6

I am studying the Abel Jacobi map $$\mathrm{Div}_{X/k} \to \mathrm{Pic}_{X/k}$$ for projective, smooth, irreducible curve $X/k$ where $k$ is algebraically closed. Let $S = \operatorname{Spec}(k)$, $T$ a scheme over $k$ and let $\mathcal{L}$ be a line bundle on the base change $X_T = X \times_S T$. As the Picard functor is representable for projective, smooth, irreducible curves, this gives rise to a morphism $T \to \mathrm{Pic}_{X/k}$. I want to understand the scheme theoretic fiber $$\mathrm{Div}_{X/k} \times_{\mathrm{Pic}_{X/k}} T.$$ I am in particular interested in the case that $T=S$. I believe that in this case, it is given by the projective space $\mathbb{P}(H^0(X,\mathcal{L})^*)$ where $H^0(X,\mathcal{L})^*$ is the dual of $H^0(X,\mathcal{L})$. I wanted to apply Proposition 8.2.7 of Bosch, Néron Models. It says that if $\mathcal{L}$ is cohomologically flat in dimension 0, then the fiber is represented by the projective $k$-scheme $\mathbb{P}((f_*\mathcal{L})^*)$ where again, $(f_*\mathcal{L})^*$ is the dual of $f_*\mathcal{L}$.

My questions are:

  1. Is the line bundle indeed cohomologically flat in dimension 0 in our setting?

  2. If the first question is true, why does $f_*\mathcal{L}$ correspond to $H^0(X, \mathcal{L})$?

  3. Why isn't it sufficient to understand the fibers of the $k$-points?

bjck
  • 81
  • Just to point out, there are typos. It should be "Let $ \mathcal{L} $ be a line bundle on $ X \times_S T $", and the scheme theoretic fiber is actually "fiber product with $ T $". Other than that, good question, I would like to know the answer too. – Cranium Clamp May 23 '22 at 04:15
  • I corrected it, thankks! – bjck May 23 '22 at 09:22

1 Answers1

2

Indeed, in the case that $T=Spec(k)$, but also in the case that $T$ is arbitratry and $\mathcal{L}$ is of degreee $d > 2g-2$, $\mathcal{L}$ is cohomologically flat of dimension zero, where $g$ is the genus of the curve. I will come to this later.

Using this, the fiber is representable by $\mathbb{P}((f_*\mathcal{L})^*)$ and as $$(f_*\mathcal{L})(Spec(k))=\Gamma(X,\mathcal{L})=H^0(X,\mathcal{L}),$$ we have $\mathbb{P}((f_*\mathcal{L})^*)=\mathbb{P}(H^0(X,\mathcal{L})^*)$. We note that $H^0(X,\mathcal{L})^*$ is a finite dimensional vectorspace over $k$. Thus, $\mathbb{P}((f_*\mathcal{L})^*)=\mathbb{P}_k^{dim H^0(X,\mathcal{L})^*-1}$.

In the case that $T=Spec (k)$, cohomological flatness follows from flat basechange.

In the more general case, we first note that the degree map $t \to \mathcal{L}_t$ is locally constant. Now, let $d = deg \mathcal{L} > 2g-2$. We assume that $T=Spec (A)$ is open affine. Let $K^{\cdot}=(K^0 \to K^1)$ be a complex of finitely generated projective $A$-modules such that for all $A$-algebras $B$, $$H^p(K^{\cdot}\otimes_{A}B) \simeq H^p(X,\mathcal{L}), \quad p \geq 0.$$ Such a complex existes by Mumford, Abelian Varieties, Chap II, §5. We want to show that $H^1(X,\mathcal{L})=0$, i.e. $coker(d^0 \otimes_A B)=0$. By Nakayama's lemma it suffices to show this claim for $B=k(p), \, p \in Spec(A)$. Thus, we may assume that $T=Spec (k)$. By Serre duality $$h^1(X,\mathcal{L})=h^0(X,\mathcal{L}^*\otimes\omega_X)$$ where $\omega_X$ is the canonical line bundle. But $deg(\omega_X)=2g-2$ and $deg(\mathcal{L}^*)<2-2g$. Thus, $h^1(X,\mathcal{L})=0$, which shows cohomological flatness in dimension $0$.

For $\mathcal{L}$ a linebundle of degree $\leq 2g-2$, I think that cohomological flatness in dimension zero does not hold in general. However, by the proposition I've mentioned in the question, the fiber is represented by a propjective space $\mathbb{P}(\mathcal{F})$ for some $\mathcal{O}_T$-module $\mathcal{F}$ locally of finite presentation. The construction of $\mathcal{F}$ is given in Bosch, Néron models, Theorem 8.1.7.

bjck
  • 81