Kodaira dimension on curve is $0$ implies elliptic curve

Question

Consider $C$ a smooth projective curve, suppose its Kodaira dimension is $0$.

What this means for me is that $\max \{ n\in \mathbb N(K_C)|\dim \overline{\phi_n(C)}\subseteq \mathbb P(H^0(C,nK_C)) \}=0$ where $\phi_n:C\to \mathbb P(H^0(C,nK_C))$ is the map induced by global generators of $H^0(C,nK_C)$ and $\mathbb N(K_C)=\{n\in \mathbb N^*|H^0(C,nK_X)\neq\{0\} \}$.

I have 3 questions:

My first question is : Why if $n$ is such that $H^0(C,nK_C)\neq0$, then it's globally generated. This is not specific to curves but since the maps we get are induced by theorem II.7.1 in Hartshorne (to my understanding) we would not only need generators but global generators, so the fact that there are global sections is not sufficient, at least in appearance, they have to generate at stalks. However this is not explicitly required in the definitions I saw.

Now I tried to show that $\dim H^0(C,nK_C)\leq 1$ by saying that otherwise, the induced morphism $C\to \mathbb P(H^0(C,nK_C))$ would be of dimension $1$, absurd since Kodaira dimension is $0$. Is this correct ?

Finally once we've shown that we want to say that $C$ is elliptic, that is of genus $1$. However I don't know how to do it. I tried to show by contradiction that $\deg K_C=0$ and use $\deg K_C=2g-2$. We could also try to show that $K_C\cong \mathcal O_C$ which will happen to be true but I don't know how to show it without knowing the degree. Any help on this ?

Answer 1

If $C$ is a smooth projective curve, the canonical bundle either has no sections (if $g=0$) or is globally generated (if $g\neq 0$):

• If $g=0$, $\omega_C\cong\mathcal{O}_{\Bbb P^1}(-2)$, which clearly has no sections.
• If $g\neq 0$, then by Riemann-Roch, for any $P\in C$, we have $l(P)-l(K-P)=2-g$, which implies $l(K-P)=g-1$ as $l(P)=1$ for any curve of positive genus. Combining this with $l(K)=g$, we see that $K$ is globally generated, as it satisfies $l(K-P)=l(K)-1$ for any $P$. This resolves the first question.

For the second question, sure, this is correct, but may require a little more justification depending on how exacting you're trying to be.

To conclude that $g=1$, note that $\deg K_C \geq 0$ since it has sections, and that if $\deg K_C > 0$, then $K_C$ is ample, and eventually $nK_C$ will be very ample, implying that $C$ has Kodaira dimension $1$. Since the Kodaira dimension of $C$ is zero, we must have $\deg K_C=0$, or $g=1$.