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Let $C \subseteq \mathbb{P}^2$ be an irreducible plane curve of degree $d$ and let $p \in C$. For any integer $n \geq 1$, let $V_n$ be the set of all forms $f$ of degree $d-1$, such that $f$ vanishes in $p$ and such that the intersection multiplicity of $C$ and the curve defined by $f$ is at least $n$. If $p$ is a smooth point of $C$, $V_n$ is a vector space. Is this also the case, if $p$ is a singularity of $C$? In particular, I am interested in the case, where the ground field are the complex numbers.

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No. We might as well work in the affine plane for simplicity, since intersection multiplicity is defined locally.

Let $C$ be defined by the equation $xy$. and let $P$ be the point $x=y=0$. Let $f_1=x+y^{n-1}$, $f_2=y+x^{n-1}$. These have intersection multiplicty $n$ with $C$, which you can compute by adding their intersection multiplicity with the curve defined by $x$ to their intersection multiplicity with the curve defined by $y$.

But $f_1+f_2=x+y+x^{n-1}+y^{n-1}$, which has an intersection multiplicity of two.

Edit: To make $C$ a curve of degree $n$, we multiply by any polynomial of degree $n-2$ that that does not vanish at the origin.

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thanks, although I wanted $C$ to be irreducible, I think I know where this is going. –  Döni Jan 29 '13 at 10:16
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Yeah, you can just deform it to something irreducible, maybe $xy+x^n+y^n$, without changing any multiplicities. –  Will Sawin Jan 29 '13 at 16:02
    
Reading the question is not my strongest suit. –  Will Sawin Jan 29 '13 at 16:05
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No, it is not always a vector space. Take $C$ a cubic with a node at $p$ and $n=3$. Then your set $V_3$ is a union of two vector spaces each corresponding to the conics that pass through $p$ and having as tangent at $p$ the tangent to one of the two branches to $C$ at $p$. The intersection of these two vector spaces is the set of multiples of the product of these two lines.

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okay thank you, this settles exactly the assumptions I made. Is it at least true, that Vn is always a finite (union) of vector spaces? –  Döni Jan 29 '13 at 10:24
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If $C$ is unibranch at $p$, $V_n$ is always a vector space. If $C$ is not unibranch, there exist $n$ such that $V_n$ is not a vector space (as noted by the previous answers).

To see the first claim, consider the normalization $\eta:\tilde C \rightarrow C$ of the curve. $C$ is unibranch at $p$ if $\eta^{-1}(p)$ is a single point $q$. Let $x,y$ be affine coordinates in a neighborhood of $p$, and $t$ a parameter of $\tilde C$ at $q$. Consider the induced map $\eta^*:\mathbb{C}[x,y]\rightarrow \mathbb{C}[[t]]$ and the ideal $I_n=(t^n)\subset \mathbb{C}[[t]]$. Then $V_n$ can be identified with $(\eta^*)^{-1}(I_n)\cap \mathbb{C}[x,y]_{\le d-1},$ where $\mathbb{C}[x,y]_{\le d-1}$ denotes the set of polynomials of degree at most $d-1$.

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Besides, there is nothing special about degree d-1 forms for this question. –  quim Jan 29 '13 at 9:48
    
Thank you, thats interesting. Does this implie, that $V_n$ is always at least some (finite) union of vector spaces? This assumption about degree $d-1$ I only stated to get no common components of $C$ and $f$. –  Döni Jan 29 '13 at 10:20
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Yes. If there are $k$ branches, the intersection multiplicity with $C$ is the sum of the intersections with each branch, so there would be a vector subspace for each partition of $n=n_1+\dots+n_k$. (In fact not all partitions need to be considered, because of the intersections of branches with one another). –  quim Jan 29 '13 at 11:44
    
And for degrees $\ge d$, the multiples of the equation of $C$ are themselves a linear subspace, which is the intersection of all $V_n$. –  quim Jan 29 '13 at 11:45
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