# Linear system on singular plane curve

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

• 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. – Evgeny Shinder Apr 22 at 20:50
• 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? – Zach Teitler Apr 22 at 22:35
• @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... – katalaveino Apr 22 at 22:58
• any idea why $\dim \vert R \vert= \dim \vert L \vert$? – katalaveino Apr 22 at 23:20
• 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 $\dim |R| = \dim |L|$. – Zach Teitler Apr 23 at 3:49