# Existence of divisor in the Jacobian of smooth curve of genus two whose intersection with theta divisor is 1

Let $$C$$ be a smooth projective curve of genus $$2$$ and $$J$$ denotes the Jacobian of $$C$$. Let $$\theta$$ be the image of $$C$$ under the abel Jacobi map. Is there exist a divisor $$D$$ in $$J$$ such that $$D.\theta=1$$?

Is it true that the condition $$D.\theta=1$$ implies that $$J$$ biregular to $$D\times \theta$$?

• Assuming $g(C)>1$, you can not have $J=D\times \theta$, because an abelian variety (such as $J$) does not dominate any curve of genus at least two. – Ariyan Javanpeykar Aug 11 at 11:50
• @AriyanJavanpeykar I have modified question now for genus $2$ case for simplicity. – PSUN Aug 11 at 12:39
• There can be no isomorphism $J = D\times \theta$. If there is, then there would be a surjective morphism $J\to \theta$, but $\theta \cong C$, so $J$ would dominate the genus two curve $C$. That is impossible. – Ariyan Javanpeykar Aug 11 at 15:38
• Is it true that the condition $D.\theta=1$ implies that $J$ biregular to $D\times \theta$? – PSUN Aug 12 at 5:51
• Please read Ariyan's answer. If an abelian surface is isomorphic to a product, both factors are elliptic curves. – abx Aug 12 at 7:30

## 1 Answer

I think I understand your question now: you want your $$D$$ to be effective, hence irreducible since $$\theta$$ is ample. Now the index theorem $$(D\cdot \theta )^2\geq D^2\cdot\theta ^2\$$ implies $$D^2=0$$, hence $$D$$ must be a smooth elliptic curve. Consider the quotient map $$p:J\rightarrow J/D$$. The condition $$D\cdot \theta=1$$ means that the restriction of $$p$$ to $$\theta$$ is one-to-one, which is impossible since $$g(\theta )=2$$.

• Thanks. Is there exist any map from $J\to D\times \theta$? Just by using the fact $D.\theta=1$, where $D$ is effective divisor. I am very sorry if I am asking very silly question. – PSUN Aug 12 at 11:30
• No, there is no nontrivial map from an abelian variety to a curve of genus $\geq 2$ (as Aryan said already). – abx Aug 12 at 11:54