Let $C \subset \mathbb{P}^2_k$ an irreducible plane curve of degree $d >1$

over algebraically closed field $k$. That is $C=V(f(x,y,z))$ where $f \in k[x,y,z]$ homogeneous of degree $d$. Let $\{p_1,...,p_n\}$ be the singular points of $C$ and let $m_i$ the multiplicity of $C$ at $p_i$.

We look at the linear system $\vert L \vert$ of all curves (in $\mathbb{P}^2$) of degree $d − 1$ that have multiplicity $m_i − 1$ at every singular point $p_i \in C$. $\vert L \vert$ is not empty since e.g. the curve $V(\frac{\partial f}{\partial x})$ is contained in $\vert L \vert$.

For any $L \in \vert L \vert$, the intersection $L \cap C$ consists of points $p_i$, each with multiplicity $\ge m_i(m_i − 1)$ and a residual $R$. These build a linear system $\vert R \vert$ on $C$.

Two questions:

Why the multiplicity of $L \cap C$ in $p_i$ satisfies only the equality $\ge m_i(m_i − 1)$? Shouldn't it be strictly equal $m_i(m_i − 1)$?

Why following equality hold?

$$\dim \vert R \vert =\dim \vert L \vert = \binom{d+1}{2}-1 - \sum_i \binom{m_i}{2}$$

Recall that the dimension $\dim \vert D \vert$ of a linear system corresponding to a divisor $D$ is defined as dimension of projective variety $V_D= (\Gamma(X, \mathcal{L}(D))-\{0\}) / k^*$.

About the second equation it is clear that the $\binom{d+1}{2}-1$ represents the linear system curves of degree $d − 1$ in $\mathbb{P}^2$. The question is why the additional condition to have multiplicity $m_i − 1$ at every singular point $p_i \in C$ is encoded in $\sum_i \binom{m_i}{2}$?

generalmember $L \in \vert L \vert$ multiplicity of $L∩C$ in $p_i$ satisfies only the equality $=m_i(m_i−1)$? I think that causes my confusion... $\endgroup$ – katalaveino Apr 22 at 22:58not$R = L - \sum \operatorname{mult}_{p_i}(L \cap C) p_i$. That is, if $L \cap C$ has multiplicity $> m_i(m_i-1)$ at a point $p_i$, then $R$ is obtained by still subtracting $m_i(m_i-1) p_i$, not the higher multiple. I hope that this sheds some light on why $\dim |R| = \dim |L|$. $\endgroup$ – Zach Teitler Apr 23 at 3:49