Holomorphic Differentials on a non-singular curve. So I've been working on this for an exam I have coming up, and I'm not sure if I really understand.
If I have a curve defined by some homogeneous polynomial $P$, I can show that the canonical divisor class on $C$ is ($\deg P = 3$) times the hyperplane class, but I'm not sure why this can help me construct the family of holomorphic differentials of $C$ forming a basis for the space of holomorphic differentials on $C$. Any help would be appreciated. 
 A: As you suggested, let's work on the affine portion $w\ne 0$ of the curve $P=0$, and let's put $p(u,v)=P(u,v,1)$. Assume $\deg P = n$. Let $Q$ be any homogeneous polynomial of degree $n-3$, and set $q(u,v) = Q(u,v,1)$. 
Because $C$ is a smooth curve, we know that $dp = \dfrac{\partial p}{\partial u}du+\dfrac{\partial p}{\partial v}dv$ does not vanish (as a $1$-form on $\Bbb C^2$) at points of $C$. Thus, at each point of $C$, either $\dfrac{\partial p}{\partial u}$ or $\dfrac{\partial p}{\partial v}$ is nonzero. Thus, the $1$-form
$$\frac{du}{\partial p/\partial v} = - \frac{dv}{\partial p/\partial u}$$
is well-defined and holomorphic on the affine portion of $C$.
What happens when we consider $\omega = q(u,v)\dfrac{du}{\partial p/\partial v}$? It is certainly holomorphic on the affine part of $C$. (In fact, this would be true for a polynomial $q$ of any degree?) But what happens at $\infty$, i.e., when $w=0$? Now we assume that $[0,1,0]\notin C$ and use $(v',w')$, with $v'=v/u$ and $w'=1/u$ as our local coordinates near such points. In other words, $u=1/w'$ and $v=v'/w'$. Thus, in these coordinates, we have
$$\omega = \frac{q(1/w',v'/w')(-1/w'^2)dw'}{\partial p/\partial v (1/w',v'/w')}=-\frac{w'^{n-3}q(1/w',v'/w')}{w'^{n-1}\partial p/\partial v (1/w',v'/w')}dw'.$$
But because $\deg q\le n-3$ and $\deg(\partial p/\partial v)\le n-1$, we see that both the numerator and denominator are polynomials. To see that $\omega$ is holomorphic at points with $w'=0$, we need to make sure the denominator doesn't vanish at such points. In these coordinates, $C$ is given by $\tilde p(v',w') = w'^np(1/w',v'/w') = 0$, and so, by the chain rule, $\dfrac{\partial\tilde p}{\partial v'} = w^{n-1}\dfrac{\partial p}{\partial v}(1/w',v'/w')$ and smoothness of the curve tells us that this is nonzero. (Well, it might be, but then switch to the other formula for $\omega$.)
If you work through this carefully, you understand that if the degree of $Q$ is any greater, we will end up with a pole at infinity. 
To say that $\kappa\sim (n-3)H$ is to say that a meromorphic $1$-form on $C$ has poles at worst of order $n-3$ on the hyperplane (line) section of $C$ at infinity.
