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Let $(x_1, \ldots, x_n) \in \Zbb^n$. The height of
it is
\[
  H(x_1, \ldots, x_n) = \max(|x_1|, \ldots, |x_n|)
.\]

\begin{fcthm}[Subspace theorem, Archimedean
  version, Schmidt]
  Assuming:
  - $n \in \Zbb_{\ge 2}$
  - $L_1, \ldots, L_n$ linearly independent linear
  forms with algebraic coefficients in
  $n$-variables
  Then:
  for all $\eps > 0$ the solutions of
  \[
    \prod_{j = 1}^{n} |L_J(x_1, \ldots, x_n)| <
    H(x_1, \ldots, x_n)^{-\eps}
    \tag{$*$}
    \label{subthm_concleq}
  ,\]
  for $(x_1, \ldots, x_n) \in \Zbb^n$ are
  contained in a finite collection of proper
  linear subspaces of $\Qbb^n$, which depend only
  on $L_1, \ldots, L_n, \eps$.
\end{fcthm}

\begin{center}
  \includegraphics[width=0.6\linewidth]{images/370604435f07457c.png}
\end{center}

The volume of the region is
\[ H(x_1, \ldots, x_n) \le H \qquad \text{and}
\qquad \prod_{j = 1}^{n}  |L_j(x_1, \ldots, x_n)|
< H^{-\eps} \]
is $\sim (\log H)^{n - 1} H^{-\eps}$. Consider the
paralellepipeds:
\[ |L_j(x_1, \ldots, x_n)| < H^{\kappa_k} \]
for some $\kappa_j \in \Rbb$ with $\sum \kappa_j =
-\eps$.

This implies Roth's theorem:

Let $\alpha \in \ol{\Qbb} \cap \Rbb$ irrational.
Consider the linear forms
\begin{align*}
  L_1(X_1, X_2)
  &= X_1 - \alpha X_2 \\
  L_2(X_1, X_2)
  &= X_2 \\
\end{align*}
Let $p, q \in \Zbb$. Then \eqref{subthm_concleq}
is equivalent to $|p - \alpha q||q| < \max(p,
q)^{-\eps}$. If $\left| \frac{p}{q} - \alpha
\right| < \frac{|\alpha|}{2}$, then this is
equivalent to $\left| \frac{p}{q} - \alpha \right|
< Cq^{-2-\eps}$. Roth's theorem is true apart from
$p, q$ contained in a finite collection of
subspaces. A subspace is of the form $p + \beta q
= 0$ for some $\beta \in \Qbb$ or maybe $q = 0$.

Obvious subspaces:
\begin{itemize}
  \item $\ker(L_j)$
  \item Example $n = 3$: $L_1 = X_1 - \sqrt2 X_2$,
    $L_2 = X_1 - \sqrt2 X_2 + X_3$, $L_2 = X_2$.
    Consider the subspace $V = \{(p, q, 0) : p, q
    \in \Qbb\}$. Now \eqref{subthm_concleq}
    becomes:
    \[
      |p - \sqrt2 q|^2 |q| < \max(p, q)^{-\eps}
    ,\]
    or alternatively
    \[
      \left| \frac{p}{q} - \sqrt2 \right|^2 <
      q^{-3} \max(p, q)^{-\eps/2}
    .\]
    This has plenty of solutions by Dirichlet if
    $\eps < 1$.
  \item A line, that is a $1$-dimensional subspace
    may contain only finitely many solution.
\end{itemize}
The places of $\Qbb$ is $M_{\Qbb}$ and it consists
of all prime numbers and $\infty$. For each $v \in
M_{\Qbb}$, we define an absolute value on $\Qbb$.
$|\bullet|_\infty$ is the ordinary absolute value.
If $v \in M_{\Qbb}$ is a prime number, this is the
$v$-adic absolute value, that is, for $a \in
\Zbb$, $|a|_v = v^{-b}$ where $b \in \Zbb$ is
maximal such that $v^b \mid a$. For $\frac{a}{b}
\in \Qbb$, we define $\left| \frac{a}{b} \right|_v
= \frac{|a|_v}{|b|_v}$. If $x, y \in \Qbb$, then:
\begin{itemize}
  \item $|x|_v|y|_v = |xy|_v$
  \item $|x + y|_v \le |x|_v + |y|_v$
\end{itemize}
When $v \neq \infty$,
\[
  |x + y| \le \max(|x|_v, |y|_v)
.\]
This is called the ultrametric inequality.

\begin{fcthm}[Subspace theorem, $p$-adic version with Q coeffs]
  Assuming:
  - $n \in \Zbb_{\ge 2}$
  - $S \subset M_{\Qbb}$ with $\infty \in S$
  - for each $v \in S$, let $L_1^{(v)}, \ldots,
  L_n^{(v)}$ be linearly independent forms with
  rational coefficients in $n$ variables
  Then:
  the solutions of
  \[
    \prod_{v \in S} \prod_{j = 1}^{n}  |L_j^{(v)}
    (x_1, \ldots, x_n)|_v < H(x_17 \ldots,
    x_n)^{-\eps}
  ,\]
  with $(x_1, \ldots, x_n) \in \Zbb^n$ are
  contained in a finite collection of proper
  subspaces of $\Qbb^n$.
\end{fcthm}

$n = 2$, $S = \{2, 3, \infty\}$, $L_j^{(v)} =
X_j$, $v \in S$, $j = 1, 2$. Consider $a \in
\Zbb$. Let $a = 2^k 3^l b$ with $b$ not divisible
by 2 or 3.
\[
  |a|_2 |a|_3 |a|_\infty = 2^{-k} 3^{-l} |a| = |b|
.\]
Consider $X_1 = 2^k$, $X_2 = 3^l$, then
\[
  \prod_{v \in S}  \prod_{ j = 1}^{2}
  |L_j^{(v)}(2^k, 3^l)|_v = 1
.\]
What happens if you replace $L_2^{(\infty)}$ with
$X_1 - X_2$?