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$\newcommand{\alg}{\overline{\mathbb{Z}}}$ Consider the integral closure $\alg$ of $\mathbb{Z}$ in $\overline{\mathbb{Q}}$. My question is:

Is there a prime ideal $P \subset \overline{\mathbb{Z}}$ such that $\bigcap\limits_{n \geq 1} P^n \neq (0)$ ?

The property $\bigcap\limits_{n \geq 1} P^n = (0)$ holds in any noetherian domain (see Zariski–Samuel, Commutative algebra, volume 1, Chap. IV, §7, corollary 1, p. 216) but $\alg$ is known to be a non-noetherian ring. However, it is a Bézout domain ; in particular it is Prüfer, so apparently such an intersection is always a prime ideal.

Thank you!

Watson
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    Since $\overline{\mathbb{Z}}$ is of Krull dimension $1$, this is equivalent (using the Bézout domain property) to $\bigcap\limits_{n \geq 1} P^n = P$, and so equivalent to $P^2 = P$. – Christopher Jun 15 '18 at 09:16
  • How about the prime ideal generated by $2^{1/2^n}, n \geq 0$ ? Edit: I'm not sure if this is prime, sorry. – cat Jun 15 '18 at 09:51
  • @cat : I was thinking the same thing. According to this comment, it is maximal in $\Bbb Z[1/2^{1/2^n} : n \geq 1]$. – Watson Jun 15 '18 at 10:16
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    Since every element in $\overline{\mathbf{Z}}$ is an $n$th power, it's obvious that $P^n = P$ for every prime ideal. (Given $\alpha \in P$, write $\alpha = \beta^n$, then $\beta \in P$, so $\alpha \in P^n$). So yes, it's true for every non-zero ideal. – Infinity Jun 15 '18 at 15:22
  • @Infinity : thank you for this elementary observation (I don't know how I missed it…). I would suggest that you can turn it into an answer as well. – Watson Jun 16 '18 at 14:02

1 Answers1

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Thanks to this helpful comment, I can provide an answer.

Proposition. Let $K$ be any algebraically closed field and $R \subset K$ be any subring. Let $A$ be the integral closure of $R$ in $K$ (we could just take an integrally closed subring $A \subset K$). Then every prime ideal $P$ of $A$ satisfies $P^n = P$ for all $n \geq 1$. In particular, $\bigcap_{n \geq 1} P^n \neq (0)$ iff $P\neq(0)$.

Proof. Clearly, $P^n \subseteq P$ holds. For the reverse inclusion, pick $x \in P$. The polynomial $T^n - x$ has a root $y$ in $A$. Then $y^n = x \in P$ implies $y \in P$ since $P$ is a prime ideal. We get $x = y^n \in P^n$, as desired. $\blacksquare$

We see that in our case $A = \overline{\Bbb Z}$ is a Prüfer domain but not a Dedekind domain. Typically, the localizations $A_P$ are valuation rings, but not DVR (since $P^n=P \leq A$, one cannot just have an integer-valued valuation, as in the case of Dedekind domains...).

user26857
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Watson
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  • In particular, one cannot define ramification subgroups of the absolute Galois group of $\Bbb Q$ by imitating the definition from the number field case. – Watson Jun 26 '18 at 13:32
  • (One can only talk about ramification subgroups of the absolute Galois group of $\Bbb Q_p$). – Watson Sep 25 '18 at 20:34
  • (The valuation on the algebraic $p$-adic integers $O = \overline{\Bbb Z_p}$ is not the valuation attached to the maximal ideal of $O$. Actually, the $p$-adic valuation on $O$ is not discrete, but takes values in $\Bbb Q$.) – Watson Nov 05 '18 at 11:26