Elliptic Curve has Canonical Bundle $K_E = \mathcal{O}_E$

Question

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 $K_E = \mathcal{O}_E$.

Answer 1

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.

Answer 2

It might be helpful to simply $\bf{copy\text{-}paste}$ materials from Silverman's book. Fix a field $K.$ Restrict ourselves to char $K\neq 2$ case with $e_{1}\neq e_{2}\neq e_{3}:$ $E:y^{2}=(x-e_{1})(x-e_{2})(x-e_{3})$ in $\Bbb{P}^{2}_{K}.$ To show $$\Omega^{1}_{E}\cong \mathcal{O}_{E},$$ we need to show canonical divisor class is same as the divisor class of $0.$ As canonical divisor class is given by $\text{div}(\omega)$ with any $0\neq\omega\in\Omega^{1}_{E}.$ Consider $dx/y.$ In fact $\text{div}(dx/y)=0,$ where $$\text{div}(dx)\overset{(1)}{=}(e_{1},0)+(e_{2},0)+(e_{3},0)-3(\infty)\overset{(2)}{=}\text{div}(y).$$ To show (1), consider $dx=d(x-e_{1})$ firstly and we have $$\text{ord}_{(e_{1},0)}(d(x-e_{1}))=\text{ord}_{(e_{1},0)}(x-e_{1})-1.$$ Admit $\text{ord}_{(e_{1},0)}(x-e_{1})=2.$ Then $\text{ord}_{(\infty)}(x-e_{1})=-2$ and $\text{div}(x-e_{i})=2(e_{i},0)-2(\infty).$ Moreover, we have $\text{ord}_{(\infty)}(d(x-e_{1}))=-3$ and (2).

To show $\text{ord}_{(e_{1},0)}(x-e_{1})=2,$ we show $x-e_{1}\in \mathfrak{m}_{(e_{1},0)}^{2}.$ Here $$\mathfrak{m}_{(e_{1},0)}=(x-e_{1},y),\,\,(x-e_{1},y)^{2}=((x-e_{1})^{2},(x-e_{1})y,y^{2}).$$ Consider $$-(x-e_{1})(e_{3}-e_{1})(e_{2}-e_{1})=y^{2}-((x-e_{1})^{2}-(e_{2}+e_{3}-2e_{1})(x-e_{1}))(x-e_{1})\equiv 0\text{ mod }(x-e_{1},y)^{2}$$ and thus $(x-e_{1})\in (x-e_{1},y)^{2}.$