%! TEX root = Ramsey.tex
% vim: tw=50
% 05/11/2024 11AM

% Example class on Monday 2:30pm MR9
% Hand in everything by Friday (10pm)

Currently looking at: single equation, i.e.
vectors $(a_1, \ldots, a_n)$, $a_i \in \Qbb
\setminus \{0\}$.

Showed that if $(a_1, \ldots, a_n)$ is \gls{pr}
then it has the \gls{cp} (recall that in this one
dimensional case, having the \gls{cp} is the same
as there exists $I \neq \emptyset$ such that
$\sum_{i \in I} a_i = 0$).

The other direction:

We want to take a vector $(a_1, \ldots, a_n)$ with
$\sum_{i \in I} a_i = 0$ and show that it is
\gls{pr}.

For $(a, 1)$ we know that it is \gls{pr} if and
only if $a = 1$.

For length $3$, $(1, \lambda, -1)$ is the only
non-trivial case with \gls{cp}. Note $\lambda = 1$
is Schur's theorem.

\begin{fclemma}[]
  \label{lemma:2.3}
  % Lemma 2.3
  Assuming:
  - $\lambda \in \Qbb$
  Then:
  for any finite colouring of $\Nbb$, there exists
  a monochromatic solution to $x + \lambda y = z$.
\end{fclemma}

\begin{remark*}
  We in fact show that whenever $[n]$ is
  $k$-coloured ($n = n(k)$), then we have a
  monochromatic solution.
\end{remark*}

\begin{proof}
  If $\lambda = 0$ then nothing to show.

  If $\lambda < 0$ then $z - \lambda y = x$, so
  these are equivalent.

  Assume $\lambda > 0$, $\lambda = \frac{r}{s}$,
  $r, s \in \Nbb$.

  Seek solution to $x + \frac{r}{s} = z$. Let $c :
  \Nbb \to [k]$ be a finite colouring. Prove this
  by induction on $k$.

  $k= 1$ trivial.

  Assume this is true for $k - 1$ and we want to
  show it for $k$. Assume $n$ is suitable for $k -
  1$. We show that $\W(k, nr + 1)ns$ is suitable for
  $k$.

  We now have $[\W(k, nr + 1)]$ $k$-coloured.
  There exists a monochromatic \gls{ap} of length
  $nr + 1$, say $a, a + d, \ldots, a + dnr$ of
  colour red. Let
  us look at $dis$, where $i \in [n]$. Note $dis
  \le \W(k, nr + 1) ns$, so ``it is in our set of
  coloured numbers''.

  If $dis$ is also red, then $a, a + ird, dis$ is
  a monochromatic solution.

  If no such $dis$ exists, then $\{ds, 2ds,
  \ldots, nds\}$ is $k - 1$ coloured. So there
  exists $i, j, k$ such that $c(ids) = c(jds) =
  c(kds)$ and $i + \lambda j = k$. Then $dsi +
  \lambda dsj = dsk$, i.e. $dsi, dsj, dsk$ is a
  monochromatic solution.
\end{proof}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item This is ``manually'' same proof for
      \nameref{svw}.
    \item $\lambda = 1$ is Schur's theorem (which
      you can prove by Ramsey). The general case
      ($\lambda \in \Qbb$) does not seem to be a
      proof ``by Ramsey''.
  \end{enumerate}
\end{remark*}

\begin{fcthm}[Rado's Theorem for single equation]
  % Theorem 2.4
  Assuming:
  - $(a_1, \ldots, a_n) \in \Qbb^n$
  Then:
  it is \gls{pr} if and only if it has the
  \gls{cp}.
\end{fcthm}

\begin{proof}
  We saw in \cref{prop:2.1} that if it is \gls{pr}
  then it has the \gls{cp}.

  For the other direction, we know that
  $\sum_{i \in I} a_i = 0$ and we need to show
  that given $c : \Nbb \to [k]$ such that there
  exists monochromatic $(x_1, \ldots, x_n)$ such
  that $\sum a_i x_i = 0$.

  Fix $i_0 \in I$ and we ``cook up'' the following
  vector:
  \[
    x_i = \begin{cases}
      x & \text{if $i = i_0$} \\
      y & \text{if $i \notin I$} \\
      z & \text{if $i \in I \setminus \{i_0\}$}
    \end{cases}
  \]
  Need $x, y, z$ monochromatic such that $(\sum_{i
  \notin I} a_i)y + x a_{i_0} + (\sum_{i \in I
  \setminus \{i_0\}} a_i)z = 0$, which is the same
  as requiring $x a_{i_0} - z a_{i_0} + (\sum{i
  \notin I} a_i) y = 0$.

  Upon dividing by $a_{i_0}$, we see that this is
  the same as
  \[
    x + \left( \frac{\sum_{i \notin I}
    a_i}{a_{i_0}} \right) y = z
  ,\]
  which is true by \cref{lemma:2.3}.
\end{proof}

\begin{remark*}
  Rado's Boundedness Conjecture: Let $A$ be an $n
  \times m$ matrix that is \emph{not} \gls{pr}. In
  other words, there exists a \emph{bad}
  $k$-colouring for some $k$. Is this $k$ bounded,
  i.e. $k = k(m, n)$?

  This is known for $1 \times 3$ matrices (Fox,
  Kleitman, 2006). $24$ colours suffices in this
  case.
\end{remark*}

Onto the general case for \nameref{rado}.

Recall that for a prime $p$ and $x \in \Nbb$,
$\e(x) = \text{last non-zero digit in base $p$}$.

Also recall $\et(x) = \text{position on which you find
$\e(x)$}$.

\begin{fcprop}[]
  \label{prop:2.5}
  % Proposition 2.5
  Assuming:
  - $A$ a matrix with entries in $\Qbb$
  - $A$ is \gls{pr}
  Then:
  $A$ has the \gls{cp}.
\end{fcprop}

\begin{proof}
  Let $C_1, \ldots, C_n$ be its columns. Fix $p$
  prime.

  Colour $\Nbb$ as we did before, by $c(x) =
  \e(x)$. By assumption there exists monochromatic
  $x_1$ such that $\sum_i x_i C_i = 0$.

  Let us partition $C_1, \ldots, C_n$ as $B_1
  \sqcup B_2 \sqcup \cdots \sqcup B_l$ where $i, j
  \in B_k$ if and only if $\et(x_i) = \et(x_j)$,
  $i \in B_{k_1}$, $j \in B_{k_2}$ for $k_1 < k_2$ if
  and only if $\et(x_i) < \et(x_j)$.

  We do this for infinitely many $p$. Because
  there exist finitely many partitions, for
  infinitely many primes $p$ we will have the same
  blocks.
  \begin{itemize}
    \item For $B_1$: $\sum_i x_i C_i = 0$, say all
      have colour $d$, i.e. $\e(x_i) = d \in [1,
      \ldots, p - 1]$. Then $\sum d C_i \equiv 0
      \pmod p$ (by collecting the right-most terms
      in base $p$). Since this holds for
      infinitely many $p$, we have that $\sum_{i
      \in B_i} C_i
      = 0 \pmod p$ for infinitely many primes (the
      large primes), and hence we have that
      $\sum_{i \in B_i} C_i = 0$.
    \item $\sum_{i \in B_{k'}} p^t d C_i + \sum_{i
      \in B_1, \ldots B_{k' - 1}} x_i C_i \equiv 0
      \pmod{p^{t + 1}}$.