t/e$.

Proof: If $x \in \mathfrak P^t \cap A$, then since $\mathfrak P^t$ is the set of $x \in B$ for which $\nu_{\mathfrak P}(x) \geq t$. Since $x \in A$, we have $\nu_{\mathfrak p}(x) \geq t/e$. Since $\nu_{\mathfrak p}(x)$ is an integer, it must be $\geq d$, so $x \in \mathfrak p^d$. Conversely if $$x \in \mathfrak p^d \subseteq \mathfrak p^d B = (\mathfrak pB)^d \subseteq (\mathfrak P^e)^d = \mathfrak P^{ed}$$ then $\nu_{\mathfrak P}(x) \geq ed \geq t$, so $x \in \mathfrak P^t$.

For example, if $B$ is the ring of integers of $\mathbb{Q}(i)$, then $B(1+i)$ is the only prime ideal lying over $2 \mathbb{Z}$, and it has ramification index $2$. Then the (rational) integers which are in $B(1+i)^5$ are exactly those integers which are divisible by $2^3 = 8$.

Now for the proof. If you write $\mathfrak J = \mathfrak P_1^{m_1} \cdots \mathfrak P_s^{m_s}$ for $m_i \geq 1$, and if $\mathfrak P_i$ lies over $\mathfrak p_i$ (the $\mathfrak p_i$ are not necessarily distinct here, but the $\mathfrak P_i$ are assumed to be), then $\mathfrak P_i^{m_i} \cap A = \mathfrak p_i^{d_i}$ for some integers $d_i \geq 1$ by the lemma. In a Dedekind domain, the product of relatively prime ideals is the same as the intersection, so $$\mathfrak J = \bigcap\limits_{i=1}^s \mathfrak P_i^{m_i}$$ which implies $$\mathfrak J \cap A = \bigcap\limits_{i=1}^s \mathfrak P_i^{m_i} \cap A = \bigcap\limits_{i=1}^s \mathfrak p_i^{d_i}$$ and now you can combine those $\mathfrak p_i$ which are the same in an appropriate way to get a factorization for $\mathfrak J$.