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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:

  1. 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 $m_i(m_i − 1)$?

  2. 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 $\sum_i \binom{m_i}{2}$?

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  • $\begingroup$ Regarding the last question: if you want a polynomial $f(x,y)$ and all its derivatives up to $m$ to vanish at a point $p$, it gives $\binom{m+2}{2}$ independent linear conditions on the coefficients. Indeed, if we assume $p = (0,0)$, then the condition is that monomials of $f$ of degrees up to $m$ vanish, and there are $1 + 2 + \dots + (m+1) = \binom{m+2}{2}$ of these. $\endgroup$
    – Evgeny Shinder
    Commented Apr 22, 2020 at 20:50
  • $\begingroup$ Regarding the first question, here is an example that might help understand why the statement is an inequality rather than an equation: the curve $C$ defined by $f = y^2 - x^2 + x^3$ has multiplicity $m=2$ at the origin and the curve (line) $L$ defined by $g = y-x$ has multiplicity $m-1=1$ at the origin. So the statement says that the multiplicity of $C \cap L$ at the origin is $\geq m(m-1) = 2$. What do you think is actually the multiplicity? $\endgroup$
    – Zach Teitler
    Commented Apr 22, 2020 at 22:35
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    $\begingroup$ @ZachTeitler: I see, it's of course $3$ here. The reason that leaded me to this wrong suspicion in question 1 on equality instead of inequality $\ge m_i(m_i − 1)$ was the caclulation of the degree $\operatorname{deg} \vert R \vert = d(d-1) + \sum_i m_i(m_i-1)$. It looks like application of Bezout’s theorem. And I think the point is that it's only true that for general member $L \in \vert L \vert$ multiplicity of $L∩C$ in $p_i$ satisfies only the equality $=m_i(m_i−1)$? I think that causes my confusion... $\endgroup$
    – user267839
    Commented Apr 22, 2020 at 22:58
  • $\begingroup$ any idea why $\dim \vert R \vert= \dim \vert L \vert$? $\endgroup$
    – user267839
    Commented Apr 22, 2020 at 23:20
  • $\begingroup$ It should be that for $L \in |L|$ (dubious notation) the corresponding $R$ is $R = L - \sum m_i(m_i-1) p_i$, not $R = L - \sum \operatorname{mult}_{p_i}(L \cap C) p_i$. 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
    Commented Apr 23, 2020 at 3:49

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