Multiplicity of Birational Map Let $\varphi\in \text{Bir}(\mathbb{P}^2)$ be a birational map of degree $d$ sending the pencil of lines passing through a point $p\in \mathbb{P}^2$ onto a pencil of lines. How can we show that the multiplicity of $\varphi$ at $p$ is equal to $d-1$? 
My attempt: Let $m_p$ denote the mult of $p$. Using Noether's inequality I know that $d-1\ge m_p$. I am not sure how to derive the other inequality. 
Any hints or links to sources proving it will be appreciated! 
 A: Maybe not the best argument but I hope you can at least find some useful information.
Since a pencil of lines is completely determined by its base locus we have that $\varphi = T \circ \phi$ where $T$ is an automorphism of $\mathbb{P}^2$ and $\phi$ fixed a pencil of lines, which is known by a "de Jonquières transformation". Moreover we may assume that $p=(1:0:0)$. Then 
$$
\phi(x:y:z) = (F_0:F_1:F_2)
$$
with $F_i$ homogeneous of degree $d$. The pencil is given by $\{ay+bz=0 \mid (a:b)\in \mathbb{P}^1\}$ hence $aF_1+bF_2 = Q(cy+dz)$ for some $(c:d)\in \mathbb{P}^1$ and $Q$ a homogeneous polynomial of degree $d-1$. Therefore, there exist a (regular) map $\mathbb{P}^1 \to \mathbb{P}^1$ given by $(y:z) \mapsto (G_1:G_2)$ such that
$$
\phi(x:y:z) = (F_0:QG_1:QG_2)
$$
In order to $\phi$ be birational we need that, for a general point $q$, $\phi^{-1}(q)$ be a single point $\tilde q$. In other words, let $L_1,L_2$ be two lines intersecting at $q$ then 
$$
\phi^{-1}(L_1) \cap \phi^{-1}(L_2) = \{ \tilde q \} \cup {\rm base \,loci }
$$
Assuming that $L_1$ is an element of the pencil we have that $\phi^{-1}(L_1) = \{Q=0\}\cup \{c_1y+d_1z =0\}$. Hence
\begin{align*}
\phi^{-1}(L_1) \cap \phi^{-1}(L_2) & = (\{Q=0\}\cup \{c_1G_1+d_1G_2 =0\}) \cap \{a_2F_0 +b_2QG_1+c_2QG_2 =0\} = \\
& =\{Q=F_0 =0\} \cup (\{c_1y+d_1z =a_2F_0 +b_2QG_1+c_2QG_2 =0\}).
\end{align*}
We have that $\{Q=F_0 =0\}$ is in the base locus and, by Bézout's theorem, $\{c_1y+d_1z = a_2F_0 +b_2QG_1+c_2QG_2 =0\}$  has $d$ points (counted with multiplicities), $\tilde q$ included. We need then to force $d-1$ of them to be in the base locus hence the multiplicity of $a_2F_0 +b_2QG_1+c_2QG_2 =0$ at $(1:0:0)$ is $d-1$. 
