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Let $A$ be a simple abelian surface over $\mathbb{C}$. Let $C\subset A$ be an irreducible and reduced one-dimensional closed subscheme. Since $A$ is simple, the normalization of $C$ is of genus at least two. Let $L=\mathcal{O}_A(C)$.

Is $L$ a big line bundle on $A$?

I tried proving that $L$ is ample using Nakai-Moishezon's criterion, but I didn't manage.

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    $\begingroup$ Yes. Since $C$ is not an elliptic curve, there are points $x, y\in C$ such that $C'=(x-y)+C$ is not equal to $C$. But $C$ and $C'$ have the same class in $NS(A)$, so $C^2 = C.C' > 0$ since $y\in C\cap C'$. By Riemann-Roch, you get $h^0(L^n) \approx n^2$ and $L$ is big. $\endgroup$
    – Piotr Achinger
    Commented Feb 17, 2018 at 19:28
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    $\begingroup$ P.S. Using a similar argument but with another curve $D$ in place of $C$ shows that $C.D>0$, so $L$ is in fact ample by Nakai-Moishezon. $\endgroup$
    – Piotr Achinger
    Commented Feb 17, 2018 at 19:32
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    $\begingroup$ @PiotrAchinger Ok, I think I understand. The "similar" argument should be the following. To check that $C\cdot D>0$, we may assume $D\neq C$. Let $y\in D$ such that $y\not\in C$. Let $x\in C$. Then $(x-y) + D$ intersects $C$, because it contains $x$. If $(x-y) +D =C$, then we can use your first comment, because we already know $C^2>0$. If $(x-y) +D \neq C$, then clearly $C\cdot D>0$ (because the intersection is a finite set and contains $x$). Is this the argument you had in mind? $\endgroup$
    – Gonal_curve
    Commented Feb 17, 2018 at 21:01
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    $\begingroup$ Yes. Or: by previous point, we can assume that $D$ is not a translate of $C$, and then we translate $C$ so that it meets $D$. $\endgroup$
    – Piotr Achinger
    Commented Feb 17, 2018 at 21:12
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    $\begingroup$ @PiotrAchinger Ok, that's even better. Thank you. $\endgroup$
    – Gonal_curve
    Commented Feb 17, 2018 at 21:59

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