Elementary reason why an effective canonical divisor is a hyperplane section I am a beginner in algebraic curves, currently studying with Frances Kirwan's text "Complex Algebraic Curves". While I was working on my term paper on theta characteristics and bitangent lines to plane quartics, the following facts appeared multiple times and seemed to be unavoidable:


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*An effective canonical divisor $D$ on a smooth (projective) plane curve $C$ is a hyperplane section, i.e., $D=L\cdot C$ for some line $L$ in $\mathbb{P}^2$.

*If $P+Q$ and $R+S$ are theta characteristics on $C$ (here $P,Q,R,S$ are some points on $C$), then $P+Q \sim R+S$ implies $P+Q=R+S$.


Unfortunately, I could not figure out by myself why these facts must hold true, and whenever I tried to find it from other sources, they were always written in the language of schemes and the notion of very ampleness, which were overwhelming for me. Won't there be a way to explain these in an elementary way, without using sophisticated languages? I learned that divisors can be regarded as line bundles and also heard of some sheaf theory, so up to this level I might be able to understand. Any help would be greatly appreciated.
 A: For the first, the fact that you are dealing with quartic curves is key. If one takes a curve $C$ of degree $n$ in the plane then the canonical divisor is cut out by the intersections with all adjoint curves of degree $n-3$. Adjoint means that if the base curve $C$ has multiplicity of $m$ at a point $P$, then the adjoint has multiplicity of $m-1$ in $P$. 
So for example a quartic curve with a double point will have its canonical divisor defined by all lines passing through the double point. 
The book Algebraic Curves by Fulton is good. The books of the same name by Walker and by Semple and Roth are also classic but require more patience to read since they are written in a very old style. 
I understand 2 now, it is an application of the residue theorem. Two divisors are coresidual written as $\mathcal{D}\sim \mathcal{E}$ if there are curves $F$ and $G$ of the same degree such that their respective intersections with $C$ are $\mathcal{D}+\mathcal{R}$ and $\mathcal{E}+\mathcal{R}$. That is both are residual to $\mathcal{R}$. Now the residue theorem says that $\mathcal{R}$ can be chosen arbitrarily. Thus take $P+Q$ and take a line passing through $P$ and $Q$. This line intersects $C$ in $P+ Q+M+N$ thus 
$P+ Q$ is residual to $M+N$. Now if $P+Q \sim R+S$ then $R+S$ must also be residual to $M+N$. However there is only one line passing through $M+N$. And its intersection is $P+ Q$. Thus $P+ Q=R+ S $. 
