%! TEX root = DA.tex % vim: tw=50 % 02/12/2024 12PM $d = \gcd(a - 1, b - 1)$ where $a, b \in \mathcal{S}$ are multiplicatively independent. We assume: $d > \max(a, b)^\eps$ for some $\eps > 0$. Our goal is to prove $d < C(\eps)$. \[ \prod_{v \in S = \{\infty, 2, 3\}} \prod_{j = 1}^{n^ - 1} |\Lambda_j^{(v)}(e / d)|_v \le da^{n - 1} b^{n - 1} d^{-n^2} \tag{$*$} .\] $e_1, \ldots, e_{n^2}$ is an enumeration of $a^k b^l$, $k, l = 0, \ldots, n - 1$. \[ (*) \le \max(a, b)^{2n - 2} \cdot \max(a, b)^{-\eps(n^2 - 1)} .\] Let's take $n > 3\eps^{-1}$, $(*) < \max(a, b)^{-n}$. \[ H(\Lambda_1^{(0)}(e / d), \ldots, \Lambda_{n - 1}^{(0)}(e / d)) \le a^{n - 1} b^{n - 1} .\] $(*) < H(\cdots)^{-\half}$. Subspace theorem applies hence there exists a linear relation between $e_1, \ldots, e_{n^2} \in \mathcal{S}$ (distinct by multiplicative independence of $a, b$). Proposition implies \[ |e_i - e_j|_\infty |e_i - e_j|_2 |e_i - e_j| < C = C(\eps) \] for some $i \neq j$. Then $e_i \neq e_j$ so $e_i - e_j \neq 0$. However, $d \mid e_i - e_j$. \[ d \le |e_i - e_j|_\infty |e_i - e_j|_2 |e_i - e_j|_3 < C .\] \end{proof} \begin{theorem}[Feldman] Let $\alpha \in \ol{\Qbb}$ of degree $d \ge 3$. Then there exists effective $C = C(\alpha) > 0$ and $\eps = \eps(\alpha) > 0$ such that for all $\frac{p}{q} \in \Qbb$, \[ \left| \alpha - \frac{p}{q} \right| > \frac{C}{q^{d - \eps}} .\] \end{theorem} \begin{remark*} This is enough to solve $P(x, y) = m$, where $P$ is a degree $d$ homogeneous polynomial without repeated factors. Thue equation. \end{remark*} \begin{proposition*} Let $K$ be a number field. Then there exists $r \in \Zbb_{\ge 0}$ and $u_1, \ldots, u_r \in \O_K^\times$ and a constant $C = C(K)$ such that $\forall \alpha \in \O_K$, there exists $\tilde{\alpha} \in \O_K$ and $b_1, \ldots, b_r \in \Zbb$ such that \begin{align*} H(\tilde{\alpha}) &\le C \cdot |N_{K / \Qbb}(\alpha)|^{\frac{1}{[K : \Qbb]}} \\ |b_1|, \ldots, |b_r| &\le C\log H(\alpha) \\ \alpha &= \tilde{\alpha} u_1^{b_1} \cdots u_r^{b_r} \end{align*} \end{proposition*} Define $\Phi : K^\times \to \Rbb^{M_{K, \infty}} : (\Phi(\alpha))_v = d_v \cdot \log|\alpha|_v$ (logarithmic embedding). Note that here $K^\times$ is the group under multiplication, while $\Rbb^{M_{K, \infty}}$ is the additive group. \begin{align*} |N_{K / \Qbb}(\alpha)| &= \exp \left( \sum_{v \in M_{K, \infty}} (\Phi(\alpha))_v \right) \\ H(\alpha)^{[K : \Qbb]} &= \exp \left( \sum_{v \in M_{K, \infty}} \max(0, (\Phi(\alpha))_v) \right) \end{align*} For $\alpha \in \O_K$, $\Sigma(\Phi(\alpha))_v \ge 0$. Then: \[ \exp(\|\Phi(\alpha)\|_1 / 2) \le H(\alpha)^{[K : \Qbb]} \le \exp(\|\Phi(\alpha)\|_1) .\] For $\alpha \in \O_K^\times$, $N_{K / \Qbb}(\alpha) = 1$. So \[ \Phi(\alpha) \in W = \{x \in \Rbb^{M_{K, \infty}} : \sum x_v = 0\} .\] Kronecker's theorem: $\Phi^{-1}(0) = \Ker\Phi$ are the roots of unity. Dirichlet's unit theorem: $\Phi(\O_K^\times)$ is a lattice in $W$ that is a $\Zbb$-module of rank $\dim W = r$ which spans $W$. \begin{center} \includegraphics[width=0.6\linewidth]{images/ef90967a273d4727.png} \end{center} Let $u_1, \ldots, u_r$ be a fundamental system of units, that is $\Phi(u_1), \ldots, \Phi(u_r)$ is a basis for the lattice $\Phi(\O_K^\times)$. Fix some $\alpha \in \O_K$. Pick some $x \in \Rbb_{\ge 0}^{M_{K, \infty}}$ such that \[ \sum x_v = \ub{\log |N_{K / \Qbb}(\alpha)|}_{\ge 0} .\] $x \in \Phi(\alpha) + W$. Then there exist $y_1, \ldots, y_r \in \Rbb$ such that \[ x = \Phi(\alpha) + y_1 \Phi(u_1) + \cdots + y_r \Phi(u_r) .\] There exists $C = C(K)$ such that $|y_j| \le C \cdot \|\Phi(\alpha)\|_1$. Let $b_j \in \Zbb$ with $|y_j - b_j| \le 1$ and $|b_j| \le |y_j|$. This gives $|b_j| \le C \cdot \|\Phi(\alpha)\|_1 \le C' \log H(\alpha)$. Take: $\tilde{\alpha} = \alpha u_1^{b_1} \cdots u_r^{b_r}$. \[ \Phi(\tilde{\alpha}) = \Phi(\alpha) + b_1 \Phi(u_1) + \cdots + b_r \Phi(u_r) = x + \ub{(b_1 - y_1) \Phi(u_1) + \cdots}_{(*)} .\] $(*)$ is in a fixed, compact region of $W$. $\|\Phi(\tilde{\alpha})\|_1 \le C + \|x\|_1$. \[ H(\tilde{\alpha})^{[K : \Qbb]} \le \exp(\|\Phi(\tilde{\alpha})\|_1) \le \exp(C) \cdot N_{K / \Qbb}(\alpha) .\]