I am following Hindry-Silverman's book "Diophantine Geometry - An Introduction". In the proof of Roth's theorem, there is a reduction step showing that the following two theorems are equivalent:

Roth's theorem: Let $K$ be a number field, let $S \subset M_K$ be a finite set of absolute values on $K$. Let $\alpha \in \bar{K}$ and $\epsilon >0$. Then there are only finitely many $\beta \in K$ satisfying $$\prod_{v \in S}\min(|\beta-\alpha\|_v,1)\leq \dfrac{1}{H_K(\beta)^{2+\epsilon}}.\tag{1}$$


Theorem 1: Let $K$ be a number field, let $S \subset M_K$ be a finite set of absolute values on $K$. Let $\alpha \in \bar{K}$ and $\epsilon >0$. Suppose that $$\xi:S\to [0,1]$$ is a function satisfying $$\sum_{v\in S}\xi_v=1.$$ Then there are only finitely many $\beta \in K$ with $$|\beta-\alpha|_v\leq \dfrac{1}{H_K(\beta)^{(2+\epsilon)\xi_v}} \tag{2}$$ for all $v\in S$.

The proof given as follows:

One direction is easy. We just look at Theorem 1 implies Roth. Suppose Thoerem 1 is true, and suppose there are infinitely many numbers $\beta \in K$ satisfying (1). Let $s=|S|$. Consider the collection of maps $$\xi:S\to [0,1]$$ of the form $\xi_v=\dfrac{a_v}{s}$ with $a_v\in \mathbb{Z}$, $a_v\geq 0$ and $\sum_{v\in S}a_v=s.$ Denote this collection by $Z$. Now suppose $\beta \in K$ satisfies (1). For each $v\in S$, define $\lambda_v(\beta)\geq 0$ by the formula $$\min(|\beta-\alpha|_v,1)=\dfrac{1}{H_K(\beta)^{(2+\epsilon)\lambda_v(\beta)}}.$$

Multiplying over $v\in S$ and comparing to (1), we get $\sum_{v\in S}\lambda_v(\beta)\geq 1$, so $$\sum_{v\in S}\lfloor 2s\lambda_v(\beta)\rfloor\geq \sum_{v\in S}(2s\lambda_v(\beta)-1)=2s\sum_{v\in S}\lambda_v(\beta)-s\geq s.$$

This implies that we can find integers $b_v(\beta)$ with $$0\leq b_v(\beta)\leq 2s\lambda_v(\beta)$$ and $$\sum_{v\in S}b_v(\beta)=s.$$ Then the function $\xi:S\to [0,1]$ sending $v$ to $\dfrac{b_v(\beta)}{s}$ is in $Z$.

Then the book claims if $\beta$ satisfies (1), then $\beta$ satisfies (2) for at least one of the functions in $Z$.

My question is: from the construction of $\xi$, I only get $\min(|\beta-\alpha|_v,1)\leq \dfrac{1}{H_K(\beta)^{(2+\epsilon)\xi_v/2}}$. How should one get rid of the $1/2$ in the denominator and get rid of the $\min$?


1 Answer 1


I remember feeling unsatisfied at this part of the book. Here’s a complete argument.

We use induction on $|S|$, where the Theorem 1 solves the case $|S|=1$.

Now assume $|S|=s >1$ and that Roth’s Theorem holds for all $\epsilon >0$ and all $S$ of cardinality less than $s$. Fix $\epsilon>0$.

Let $B$ be the set of $\beta \in K$ such that (1) holds for $S$ and $\epsilon$. Let $\eta=\epsilon/2$.

Let, for each $w \in S$, $B_w$ be the set of $\beta \in B$ such that $|\beta-\alpha|_w \geq 1$. Then, if $\beta \in B_w$, $\beta$ verifies (1) for $\epsilon$ and $S \backslash \{w\}$, and thus $B_w$ is finite.

So all we need to show is that $B’ =B \backslash \cup_{w \in S}{B_w}$ is finite. By Theorem 1, it is enough to find a finite set $F$ of functions $S \rightarrow [0,1]$ with sum one, such that for all $\beta \in B’$, there is a $\xi \in F$ such that $|\beta-\alpha|_w \leq H_K(\beta)^{-(2+\eta)\xi(w)}$ for every $w \in S$ (ie $\beta$ verifies (2) for $\eta$ and $\xi$).

Choose an integer $N>0$ such that $\frac{2+\epsilon}{2+\eta} > 1 +\frac{s}{N}$. Let $F$ be the set of functions $f: S \rightarrow [0,1]$ adding up to one such that $Nf$ is integer-valued. (Clearly, $F$ is finite).

So, let $\beta \in B’$. We define $\lambda: S \rightarrow [0,\infty)$ such that $|\beta-\alpha|_v=H_K(\beta)^{-(2+\eta)\lambda(v)}$. Then $\sum_{v \in S}{\lambda(v)}=\frac{2+\epsilon}{2+\eta} > 1+s/N$. In particular, $F$ contains a function $\mu \leq \lambda$ (consider $1/N$ times the integer part of $N\lambda$, then remove multiples of $1/N$ as needed), then $\beta$ verifies (2) with $\eta$ and $\mu$. QED.

  • $\begingroup$ For $|S|=1$ and $|\beta-\alpha|_v \geq 1$ case, do we use Northcott property to show there are only finitely many $\beta$'s? $\endgroup$
    – finiteness
    Sep 10, 2022 at 4:33
  • $\begingroup$ If $|\beta-\alpha|_v \geq 1$ and $\beta$ satisfies (1), then $H_K(\beta) \leq 1$, thus $\beta$ is zero or a root of unity, so there are finitely many such $\beta$. Good catch! $\endgroup$
    – Aphelli
    Sep 10, 2022 at 8:22

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