# Dimension of a linear system of divisors on singular curve

Consider an singular irreducible plane curve $$C \subset \mathbb{P}^2_k$$ of degree $$d>1$$ over algebraically closed field $$k$$ which is given as vanishing locus $$C=V(f(x,y,z))$$ of a $$f \in k[x,y,z]$$ homogeneous of degree $$d$$. Let $$\{p_1,...,p_n\}$$ be the singular points of $$C$$. Assume $$\vert D \vert$$ be a linear system of divisors; that is it consist of all $$D' \in \vert D \vert$$ are divisors $$D' \subset C$$ such that there exist a $$f \in K(C)$$ with $$D' = \operatorname{div}(f) +D$$.

We assume that every member of $$\vert D \vert$$ has every singular point $$p_i$$ the multiplicity $$\ge a_i$$ ( where $$a_i \in \mathbb{N}$$ with $$a_i \ge 1$$)

That is we can build another linear system $$\vert L \vert$$ consists of all $$L':= D' - \sum_i a_i \cdot (p_i)$$ where $$D' \in D$$.

Question: Why and how to see that $$\dim_k \vert D \vert= \dim_k \vert L \vert$$?

This question is closely related to my other question Linear system on singular plane curve

• What do you call a linear system on a singular curve? $L'$ is not a Cartier divisor. – abx May 18 at 17:11
• The question arised from a proof in Janos Kollar's "Lectures on Resolution of Singularities" (page 39). The proof is also quoted literally here: mathoverflow.net/questions/358245/… The point was that he started with a certain linear system on $\mathbb{P}^2$, then pulled it back to $C$ and obtained in a way I explaned above (by throwing away singular points with certain multiplicity from the pullbacks) "linear system"(at least he called it so) on $C$ of residuals. – katalaveino May 19 at 1:22
• You are right, the terminology "linear system" isn't common on singular curves, do you have an idea what Kollar there had mind? – katalaveino May 19 at 1:23