Consider a number field $K$, and let $v_1, \cdots v_n$ ($n \in \mathbb N$) be some finite (i.e. non-archimedean) places of $K$. Is the following true?

For every $\alpha \in K^\times$ there exists $\beta \in \mathcal{O}_K$ for which $\alpha\beta \in \mathcal{O}_K$ and $$|\beta|_{v_j} = \frac{1}{\max \{1, |\alpha|_{v_j} \}} \text{ for every }j \in \{1, \cdots ,n\} \hspace{10mm} \cdots (1)$$

I could see that this was immediate if $\mathcal{O}_K$ is a UFD (or equivalently a PID) for in that case, I could simply write $\alpha$ uniquely as $\alpha_1/\alpha_2$ where $\alpha_1$ and $\alpha_2$ are algebraic integers sharing no common prime factor and let $\beta:=\alpha_2$. However, I couldn't verify this in the general case. Two things I tried are the following:

  1. I tried to naturally generalize the above approach for UFD's, by factoring the principal fractional ideal $\alpha \mathcal O_K$ uniquely into prime ideals and consider the "denominator" ideal of $\alpha \mathcal O_K$ (that is, if $\alpha \mathcal O_K = \prod_{i=1}^k \mathfrak{p}_i^{a_i} \prod_{j=1}^l \mathfrak{q}_j^{-b_j}$ where $\mathfrak{p}_i$ and $\mathfrak{q}_j$ are all distinct prime ideals and $a_i, b_j \in \mathbb N$ for all $i \in [k], j \in [l]$, then the ideal I'm talking about is $\mathfrak{a} := \prod_{j=1}^l \mathfrak{q}_j^{b_j}$). This need not be a principal ideal but I could raise it to the power of its order in the ideal class group. However that would disturb the exponents, violating my requirement (1). Not sure if there's a work-around.....

  2. I tried using the Strong Approximation Theorem, in an attempt to obtain $\beta$ so as to make the $v$-adic absolute values of the difference $\beta-\alpha^{-1}$ sufficiently small for $v \in \{v_1, \cdots v_n\}$ (so the set of absolute values $w$ for which I'm trying to make $|\beta-\alpha^{-1}|_w$ sufficiently small are supersets of $\{v_1, \cdots v_n\}$), but that hasn't worked out so far.....

I haven't had any luck in finding a counterexample either. I would really appreciate any help, and would also like to know if there is any similar result along these lines.

Edit 1: Another thing I tried along the lines of Approach 1 was to write $\alpha$ as $\beta / \gamma$ (where $\beta$ and $\gamma$ are algebraic integers) and compare the aforementioned prime factorization of $\alpha \mathcal{O}_K$ with those of $\beta \mathcal{O}_K$ and $\gamma \mathcal{O}_K$. What I obtained (after some careful exponent comparison) was the following: $$\beta\mathcal{O}_K = \big(\prod_{i=1}^k \mathfrak{p}_i^{\alpha_i} \big) \mathfrak{a_1}\mathfrak{a_2}\mathfrak{a_3}$$ $$\gamma \mathcal{O}_K = \big( \prod_{j=1}^l \mathfrak{q}_j^{b_j} \big) \mathfrak{a_1}\mathfrak{a_2}\mathfrak{a_3}$$ where $\mathfrak{a_1}$ is made up of prime factors among the $\mathfrak{p}_i$'s, $\mathfrak{a_2}$ is made up of prime factors among the $\mathfrak{q}_j$'s and $\mathfrak{a_3}$ is made up of primes not in the set $\{\mathfrak{p}_i : 1 \leq i \leq k\} \cup \{\mathfrak{q}_j : 1 \leq j \leq l\}$. Of course, one or more of the $\mathfrak{a}_i$ could be trivial (i.e. the unit ideal) but I don't think that's necessary.

And there seems to lie the source of my problem in Approach 1 - I couldn't seem to get rid of the common prime factors between $\beta\mathcal{O}_K$ and $ \gamma \mathcal{O}_K$, in order to be able to generalize the PID approach. Anyway I don't expect that the "numerator" and "denominator" ideals $\prod_{i=1}^k \mathfrak{p}_i^{a_i}$ and $\prod_{j=1}^l \mathfrak{q}_j^{b_j}$ of $\alpha \mathcal{O}_K$ to be principal, if they did, then I should've been able to obtain algebraic integers $\beta$ and $\gamma$ for which $\alpha = \beta / \gamma$ and the principal ideals $\beta\mathcal{O}_K$ and $\gamma \mathcal{O}_K$ would factor as $\prod_{i=1}^k \mathfrak{p}_i^{a_i}$ and $\prod_{j=1}^l \mathfrak{q}_j^{b_j}$ respectively, and life would've been a lot easier.

