Set of plane curves which intersect a fixed curve with given multiplicity

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.

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.

• thanks, although I wanted $C$ to be irreducible, I think I know where this is going.
– Döni
Commented Jan 29, 2013 at 10:16
• Yeah, you can just deform it to something irreducible, maybe $xy+x^n+y^n$, without changing any multiplicities. Commented Jan 29, 2013 at 16:02
• Reading the question is not my strongest suit. 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.

• 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
Commented 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 $(\eta^*)^{-1}(I_n)\cap \mathbb{C}[x,y]_{\le d-1},$ where $\mathbb{C}[x,y]_{\le d-1}$ denotes the set of polynomials of degree at most $d-1$.

• Besides, there is nothing special about degree d-1 forms for this question.
– quim
Commented Jan 29, 2013 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
Commented Jan 29, 2013 at 10:20
• 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
Commented Jan 29, 2013 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
Commented Jan 29, 2013 at 11:45