Elliptic Curve has Canonical Bundle $K_E = mathcal{O}_E$
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Let $E$ be an elliptic curve over a field $k$. I'm looking for a proof that for the canonic divisor $K_E$ we have $K_E = mathcal{O}_E$? RR says $h^0(K_E) = h^0(mathcal{O}_E)+deg(K_E) +g-1= h^0(mathcal{O}_E)=1$ since $E$ elliptic. I don't see wy this already imply $K_E = mathcal{O}_E$.
algebraic-geometry elliptic-curves
asked Jan 14 at 3:57
KarlPeterKarlPeter
5341315
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add a comment |
$begingroup$
Let $E$ be an elliptic curve over a field $k$. I'm looking for a proof that for the canonic divisor $K_E$ we have $K_E = mathcal{O}_E$? RR says $h^0(K_E) = h^0(mathcal{O}_E)+deg(K_E) +g-1= h^0(mathcal{O}_E)=1$ since $E$ elliptic. I don't see wy this already imply $K_E = mathcal{O}_E$.
algebraic-geometry elliptic-curves
asked Jan 14 at 3:57
KarlPeterKarlPeter
5341315
$endgroup$
add a comment |
$begingroup$
Let $E$ be an elliptic curve over a field $k$. I'm looking for a proof that for the canonic divisor $K_E$ we have $K_E = mathcal{O}_E$? RR says $h^0(K_E) = h^0(mathcal{O}_E)+deg(K_E) +g-1= h^0(mathcal{O}_E)=1$ since $E$ elliptic. I don't see wy this already imply $K_E = mathcal{O}_E$.
algebraic-geometry elliptic-curves
asked Jan 14 at 3:57
KarlPeterKarlPeter
5341315
$endgroup$
Let $E$ be an elliptic curve over a field $k$. I'm looking for a proof that for the canonic divisor $K_E$ we have $K_E = mathcal{O}_E$? RR says $h^0(K_E) = h^0(mathcal{O}_E)+deg(K_E) +g-1= h^0(mathcal{O}_E)=1$ since $E$ elliptic. I don't see wy this already imply $K_E = mathcal{O}_E$.
algebraic-geometry elliptic-curves
algebraic-geometry elliptic-curves
asked Jan 14 at 3:57
KarlPeterKarlPeter
5341315
asked Jan 14 at 3:57
KarlPeterKarlPeter
5341315
edited Jan 14 at 4:12
KarlPeter
asked Jan 14 at 3:57
KarlPeterKarlPeter
5341315
asked Jan 14 at 3:57
KarlPeterKarlPeter
5341315
asked Jan 14 at 3:57
KarlPeterKarlPeter
5341315
5341315
add a comment |
add a comment |
1 Answer
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Let $L(D)$ be a line bundle on a (smooth projective) curve and suppose $D$ has degree zero. If there is a global section of $L(D)$, then there is a function $f$ whose zeros and poles are exactly described by $D$ (the function must have zeros along the zeros of $D$ and thus can only have correpsonding poles along the poles of $D$.) But now division by $f$ gives an isomorphism between $L(D)$ and the trivial bundle.
answered Jan 14 at 4:07
hunterhunter
14.5k22438
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1 Answer
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1 Answer
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active
oldest
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active
oldest
votes
$begingroup$
Let $L(D)$ be a line bundle on a (smooth projective) curve and suppose $D$ has degree zero. If there is a global section of $L(D)$, then there is a function $f$ whose zeros and poles are exactly described by $D$ (the function must have zeros along the zeros of $D$ and thus can only have correpsonding poles along the poles of $D$.) But now division by $f$ gives an isomorphism between $L(D)$ and the trivial bundle.
answered Jan 14 at 4:07
hunterhunter
14.5k22438
$endgroup$
add a comment |
$begingroup$
Let $L(D)$ be a line bundle on a (smooth projective) curve and suppose $D$ has degree zero. If there is a global section of $L(D)$, then there is a function $f$ whose zeros and poles are exactly described by $D$ (the function must have zeros along the zeros of $D$ and thus can only have correpsonding poles along the poles of $D$.) But now division by $f$ gives an isomorphism between $L(D)$ and the trivial bundle.
answered Jan 14 at 4:07
hunterhunter
14.5k22438
$endgroup$
add a comment |
$begingroup$
Let $L(D)$ be a line bundle on a (smooth projective) curve and suppose $D$ has degree zero. If there is a global section of $L(D)$, then there is a function $f$ whose zeros and poles are exactly described by $D$ (the function must have zeros along the zeros of $D$ and thus can only have correpsonding poles along the poles of $D$.) But now division by $f$ gives an isomorphism between $L(D)$ and the trivial bundle.
answered Jan 14 at 4:07
hunterhunter
14.5k22438
$endgroup$
Let $L(D)$ be a line bundle on a (smooth projective) curve and suppose $D$ has degree zero. If there is a global section of $L(D)$, then there is a function $f$ whose zeros and poles are exactly described by $D$ (the function must have zeros along the zeros of $D$ and thus can only have correpsonding poles along the poles of $D$.) But now division by $f$ gives an isomorphism between $L(D)$ and the trivial bundle.
answered Jan 14 at 4:07
hunterhunter
14.5k22438
answered Jan 14 at 4:07
hunterhunter
14.5k22438
answered Jan 14 at 4:07
hunterhunter
14.5k22438
answered Jan 14 at 4:07
hunterhunter
14.5k22438
14.5k22438
add a comment |
add a comment |
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