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I want to complete the proof of the following theorem. Here is what I have got so far:

Theorem Every non-euclidean valuation $v$ on a number field $K$ is equivalent to $v_{\mathfrak p}$ for some prime ideal $\mathfrak p$ of $\mathbb O_K.$

proof: Let $R_v$ be the valuation ring of $K$ w.r.t $v$ is integrally closed (by lemma 1) so it contains $\mathbb O_K$. The maximal idea $m_v$ is prime so $\mathfrak p = m \cap \mathbb O_K$ is a prime ideal of $\mathbb O_K$ by lemma 2. To show $v$ equal to $v_{\mathfrak p}$ we will show that $R_{v_{\mathfrak p}} \subseteq R_v$ which gives the equivalence by lemma 3. (But here I get stuck.. please tell me if you know how to finish or another way to prove this)


  • $\mathbb O_K$ is the ring of integers of $K$, those elements of $K$ which satisfy a monic irreducible polynomial with integer coefficients.
  • valuation ring: $R_v = \{ x \in K | v(x) \ge 0 \}$
  • maximal ideal: $m_v = \{ x \in K | v(x) > 0 \}$

Definition For a prime ideal $\mathfrak p$ of the ring of integers of a number field $K$ we define the valuation $v_{\mathfrak p}(x) =$ the exponent of $\mathfrak{p}$ in the factorization of the fractional ideal $x \mathbb O_K$ this is well defined by unique factorization of ideals.

Lemma (1) If $R$ is a valuation ring of $K$ then $R$ is integrally closed in $K$. (proof here)

Lemma (2) If $R$ is a subring of $R'$ and $m$ an ideal of $R'$ then $m \cap R$ is an ideal of $R$. Furthermore if $m$ is prime in $R'$ then $m \cap R$ is prime in $R$. (This is not hard if you just write out the definition of ideal and prove each property)

Lemma (3) if $R_u \subseteq R_v$ are valuation rings of $u,v$ then $u \equiv v$ (equivalence of valuations is defined here, I omitted the proof but I can add it if people ask).


I think the idea of the proof is to relate the valuations by relating the rings via their ideals. I was trying to do this by starting with $x \in \mathbb O_K$ we have $v(x) = 0$ implies $x \not \in m_v$ so $x \not \in \mathfrak p$ so $v_{\mathfrak p}(x) = 0$, but I need to prove something like this for all $v(\frac{x}{y})$.

Thanks for any hints or advice how to complete this proof.

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What is exactly the question ? Proof of Lemma 3 or the inclusion $R_{v_p}\subseteq R_v$ ? –  user18119 Nov 2 '12 at 22:09
    
@QiL, I have a proof lemma 3 and will post it if anyone wants. It's finishing the proof of the theorem that I cannot do. $R_{v_{\mathfrak p}} \subseteq R_v$ is just my idea so it might not work. (I couldn't make it work). –  sperners lemma Nov 2 '12 at 22:11
    
Thanks for asking because other people might have skipped my question because of that confusion. –  sperners lemma Nov 2 '12 at 22:11
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1 Answer

up vote 1 down vote accepted

Note that $R_{v_{\mathfrak p}}$ is just the localization of $O_K$ at the maximal ideal $\mathfrak p$. So each of its element $x$ can be writen as $$x=a/s, \quad a\in O_K, s\in O_K\setminus \mathfrak p.$$ We have $v(s)\le 0$ because otherwise $s\in m_v\cap O_K=\mathfrak p$. So $v(x)\ge 0$ (because $a\in O_K\subseteq R_v$) and $x\in R_v$.

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Thank you very much but how you know $v(x) = v(a/s) = v(a) - v(s) \ge 0$? Do we somehow know that $v(a) \ge 0$? (how?) –  sperners lemma Nov 2 '12 at 22:27
1  
@spernerslemma: see the edited answer. –  user18119 Nov 2 '12 at 22:29
    
thank you again!! –  sperners lemma Nov 2 '12 at 22:29
    
could we also say that $v(s) = 0$? –  sperners lemma Nov 2 '12 at 22:32
1  
@spernerslemma: yes, absolutely. –  user18119 Nov 2 '12 at 22:37
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