Let $Y=Z(g)$ and $H=Z(f)$ be projective plane curves in $\mathbb{P}^2$ given by irreducible homogeneous polynomials $f,g\in K[T_0,T_1,T_2]$ of degrees deg$(f)=d$ and deg$(g)=1$. $Y$ is a projective line and we assume $Y\neq H$.
I have showed that there exists the parametrization of $Y$, i.e. the isomorphism $\varphi:\mathbb{P}^1\rightarrow Y$.
To show:
(i) $p:=f\circ \varphi$ is a non zero polynomial $p\in K[S_0,S_1]$. Explain why you can view $p$ as a polynomial of the same degree in one variable.
(ii) For $P\in Y\cap Z$ and $Q:=\varphi^{-1}(P)$ prove that the intersection multiplicity $i(Y,H,P)$ is the multiplicity of the zero $Q$ of the polynomial $p$.
Then I have to prove Bézout's theorem in this special case using the statements above.
I'm very stack and any suggestion would be helpful
Best Answer
Since $Y$ is a line, we may apply an automorphism of $\mathbb{P}^2$ to assume this line is given by $\{x = 0\}$, where we consider $x, y, z$ the homogeneous coordinates of $\mathbb{P}^2$. (As an exercise you should explicitly describe this automorphism!) We then consider $\varphi$ to be the inclusion of this closed subset.
We can then consider $Y$ to be $\mathbb{P}^1$ with coordinates $y$ and $z$, so that with this identification, $Y \cap H \subset Y$ is the zero set of $\overline{f}(y,z) := f(\varphi(y,z)) = f(0, y, z)$. Since $H$ is irreducible $f$ has a monomial not involving $x$, so $\overline{f}$ a homogeneous form of degree $d$ in $y$ and $z$. Hence if $K$ is algebraically closed we can factor $\overline{f} = l_1^{m_1} \cdots l_r^{m_r}$ where $l_i$ are linear forms and $\sum m_i = d$. Each of these linear forms corresponds to a point $\{l_i =0\} = \{P_i\}$. We claim that the intersection multiplicity $I(Y \cap H; P_i) = m_i$.
Given $P_i$, the usual definition is $$I(Y \cap H; P_i) = \dim_{K} \mathcal{O}_{\mathbb{P}^2, P_i}/(f, g) = \dim_K \mathcal{O}_{\mathbb{P}^1, P_i}/(\overline{f}).$$
Now, to conclude, since $l_j$ is invertible in $\mathcal{O}_{\mathbb{P}^1, P_i}$ for $j \neq i$, it follows that the quantity on the right above is the same as $\dim_K \mathcal{O}_{\mathbb{P}^1, P_i}/(l_i^{m_i})$.
Since we're looking at local rings, there's no harm in dehomogenizing and moving $P_i$ to the origin of $\mathbb{A}^1$, so that we're looking to determine $\dim_{K} K[x]_{(x)}/x^{m_i}$. But now, $K[x]_{(x)}/x^{m_i} \cong K[x]/x^{m_i}$ has $K$ basis given by $1, x, x^2, \dots, x^{m_i - 1}$, so the dimension of this vector space is $m_i$ and we're done.