# Reduction step in the proof of Roth's theorem

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 $$\prod_{v \in S}\min(|\beta-\alpha\|_v,1)\leq \dfrac{1}{H_K(\beta)^{2+\epsilon}}.\tag{1}$$

and

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 $$\min$$?

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.

• For $|S|=1$ and $|\beta-\alpha|_v \geq 1$ case, do we use Northcott property to show there are only finitely many $\beta$'s? Sep 10, 2022 at 4:33
• 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! Sep 10, 2022 at 8:22