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Let $K$ be a number field, $p \in \Bbb Z$ be a prime such that $p\mathcal{O}_K = P_1 \cdots P_r$ is totally split (here $P_i \trianglelefteq \mathcal{O}_K$ are distinct prime ideals).

Is it true that for any $n \geq 0$, we have $\Bbb Z \cap P_1^n = p^n \Bbb Z$ ?


My thoughts : this is trivially true for $n=0$. For $n=1$, this is a general result : if $R (=\Bbb Z) \subset S = \mathcal{O}_K$ is a extension of domains such that $\dim(R)=1$ and $\mathfrak{p}S = P_1^{e_1} \cdots P_r^{e_r}$ (where $0 \neq \mathfrak p \subset R, P_j \subset S$ prime ideals), then $P_1 \cap R$ is a prime of $R$ containing $\mathfrak p$, so it is $\mathfrak p$ by dimension.

I was not able to conclude for $n \geq 2$. Notice that the claim is not true if $p$ is not totally split, e.g. $p\mathcal O_K = P^2$, because then for $n=2$, we have (by this) : $\Bbb Z \cap P^2 = p\Bbb Z \neq p^2 \Bbb Z$.

I believe that my claim should hold for finite-type extension of arbitrary Dedekind domains. The case $r=1$ (arbitrary $n \geq 0$) is discussed here, I think.

user26857
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Alphonse
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1 Answers1

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Yes, this follows by properties of discrete valuations. Let $v_p$ and $\newcommand{\P}{\mathfrak P} v_\P$ be the (normalized) discrete valuations associated to the local rings $\mathbb{Z}_{(p)}$ and $\newcommand{\O}{\mathcal O} (\O_K)_\P$, respectively, where $\P = P_1$. (By normalized, I mean that $v_\P(a) = 1$ for all $a \in \P \setminus \P^2$, and similarly for $(p)$.) In general, letting $e = e_{\P|(p)}$ be the ramification index of $\P$ over $(p)$, we have $v_\P|_\mathbb{Q} = ev_p$ since $v_\P(p) = e$.

In your case $e = 1$, so $v_\P|_\mathbb{Q} = v_p$. Now $$ p\mathbb{Z} = \{a \in \mathbb{Z} : v_p(a) \geq 1\} $$ and in general $$ p^n \mathbb{Z} = \{a \in \mathbb{Z} : v_p(a) \geq n\} \, . $$ Similarly, $ \P^n = \{a \in \O_K : v_\P(a) \geq n\} \, . $ Since $v_p = v_\P$, then $$ \P^n \cap \mathbb{Z} = \{a \in \mathbb{Z} : v_\P(a) \geq n\} = \{a \in \mathbb{Z} : v_p(a) \geq n\} = p^n \mathbb{Z} \, . $$

I think this argument works for general Dedekind domains, too, since Dedekind domains are locally discrete valuation rings.

You can also see how this applies to your non-example, where $p \O_k = \P^2$. Then we have $v_\P|_\mathbb{Q} = 2 v_p$, so $$ a \in \P^2 \cap \mathbb{Z} \iff 2 \leq v_\P(a) = 2v_p(a) \iff v_p(a) \geq 1 \iff a \in p \mathbb{Z} \, . $$ Interestingly we also have $\P \cap \mathbb{Z} = p \mathbb{Z}$, since by the same logic as above, an element $a \in \P \cap \mathbb{Z}$ must satisfy $v_p(a) \geq 1/2$. But since $v_p$ takes values in $\mathbb{Z}$, this means that $v_p(a) \geq 1$. Thus in general, it looks like $$ \P^n \cap \mathbb{Z} = p^{\lceil n/e \rceil} \mathbb{Z} \, . $$

Addendum: The inclusion $\mathbb{Z} \hookrightarrow \O_K$ induces an inclusion of local rings $\mathbb{Z}_{(p)} \hookrightarrow (\O_K)_\P$. Let $\mathfrak{m} = p \mathbb{Z}_{(p)}$ and $\mathfrak{M} = \P (\O_K)_\P$ be their unique maximal ideals. Since $\mathbb{Z}$ and $\O_K$ are Dedekind domains, then $\mathbb{Z}_{(p)}$ and $(\O_K)_\P$ are discrete valuation rings. This means that $\newcommand{\m}{\mathfrak{m}} \newcommand{\M}{\mathfrak{M}} \m$ and $\M$ are principal, so $\M = (\pi)$ for some $\pi \in (\O_K)_\P$. Every element of $\mathbb{Z}_{(p)}$ can be written in the form $u p^a$ where $u \in (\mathbb{Z}_{(p)})^\times$ and $a \in \mathbb{Z}_{\geq 0}$ and every element of $(\O_K)_\P$ can be written as $v \pi^b$ with $v \in ((\O_K)_\P)^\times$ and $b \in \mathbb{Z}_{\geq 0}$. The discrete valuations associated to these are the exponents in these expressions, i.e., $v_\P(v \pi^b) = b$.

One definition of the ramification index $e = e_{\P|(p)}$ (when the extension of residue fields is separable, at least) is that it is the exponent such that $v_{\P}(p) = e$. Given $u p^a \in \mathbb{Z}_{(p)}$, then $$ v_\P(u p^a) = a v_\P(p) = ae = e v_p(u p^a) $$ which shows that $v_\P = e v_p$ on $\mathbb{Z}_{(p)}$, and since elements of $\mathbb{Q}$ can be written as quotients of elements of $\mathbb{Z}_{(p)}$, the same is true on $\mathbb{Q}$.

For more on discrete valuations, I recommend Chapter 1 of Serre's Corps Locaux and Chapter 7 of Rosen's Number Theory In Function Fields.

Viktor Vaughn
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    This is a nice proof, thank you! :-) – Alphonse Dec 23 '17 at 13:09
  • This question may also be of interest. – Viktor Vaughn Dec 24 '17 at 17:24
  • Just a question : in a general case of number fields $K \supset L$ (not only $L = \Bbb Q$), how would you show that $v_{P} = e(P/\mathfrak p) v_{\mathfrak p}$ ? Thank you ! – Alphonse Dec 26 '17 at 17:26
  • I would know how to prove $v_P = e(P/p) v_p$ in the case where $\mathfrak p = (p)$ is principal (which is OK for $L = \Bbb Q$), but I think there is a better and more general way to prove this. For instance, I don't quite see why "since $v_P(p) = e$" is sufficient to conclude $v_P = e(P/p) v_p$. I would like more explanations on these valuations. – Alphonse Dec 26 '17 at 17:47
  • @Alphonse I've added some more explanation. You may find this post useful, too. – Viktor Vaughn Dec 27 '17 at 17:22
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    Thank you so much for the addendum, and for the link! I will read it carefully, and if I have no more question, I will accept this very good answer again :-) – Alphonse Dec 27 '17 at 17:28