Let $C$ be an elliptic cure, i.e. a projective curve over $\Bbb{C}$ having genus $1$.
I'm trying to understand why the Kodaira dimension of $C$ is zero (according to this article).
I know that the canonical divisor $K_C$ is trivial, so that $H^0(C,\mathcal{O}_C(nK_C))=H^0(C,\mathcal{O}_C)=\Bbb{C}$ for all $n\geq 0$.
This way, the canonical ring $\bigoplus_{n\geq 0}H^0(C,\mathcal{O}_C(nK_C))$ is just $\Bbb{C}$, so its field of fractions is also $\Bbb{C}$.
By definition of Kodaira dimension, $\kappa(C)=\text{trdeg}_{\Bbb{C}}\Bbb{C}-1=0-1=-1$.
What am I missing?
The canonical ring $R=\bigoplus_{n\geq 0}H^0(C,\mathcal{O}_C(nK_C))$ is not just $\Bbb{C}$, because it has a copy of $\Bbb{C}$ at every (positive) grading, i.e. it is the graded ring $R\cong\Bbb{C}[x]$ with $deg(x)=1$.
Thus $\kappa(C)=\text{trdeg}_{\Bbb{C}}\Bbb{C}(x)-1=1-1=0$.