3

Vakil's Foundations of Algebraic Geometry, Proposition 20.1.4 is that intersection multiplicity depends only on numerical equivalence classes. The proof uses base change to an algebraically closed field. I'm not sure why we can do this.

Let $X$ be a projective variety over a field $k$, not necessarily algebraically closed. Let $L_1, L_2$ be invertible sheaves on $X$ that are numerically equivalent. Let $M_1, M_2$ be the pullbacks on $X \times_k \overline{k}$. Are $M_1$ and $M_2$ also numerically equivalent?

This is equivalent to assuming $L_1 \otimes L_2^\vee $ is numerically trivial, and concluding $M_1 \otimes M_2^\vee$ is numerically trivial. Therefore, I'll just write $L$ and $M$.

Let $i : C \rightarrow X \times_k \overline{k}$ be a closed immersion where $C$ is a reduced curve. Then, we want to prove $deg(i^* M) =0$. However, since $C$ is over $\overline{k}$ and $X \otimes_k \overline{k} \rightarrow X$ is not necessarily a closed immersion, I don't know how to continue.

I tried to take the scheme-theoretic closure $i' : C' \rightarrow X$ of $\pi_1 \circ i : C \rightarrow X$. But I'm not sure how to connect $deg(i'^*(L))$ to $deg(i^*(M))$. I'm also not sure if $i'$ still satisfies the conditions necessary to apply that $L$ is numerically trivial. Especially since $C$ might not have any $k$-valued points.

KReiser
  • 65,137
David Lui
  • 6,295
  • Exercise 18.4.T is to show that numerical equivalence is preserved under proper pull-back, but the field extension $\operatorname{Spec} \overline k \to \operatorname{Spec} k$ need not be proper (e.g. if it is not of finite type). – red_trumpet May 25 '22 at 13:57
  • So, it sounds like you are really asking about exercise 20.1.D – Karl Kroningfeld May 30 '22 at 18:48
  • I understand 20.1.D, it's numerical equivalence that I'm stuck on. – David Lui May 31 '22 at 06:22

1 Answers1

3

An answer:

Numerical equivalence is indeed preserved by base change to algebraically closed field. Here's a sketch of why. We start over a field $k$. We'll take $k$ to be perfect; the case where inseparable extensions intervene is a small twist to the story. Suppose you have a line bundle $L$ that is meets all curves with multiplicity 0. Now pull back to the algebraic closure, and choose a curve there. Suppose $L$ meets the curve with multiplicity $n$. We wish to show that $n=0$. The key is to show that $L$ meets each of its (finite number of) Galois conjugates in the same number. So $L$ meets the union of its conjugates in a positive multiple of $n$. Then relate this to the intersection over $k$ of $L$ with a curve there.

Answer to the question behind your question:

In those notes, I didn't realize this issue, so thanks for noticing it! I didn't want the reader to have to construct the above argument, and intend to rewrite it so that we don't have to go to the algebraic closure.

Ravi Vakil
  • 131