MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

3 corrected mistake; edited body

Since my comment was too cryptic, I will spell it out as an answer.

A polarization on an abelian variety $A$ is an ample divisor $D$ (modulo linear equivalence). A polarization is principal if the self-intersection $D^g$ is equal to $g!$, where $g = \dim A$. It is well-known that the theta divisor is a principal polarization of the Jacobian. I am not sure what is a good reference, maybe Griffiths-Harris.

The Riemann-Roch theorem for abelian varieties (proved e.g. in Mumford) says that the Euler characteristic of the line bundle corresponding to a divisor $D$ is $D^g/g!$, so it is $1$ in the case of a principal polarization. Now, it is also shown in Mumford that only one $\dim H^i$ is non-zero, so if the divisor is ample it has to be $H^0$. Putting it all together $\dim H^0 =1$.

Edit: I was informed by email that my definition of polarization is too restrictive, so not quite right. The theta divisor defines a principal polarization anyway and this is proved in some of the other answers.

2 grammar

Since my comment was too cryptic, I will spell it out as an answer.

A polarization on an abelian variety $A$ is an ample divisor $D$ (modulo linear equivalence). A polarization is principal if the self-intersection $D^g$ is equal to $g!$, where $g = \dim A$. It is well-known that the theta divisor is a principal polarization of the Jacobian. I am not sure what is a good reference, maybe Griffiths-Harris.

The Riemann-Roch theorem for abelian varieties (proved e.g. in Mumford) says that the Euler characteristic of the line bundle corresponding to a divisor $D$ is $D^g/g!$, so it is $1$ in the case of a principal polarization. Now, it is also shown in Mumford that only one $\dim H^i$ is non-zero, so if the divisor is ample it has to be $H^0$. Putting it all together $\dim H^0 =1$.

1

Since my comment was too cryptic, I will spell it out as an answer.

A polarization on an abelian variety $A$ is an ample divisor $D$ (modulo linear equivalence). A polarization is principal if the self-intersection $D^g$ is equal to $g!$, where $g = \dim A$. It is well-known that the theta divisor is a principal polarization of the Jacobian. I am not sure what is a good reference, maybe Griffiths-Harris.

The Riemann-Roch theorem for abelian varieties (proved e.g. in Mumford) says that the Euler characteristic of the line bundle corresponding to a divisor $D$ is $D^g/g!$, so it $1$ in the case of a principal polarization. Now, it also shown in Mumford that only one $\dim H^i$ is non-zero, so if the divisor is ample it has to be $H^0$. Putting all together $\dim H^0 =1$.