MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to understand the construction of the Jacobian of a curve following the notes of J. S. Milne

The question is going to be about a particular step in the proof of Proposition 4.2b in Chapter III, but I will first briefly recall the setup.

Let $X$ be a scheme flat over $T$, a divisor $D$ on $X$ is called relative effective divisor on $X/T$ if it is effective and flat over $T$ as a subscheme of $X$ (definition 3.4). There is a one-to-one correspondence benween relative effective divisors and sheaves $\mathcal L$ over $X$ with a global section $s$ such that $\mathcal{L}/s\mathcal{O}_X$ is flat over $T$.

We are working over a field. Let $C$ be a non-singular curve of genus $\geq 2$.

We are trying to construct a section of the natural map of functors $Div^r_C \to P^r_C$ where the first functor is the functor of relative effective divisors on $C\times T/T$ of degree $r$, and is represented by the $r$-fold symmetric product of $C$, and the second functor is the functor of families of degree $r$ invertible sheaves on $C$ parametrised by $T$, modulo trivial families:

$$ P^r_C(T) = \{ \mathcal{L} \in Pic(C \times T) \mid deg\ \mathcal{L}_t=r \textrm{ for all }\ t \in T\} / q^* Pic(T) $$

(the natural projections are denoted $p: C \times T \to C$, $q: C\times T \to T$.)

Proposition 4.2 deals with subfunctors of $Div^r_C$ and $P^r_C$. We pick an effective degree $(r-g)$ divisor $D_\gamma$ and define

$$ C^\gamma(T) = \{ D \in Div^r_C(T) \mid h^0(D_t-D_\gamma)=1\ \textrm{ for all }\ t \in T\} $$ $$ P^\gamma(T) = \{ \mathcal{L} \in Pic^r_C(T) \mid h^0(\mathcal{L}_t \otimes \mathcal{L} _\gamma^{-1})=1\ \textrm{ for all }\ t \in T\} $$

part b) constructs a section $P^\gamma \to C^\gamma$. Take $\mathcal{L} \in P^\gamma(T)$. By definition of $P^\gamma$ and by Riemann-Roch, $h^1 (\mathcal{L}_t \otimes \mathcal{L}^{-1}_\gamma)=0$. This allows us to apply a base change theorem and coclude that $q_*(\mathcal{L} \otimes p^* \mathcal{L}^{-1} _\gamma)$ is locally free and thus an invertible sheaf on $T$ (call it $\mathcal{M}$). The proof then proceeds to construct a section of $\mathcal{L} \otimes (q^* q_*(\mathcal{L} \otimes p^* \mathcal{L} _\gamma^{-1}))^{-1}$.

In particular, as there is a natural map $q^* q_*(\mathcal{L} \otimes p^* \mathcal{L} _\gamma^{-1}) \to \mathcal{L} \otimes p^* \mathcal{L} _\gamma^{-1}$, one has a canonical global section of $\mathcal{L} \otimes p^* \mathcal{L} _\gamma^{-1} \otimes (q^*\mathcal{M})^{-1}$, and by composing it with the natural map $p^* \mathcal{L} _\gamma^{-1} \to \mathcal{O}_{C\times T}$ one gets the desired.

We did obtain a section $s_\gamma$ of a sheaf equivalent to $\mathcal{L}$ (in the sense of the definition of $Pic^r_C$)

My question is: why does this section give rise to a relative effective divisor, that is, why would $\mathcal{L} \otimes (q^* \mathcal{M})^{-1}/s_\gamma \mathcal{O}_{C\times T}$ be flat over $T$?

share|cite|improve this question
up vote 2 down vote accepted

Let $N = L \otimes (q^\ast q_\ast (L\otimes L_\gamma^{-1}))^{-1}$. It suffices to show that the zero locus $D \subset C \times T$ of $s \in \Gamma(N)$ is flat over $T$. If $T$ is nice (Noetherian, blah, blah), it then suffices to show that the fiberwise degree of $D$ is constant. Note that the restriction of $N$ to $C \times \{ t \}$ is isomorphic to the restriction of $L$ to $C \times \{ t \}$. Since $L$ is fiberwise degree $r$ by assumption, it follows that $D$ is fiberwise degree $r$.

Oh and you need to check that the section $s$ is fiberwise nonzero.

Well, here's how you do that --- first look at the map $\phi : q^\ast q_\ast (L \otimes p^\ast L_\gamma^{-1}) \to L \otimes p^\ast L_\gamma^{-1}$. What does this map look like on fibers? Well, note that $H^0 (C \times \{ t \} , (L \otimes p^\ast L_\gamma^{-1})|\_{C \times \{ t \}})$ is by assumption 1 dimensional. Hence the restriction of $q^\ast q_\ast (L \otimes p^\ast L_\gamma^{-1})$ to $C \times \{ t \}$ is a trivial line bundle. So, the map $\phi$ on the fiber $C \times \{ t \}$ looks like the map $\mathcal{O}\_{C \times \{ t \}} \to (L \otimes p^\ast L_\gamma^{-1})|\_{C \times \{ t \}}$ corresponding to the one nonzero global section in $H^0 (C \times \{ t \} , (L \otimes p^\ast L_\gamma^{-1})|\_{C \times \{ t \}})$. Clearly this map is not zero.

I'll let you do the rest...

share|cite|improve this answer
"check that the section s is fiberwise nonzero" - this is actually my problem. – Dima Sustretov Aug 24 '11 at 14:36
oops, just have noticed that I didn't accept the answer, sorry for being slow. – Dima Sustretov Oct 25 '11 at 16:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.