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\begin{proof}
  Let $j = 1$, and by taking the substitution $X -
  \alpha_1 Y \mapsto X$, we may assume $\alpha_1 =
  0$.

  First, we show $X$ is bounded on $\Gamma_1$. To
  this end:
  \[
    X = \frac{G(X, Y)}{a(X - \alpha_2 Y)
    \cdots (X - \alpha_d Y)}
  .\]
  Note
  \[
    a(X - \alpha_2 Y) \cdots (X - \alpha_d Y) \ge
    cY^{d - 1}
  \]
  on $\Gamma$, with some $c = c(P) > 0$. We may
  write $P = 0$ as:
  \[
    aXY^{d - 1} + bY^{d - 1} + \tilde{P}(X, Y)
  \]
  ($\tilde{P}$ of degree $\le d - 2$ in $Y$). $a$
  is not the same as in the factorisation of $F$
  and $a \neq 0$, but $b$ may be $0$.

  This gives:
  \[
    X = \frac{-b}{a} + Y^{-1} \cdot \ub{Q(X,
    Y^{-1})}_{\text{bounded}}
    \tag{$**$}
    \label{lec19_eq1}
  .\]
  For some polynomial $Q$.
  Then $\lim X = \frac{-b}{a}$ on $\Gamma$.

  Proving the first claim, suppose we can write
  \[
    f(X, Y) = R_1(X) Y^k + R_2(X) Y^{k - 1} +
    \cdots
    \tag{$**$}
    \label{lec19_eq2}
  .\]
  Here, negative exponents of $Y$ are allowed, but
  the sum must be finite. You can always do this
  with $k = D$ if $R_1 \left( -\frac{b}{a}
  \right) \neq 0$. Then $f(X, Y) \cdot Y^{-k} \to
  R \left( -\frac{b}{a} \right) \neq 0$ and
  $\ord_{p_1}(f) = -k$ and the claim holds.

  If $R_1 \left( -\frac{b}{a} \right) = 0$, then
  use \eqref{lec19_eq1} to write \eqref{lec19_eq2}
  with $k$ replaced by $k - 1$.

  Iterate this.
\end{proof}

\begin{lemma*}
  For each $j = 1, \ldots, d$, there is a basis
  $l_1, \ldots, l_n$ ($n = \dim V$) of $V$ such
  that
  \[
    \ord_{p_j}(l_i) \le -D + i - 1
  .\]
\end{lemma*}

\begin{proof}
  By induction, we show that there $l_1, \ldots,
  l_{i - 1}$ and $V_i \subset V$ such that
  \begin{align*}
    V
    &= \ol{Qbb} l_1 \oplus \cdots \oplus
    \ol{\Qbb} l_{i - 1} \oplus V_i \\
    \ord_j(l_k)
    &\le -D + k - 1
    &&\text{for $k = 1,\ldots, i - 1$} \\
    \ord_{p_j}(f)
    &\le -D + i
    &&\text{for $f \in V_i$}
  \end{align*}
  $i = 1$ is trivial: $V = V_1$.

  So suppose $i > 1$ and the claim holds for $i -
  1$. We define: $l_{i - 1}$ to be an element in
  $V_{i - 1}$ of minimal order at $p_j$.
  Let $V_i = \{f \in V_{i - 1} : \ord_{p_j}(f) >
  \ord_{p_j}(l_{i - 1})\}$.

  Just need to show: $V_{i - 1} = l_{i - 1}
  \ol{\Qbb} \oplus V_i$. To this end, let $g \in
  V_{i - 1}$. Write $m = \ord_{p_j}(l_{i - 1})$.
  Then
  \[
    \lim_{\Gamma_j} g \cdot V_1^m \eqdef b <
    \infty
  .\]
  \[
    f = g - \frac{b}{\lim_{\Gamma_j} l_{i - 1}
    \cdot Y^m} l_{i - 1}
  .\]
  Then
  \[
    \lim_{\Gamma_j} f Y^m = 0
  \]
  so by the previous lemma, $\ord_{p_j} f > m$. So
  $f \in V_i$.
\end{proof}

For this to be useful, we need $n$ to be large.
(We need $n \ge 2D + 2$).

\begin{lemma*}
  \[
    \dim V \ge dD - d(d - 1)
  .\]
\end{lemma*}

\begin{remark*}
  Thinking about $\Gamma$ as a projective curve,
  $V$ is the space of rational functions with
  poles of order at most $D$ at each point at
  $\infty$. By Riemann-Roch: $\dim V = dD - g +
  1$, provided $D$ is large enough.
\end{remark*}

\begin{proof}
  Let $R(X, Y) = \prod (X - \alpha_j Y)$ ($=
  \frac{F(X, Y)}{a}$). The point is that the
  polynomials
  \[
    Q_{jl}(X, Y) = \frac{R(X, Y)}{X - \alpha_j Y}
    \cdot Y^l \in V
  \]
  are linearly independent in $V$. $j = 1, \ldots,
  d$, $l = 1, \ldots, D - d + 1$. Suppose $Q =
  \sum_{j, l} \beta_{jl} Q_{jl}$ for some
  $\beta_{jl} \in \ol{\Qbb}$ not all $0$.

  Want to show $Q \neq 0$. To that end, let
  $\beta_{j', l'} \neq 0$ such that $l'$ is
  maximal with this property.

  We can show that:
  \[
    \lim_{\Gamma_j} Q(X, Y) \cdot Y^{-l' - d + 1}
    = \beta_{j', l'} \prod_{i = \{1, \ldots, d\}
    \setminus \{j'\}} {(\alpha_j' - \alpha_i)}
  .\]
  Uses the first lemma today.
\end{proof}