I was reading Miranda's Riemann surface book, there is a key intermediate lemma in proving the Bezout theorem in page 142:
Let $X$ be a smooth projective curve in $\Bbb{P}^2$ with defining homogenuous polynomial $F$ of degree $d$, assume further that $[0:0:1] \notin X$, therefore $[x:y:z]\in X$ can not have $x,y$ vanish simultaneously, we can define the meromorphic function $h([x:y:z]) = x/y$ on $X$, which is the same as a holomorphic function $H:X\to \Bbb{C}_\infty$.
the question is about the degree of the map $H$ above. I want to prove the degree is $d$.
consider $\lambda \in \Bbb{C}$ such that $\lambda \ne 0$, then the points on the preimage $H^{-1}(\lambda)\subset X$ can not have both $x$ and $y$ component being zero, therefore the preimage are set of points $[\lambda:1:w]\in \Bbb{P}^2$ satisfies the equation $F(\lambda,1,w) =0$.
By the fundamental theorem of algebra, we know it has $d$ solution, then Miranda says that for a general $\lambda$ these solutions are distinct, and $H$ has multiplicity one at all of those solutions.
I know since $X$ is compact the branch points are discrete, therefore for general $\lambda$ they have multiplicity 1, however, I don't understand why the solution are distinct.
I found the same question in this post and the solution seems does not answer the question ?
Claim: $F(x,1,z)$ is irreducible in $\mathbb{C}[x,z]$.
Proof: if $F(x,1,z)=Q(x,z)R(x,z)$ is a nontrivial factorization write $\tilde{Q}(X,Y,Z)=Y^qQ(X/Y,Z/Y)$ where $q$ is the degree of $Q$ and same for $\tilde{R}$. Then we formally have $F=\tilde{Q}\tilde{R}$ with $\tilde{Q},\tilde{R}$ homogeneous. As $F$ is smooth, it is irreducible, and we get a contradiction.
Now, by assumption, $F(\lambda,1,X)=cX^d+\sum_{0 \leq k < d}{P_k(\lambda)X^k}$ where $P_k$ has degree at most $d-k$ and $c \neq 0$ is a constant.
It follows (assuming $d \geq 2$) that $F_X=\partial_X F(\lambda,1,X)$ is not a unit and of lower degree with respect to $X$ than $F_1=F(\lambda,1,X)$, so these polynomials have no common divisor in $\mathbb{C}[\lambda,X]$.
By Gauss lemma, these polynomials as coprime (where the field of coefficients is $\mathbb{C}(\lambda)$): it follows that there are polynomials $A(\lambda,X), B(\lambda,X), C(\lambda) \neq 0$ such that $AF_1+BF_X=C$.
In particular, if $C(\lambda) \neq 0$ (which happens for all but finitely many $\lambda$) $F(\lambda,1,X)$ is coprime with its derivative (wrt $X$) and has degree $d$, so it has exactly $d$ pairwise distinct roots.