Remembering Riemann-Roch Embarrassingly, I've always struggled to remember the form of the Riemann-Roch theorem for curves. Does anyone have any intuition to share about how to remember the some of the terms in the formula? 
Recall that for $C$ a Riemann surface and $D$ a divisor on $C$, the Riemann-Roch theorem says that:
\begin{equation}
h^0(D) - h^0(K-D) = \mathrm{deg}(D) + 1  - g
\end{equation} 
where $K$ is the canonical divisor on $C$. I'm happy with the interpretation of the terms on the left hand side (it's some kind of Euler characteristic), but is anyone able to give an informal explanation for the quantity on the right hand side? Why $\mathrm{deg}(D) + 1  - g$? In particular, why should I expect the left hand side to grow like $\mathrm{deg}(D)$, with a correction of $1-g$? I understand that there's some very classical way to think about this, but I've never seen it explained anywhere.
 A: Not sure if that helps you, but I remember it as follows:
Riemann-Roch for a curve $C$ (or  Riemann surface if you prefer) says, that given some line bundle $\mathscr{L} \in \text{Pic}(C)$ (or some divisor), we have the equality
$$\chi(\mathscr{L}) = \text{deg}(\mathscr{L}) + \chi(\mathscr{O}_C).$$ Actually I prefer to state it as $$\text{deg}(\mathscr{L}) = \chi(\mathscr{L}) - \chi(\mathscr{O}_C)$$ since I can more easily remember the sign and to sort of use it as the definition of the degree as well (it somehow also seems more aesthetic to have all the Euler characteristics on one side).
Now under the usual mild assumptions, we have that $h^0(\mathscr{O}_C) = 1$ and we have that $h^1(\mathscr{O}_C) = g$ is the genus of the curve. Plugging that in the first equation yields the Riemann-Roch that you stated. So I guess what I am trying to say is that rephrasing both the left hand side and the $1 - g$ by some Euler characteristic makes it quite easy to remember.
For your other concerns, I would take a look at the comment of KReiser. I think the classical proof is what at least gives me the intuition.
