Let $X$ be a compact Riemann surface and the canonical divisor $K_X:=(\Omega)$, where $\Omega$ is a meromorphic top form on $X$.
Question 1: Can I show the canonical divisor is the corresponding divisor of the canonical line bundle by the following:
For the canonical (line) bundle on $X$, say $L$, we can find a corresponding global meromorphic section $s\in H^0(X,\mathcal M^*/\mathcal O^*)$, which defines a meromorphic 1-form $\omega$. And $\omega$ defines a divisor $(\omega)$, which by definition is the canonical divisor.
I feel confused because given any line bundle $L'$ over $X$, whose associated global meromorphic section $s'\in H^0(X,\mathcal M^*/\mathcal O^*)$ also defines a meromorphic 1-form $\omega'$ and $\deg(L')=(\omega')$. By Otto's Lectures on Riemann surfaces, p.139, the divisor of a non-vanishing meromorphic 1-form on a compact Riemann surface of genus $g$ satisfies $\deg(\omega)=2g-2$, we get $2g-2=\deg(\omega')=\deg(L')$ So I just get any line bundle is of degree $2g-2$, which is a contradiction.
Question 2: If the way, $s\to \omega\to (\omega)$ is not the right way to show the canonical divisor is the corresponding divisor of the canonical line bundle, then how to see the canonical divisor is the corresponding divisor of the canonical line bundle?