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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$?

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    $\begingroup$ 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. $\endgroup$
    – Ariyan Javanpeykar
    Commented Aug 11, 2019 at 11:50
  • $\begingroup$ @AriyanJavanpeykar I have modified question now for genus $2$ case for simplicity. $\endgroup$
    – PSUN
    Commented Aug 11, 2019 at 12:39
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    $\begingroup$ 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. $\endgroup$
    – Ariyan Javanpeykar
    Commented Aug 11, 2019 at 15:38
  • $\begingroup$ Is it true that the condition $D.\theta=1$ implies that $J $ biregular to $ D\times \theta$? $\endgroup$
    – PSUN
    Commented Aug 12, 2019 at 5:51
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    $\begingroup$ Please read Ariyan's answer. If an abelian surface is isomorphic to a product, both factors are elliptic curves. $\endgroup$
    – abx
    Commented Aug 12, 2019 at 7:30

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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 $g(\theta )=2$.

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  • $\begingroup$ 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. $\endgroup$
    – PSUN
    Commented Aug 12, 2019 at 11:30
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    $\begingroup$ No, there is no nontrivial map from an abelian variety to a curve of genus $\geq 2$ (as Aryan said already). $\endgroup$
    – abx
    Commented Aug 12, 2019 at 11:54

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