1

Is there an example of a finite extension of Dedekind domains $R \subset S$ and a principal ideal $I \subset S$ such that $I \cap R$ is not a principal ideal of $R$ ?

I don't have good ideas for solving this. This question is a bit related, since any localization $R_p$ at a prime $p \subset R$ is a DVR.

Thank you.

Alphonse
  • 6,462
  • Related: https://math.stackexchange.com/questions/2980727/contraction-of-non-prime-ideals-in-integral-extensions/2980788#comment6152419_2980788 – Alphonse Nov 23 '18 at 09:53

2 Answers2

0

Let $L$ be a number field of class number 1, and $K$ a subfield with nontrivial class group. Let $\mathfrak{p}$ be a non-principal prime in $K$ and $\mathfrak{P}$ a prime of $L$ lying over $\mathfrak{p}$. Then $\mathfrak{P}$ is principal, but $\mathfrak{P} \cap \mathcal{O}_K = \mathfrak{p}$ is not.

bzc
  • 8,778
0

There's a theorem that says that every ideal of a field $K$ becomes principal in its Hilbert Class Field. So to get an example, we take $K=\mathbb{Q}(\sqrt{-5})$ and ideal $(2, 1+\sqrt{-5})$ which is nonprincipal, and look at it in its HCF which is $\mathbb{Q}(\sqrt{-5}, i)$. This ideal becomes the principal ideal $(1+i)$. This takes some work to check, obviously: the identities

$2=(1+i)(1-i)$

$1+\sqrt{-5}=(1+i)\left(\frac{1+\sqrt{5}}{2}-i \cdot\frac{1-\sqrt{5}}{2}\right)$

$1+i=\left(i\cdot\frac{1-\sqrt{5}}{2}\right)\cdot 2+(1+\sqrt{-5})$

show the two inclusions. So this is an example of the theorem, and an example of your request.

Bob Jones
  • 2,265
  • @GeorgesElencwajg Do you mean my second equality or my third equality? I'm not sure what you mean. – Bob Jones Feb 24 '18 at 23:32
  • Sorry, I was confused because you sometimes write $\sqrt{-5}$ and sometimes $i\sqrt{5}$. Your answer might be clearer if you only used one of those. – Georges Elencwajg Feb 24 '18 at 23:49
  • @GeorgesElencwajg You're probably right, but I wrote it the way I did because it's clearer this way that the things in the parentheses are algebraic integers, so that the ideals truly are equal. It's less clear to me that $\frac{i+\sqrt{-5}}{2}$ is an integer than $i\cdot\frac{1+\sqrt{5}}{2}$. – Bob Jones Feb 25 '18 at 00:26
  • I guess that $(2,1+i\sqrt 5)$ is not principal in the ring $\mathbb Z[i\sqrt 5]$ because $2$ is irreducible (but not prime) in that ring, so that the only generator of the ideal can only be an associate of $2$, which cannot have as multiple $1+i\sqrt 5$ in $\mathbb Z[i\sqrt 5]$. – Georges Elencwajg Feb 25 '18 at 09:53
  • Here you showed that the extension $I O_H$ of $I = (2, 1+\sqrt 5)$ in the ring of integers of the HCF $H$ is principal $(1+i)$, but why is $$I O_H \cap O_K = I$$ (which is my main question) ? – Alphonse Feb 25 '18 at 10:13
  • Maybe if $I = P$ is prime then $P^ec = P$ holds (in extension of Dedekind domains?). However, in contrast to Brandon Carter's answer, $H$ might not have class number $1$, right? – Alphonse Feb 25 '18 at 10:16
  • 1
    Alphonse: yes, the equality $P=P^{ec}$ holds in an integral extension (for example a finite extension ) $R\subset S$ of arbitrary rings , Dedekind or not : Atiyah-Macdonald, 3.16+5.10 . – Georges Elencwajg Feb 25 '18 at 10:59
  • 1
    Alphonse: also, notice that Bob takes $I=(2,1+i\sqrt 5)$ and not, as you wrote in your first comment, $(2,1+\sqrt 5)$. It would be good to prove, or at least state, that the ring of integers of $K$ is $\mathbb Z[i, \frac {1+\sqrt 5}{2}]$ – Georges Elencwajg Feb 25 '18 at 11:05