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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!

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1 Answer 1

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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.

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