Do curves generate their Jacobians? Let $A$ be an abelian variety of dimension $g$ over an algebraically closed field, and let $C\subset A$ be a curve, not necessarily smooth. We say that $C$ generates $A$ if every point $a\in A$ can we written under the group law as a sum of $g$ points on $C$. This is also equivalent to $A$ being the smallest abelian variety containing $C$. I think the following is true but I do not know how to prove it:

If $C$ is a smooth genus $g$ curve and $A=J(C)$ is its Jacobian, then $C$ generates $A$.

The curve is thought of as embedded via an Abel-Jacobi mapping in $J(C)$, here. Does anyone know if this is true, and how to prove it?
Thanks a lot!
 A: By the Abel-Jacobi theorem, $J(C) \cong Pic^0$ where $Pic^0$ is parametrizing degree $0$ divisors modulo linear equivalence on $C$. The embedding of $C$ is given by $p \rightarrow p - p_0$ for some fixed $p_0 \in C$. Then every degree zero divisor is linearly equivalent to the sum of $g$ such elements $p_i - p_0$, i.e. to $p_1 + ... + p_g - g p_0$ for some $p_i \in C$. 
Indeed, given a general degree zero divisor $\sum_{i = 1}^{n} p_i - \sum_{i =1}^{n} q_i$, we can assume $n \geq g$ by otherwise filling up the sum with $p_0 - p_0$. By Riemann Roch, we know that a divisor of degree greater or equal to $g$ is effective, thus if $n > g$ we can replace $\sum_{i =1}^n p_i - \sum_{i=1}^{n-g} q_i$ by an effective divisor. In other words, we can assume $n = g$. 
So we have reduced to the case $\sum_{i = 1}^{g} p_i - \sum_{i =1}^{g} q_i$. We now replace the $p_i$ by $p_i' = p_0 + p_i - q_i$ (which are of degree one, possibly not effective) and get $\sum_{i = 1}^{g} p_i - \sum_{i =1}^{g} q_i = \sum_{i = 1}^{g} p_i' - g p_0$. Finally, $\sum_{i = 1}^{g} p_i'$ is of degree $g$, so as above we can replace it by something effective and get the representation we wanted. 
