Let $k$ be an algebraically closed field with positive characteristic $p>0$ and let $1<q_1<q_2<q_3$ be its powers. Let $X\subset \mathbb{P}^4$ be given by $(1:t:t^{q_1}:t^{q_2}:t^{q_2+1}+t^{q_1+q_3})$. Then $X$ has orders $0,1,q_1,q_2,q_2+1$ i.e. there exist hyperplanes that have this intersection multiplicity with the curve.

The fact that the orders $0,1$ are correct comes from the fact that every curve has such intersection multiplicities. My problem is that I am unable to prove the rest formally. My guess would be that the hyperplane $X_2=0$ has intersection multiplicity $q_1$ with $X$ at the point $(1:0:0:0:0)$, however, I can't come up with a proof.

I would greatly appreciate your help!


1 Answer 1


The intersection multiplicity of a curve $C\subset\Bbb P^n$ and a hyperplane $H$ at a point $c\in C$ is defined to be the length of $\mathcal{O}_{C,c}/(h|_C)$ as a module over the local ring $\mathcal{O}_{C,c}$, where $h$ is a local equation for $H$ and $h|_C$ is its restriction to $C$. In your case, taking $c$ to be the point corresponding to $t=0$, the local ring $\mathcal{O}_{C,c}$ is a DVR with uniformizer $t$, and the hyperplane $V(x_i)$ restricts to the $i$th coordinate of your embedding. This means you can just read off the length from the lowest exponent present in each coordinate:

  • $V(x_0)$ gives $\mathcal{O}_{C,c}/(1)$ which has length 0.
  • $V(x_1)$ gives $\mathcal{O}_{C,c}/(t)$ which has length 1.
  • $V(x_2)$ gives $\mathcal{O}_{C,c}/(t^{q_1})$ which has length $q_1$.
  • $V(x_3)$ gives $\mathcal{O}_{C,c}/(t^{q_2})$ which has length $q_2$.
  • $V(x_4)$ gives $\mathcal{O}_{C,c}/(t^{q_2+1}+t^{q_1+q_2})$ which has length $q_2+1$.

Note that these are local invariants associated to a specific point in a specific embedding - at other points or under different embeddings of the same abstract curve, you might get different numbers.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .