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 $d1$, 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.
3 Answers
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 $f_1=x+y^{n1}$, $f_2=y+x^{n1}$. 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 $f_1+f_2=x+y+x^{n1}+y^{n1}$, which has an intersection multiplicity of two.
Edit: To make $C$ a curve of degree $n$, we multiply by any polynomial of degree $n2$ that that does not vanish at the origin.

$\begingroup$ thanks, although I wanted $C$ to be irreducible, I think I know where this is going. $\endgroup$– DöniCommented Jan 29, 2013 at 10:16

1$\begingroup$ Yeah, you can just deform it to something irreducible, maybe $xy+x^n+y^n$, without changing any multiplicities. $\endgroup$ Commented Jan 29, 2013 at 16:02

$\begingroup$ Reading the question is not my strongest suit. $\endgroup$ Commented Jan 29, 2013 at 16:05
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.

$\begingroup$ 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? $\endgroup$– DöniCommented Jan 29, 2013 at 10:24
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 $(\eta^*)^{1}(I_n)\cap \mathbb{C}[x,y]_{\le d1},$ where $\mathbb{C}[x,y]_{\le d1}$ denotes the set of polynomials of degree at most $d1$.

$\begingroup$ Besides, there is nothing special about degree d1 forms for this question. $\endgroup$– quimCommented Jan 29, 2013 at 9:48

$\begingroup$ Thank you, thats interesting. Does this implie, that $V_n$ is always at least some (finite) union of vector spaces? This assumption about degree $d1$ I only stated to get no common components of $C$ and $f$. $\endgroup$– DöniCommented Jan 29, 2013 at 10:20

1$\begingroup$ 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). $\endgroup$– quimCommented Jan 29, 2013 at 11:44

$\begingroup$ And for degrees $\ge d$, the multiples of the equation of $C$ are themselves a linear subspace, which is the intersection of all $V_n$. $\endgroup$– quimCommented Jan 29, 2013 at 11:45