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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

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    $\begingroup$ What do you call a linear system on a singular curve? $L'$ is not a Cartier divisor. $\endgroup$
    – abx
    Commented May 18, 2020 at 17:11
  • $\begingroup$ 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. $\endgroup$
    – user267839
    Commented May 19, 2020 at 1:22
  • $\begingroup$ You are right, the terminology "linear system" isn't common on singular curves, do you have an idea what Kollar there had mind? $\endgroup$
    – user267839
    Commented May 19, 2020 at 1:23

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