# Existence of algebraic integer with absolute value equal to reciprocal of maximum of $1$ and absolute value of a given algebraic number

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.

• I guess it doesn't work to take $\beta$ to be the smallest positive integer $n$ such that $n\alpha$ is an algebraic integer? Jun 9, 2020 at 3:30
• I would think that your approach 2. with strong approximation should work. Why didn't it work out? Jun 9, 2020 at 5:29
• @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. Jun 9, 2020 at 12:32
• @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? Jun 9, 2020 at 12:36
• 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. Jun 9, 2020 at 14:14

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 $$\alpha\beta\in\mathcal{O}_K$$.