Edit 2 (More details on Approach 2): As asked in a comment by @Arno Fehm, here are some more details on my second approach. We know that the number of places $w$ of $K$ for which $|\alpha|_w>1$ or $|\alpha|_w<1$ are both finite. As such, I can fix some $\epsilon \in (0, \min\{1, |\alpha|_w^{-1} : w \in N_K\})$ (where I use $N_K$ to denote the set of non-archimedean places of $K$), and then use SAT to obtain a $\beta \in K$ such that $|\beta - \alpha^{-1}|_w < \epsilon$ for all $w \in S:= \{v_1, \cdots , v_n\} \cup \{w \in N_K: |\alpha|_w>1\}$ and $|\beta|_w \leq 1$ for all the other (remaining) non-archimedean places $w$ of $K$. This ensures that $|\alpha\beta|_w, |\beta|_w \leq 1$ for all the places $w \in N_K \setminus S$, whereas for places $w \in S$, I have $$|\beta - \alpha^{-1}|_w < \epsilon < \min\{1, |\alpha|_w^{-1}\} = \frac{1}{\max\{1, |\alpha|_w\}} \hspace{2mm} \cdots (2)$$ Now for the places $w \in S$ for which $|\alpha|_w \geq 1$, I could show by means of the ultrametric inequality that (2) forces $$|\beta|_w = \frac{1}{|\alpha|_w} = \frac{1}{\max\{1, |\alpha|_w\}} \leq 1$$ Problems start occurring for those places $w \in S$ for which $|\alpha|_w<1$ (so $w$ has to be one of $v_1, \cdots , v_n$). In this case, (2) yields $|\beta - \alpha^{-1}|_w<1$ which in fact again forces $|\beta|_w = |\alpha|_w^{-1}>1$, for otherwise the ultrametric inequality leads to the following contradiction $$1>|\beta - \alpha^{-1}|_w = \max\{|\beta|_w, |\alpha^{-1}|_w\} \geq |\alpha^{-1}|_w = |\alpha|_w^{-1} > 1$$ This means that $\beta$ in fact cannot be an algebraic integer if $|\alpha|_{v_j}<1$ for one of the $j \in [n]$ (not to mention that (1) clearly fails for such $j$ as well).

If this approach looks promising, I would really like to know how I should choose my parameters $\epsilon$, $S$ etc. to make it work.

  • 1
    $\begingroup$ I guess it doesn't work to take $\beta$ to be the smallest positive integer $n$ such that $n\alpha$ is an algebraic integer? $\endgroup$ Jun 9, 2020 at 3:30
  • 2
    $\begingroup$ I would think that your approach 2. with strong approximation should work. Why didn't it work out? $\endgroup$
    – Arno Fehm
    Jun 9, 2020 at 5:29
  • $\begingroup$ @Gerry Myerson Thanks for the suggestion. Unfortunately, I've already tried writing $\alpha$ as the ratio of two algebraic integers and comparing the factorizations of all the principal ideals involved (I have added details in Edit 1 above). Even if I use the stronger result and restrict $\gamma$ to be a natural number $n$, I don't see a way of going around looking at the factorization of $n \mathcal{O}_K$. I would really like to know if there's an alternative approach besides looking at factorizations or if there's something that I am missing. $\endgroup$
    – asrxiiviii
    Jun 9, 2020 at 12:32
  • $\begingroup$ @Arno Fehm Thanks for your suggestion. I also added my approach using SAT and where it runs into problems. Is there some way of making it work? $\endgroup$
    – asrxiiviii
    Jun 9, 2020 at 12:36
  • 1
    $\begingroup$ Thanks for the details. I might miss something subtle, but as far as I understand you simply want to get $\beta\in\mathcal{O}_K$ with $|\beta|_v=(\max\{1,|\alpha|_v\})^{-1}$ for all $v$ in the set $S$. So you want $|\beta-\alpha^{-1}|_v$ to be small for $v\in S$ except at those $v$ where $|\alpha|_v<1$, where instead you demand that $|\beta-1|_v$ is small. Strong approximation gives that. $\endgroup$
    – Arno Fehm
    Jun 9, 2020 at 14:14

1 Answer 1


The answer is already in the comments, but here again for completeness: It indeed follows easily from strong approximation:

Let $S=\{v_1,\dots,v_n\}\cup\{v:|\alpha|_v>1\}$. By the strong approximation theorem one can find $\beta\in\mathcal{O}_K$ that is close to $1$ at those $v\in S$ with $|\alpha|_v<1$ and close to $\alpha^{-1}$ at the other $v\in S$. In particular, $|\beta|_v=1$ for those $v\in S$ with $|\alpha|_v<1$ and $|\beta|_v=|\alpha|_v^{-1}$ at the other $v\in S$.

Thus $|\beta|_v=\max\{1,|\alpha|_v\}^{-1}$ for $v\in S\supseteq \{v_1,\dots,v_n\}$, and $|\alpha\beta|_v\leq 1$ for all finite $v$, hence $\alpha\beta\in\mathcal{O}_K$.


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