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rita
rita
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I think your claim is not true.

Let $H$ be the hyperplane section of $X$ and assume that $C$ is not linearly equivalent to a multiple of $H$. Then $h^0(X, C+rH)\to\infty$ for $r\to\infty$ (Rieman-Roch + Serre duality). On the other hand, $C+rH$ is not linearly equivalent to a multiple of $H$, so it is not a complete intersection.

I think your claim is not true.

Let $H$ be the hyperplane section and assume that $C$ is not linearly equivalent to a multiple of $H$. Then $h^0(X, C+rH)\to\infty$ for $r\to\infty$ (Rieman-Roch + Serre duality). On the other hand, $C+rH$ is not linearly equivalent to a multiple of $H$, so it is not a complete intersection.

I think your claim is not true.

Let $H$ be the hyperplane section of $X$ and assume that $C$ is not linearly equivalent to a multiple of $H$. Then $h^0(X, C+rH)\to\infty$ for $r\to\infty$ (Rieman-Roch + Serre duality). On the other hand, $C+rH$ is not linearly equivalent to a multiple of $H$, so it is not a complete intersection.

Source Link
rita
rita
  • 6.3k
  • 1
  • 29
  • 39

I think your claim is not true.

Let $H$ be the hyperplane section and assume that $C$ is not linearly equivalent to a multiple of $H$. Then $h^0(X, C+rH)\to\infty$ for $r\to\infty$ (Rieman-Roch + Serre duality). On the other hand, $C+rH$ is not linearly equivalent to a multiple of $H$, so it is not a complete intersection.