= Solution
For a <divisor on an algebraic curve> $D=\sum_pa_pp$ over an algebraically closed field, its <degree of a divisor>[degree] is
$$
\deg D=\sum_pa_p.
$$
For a nonzero <rational function> $f$, its <principal divisor on an algebraic curve>[principal divisor] is
$$
(f)=\operatorname{div}(f)=\sum_p\operatorname{ord}_p(f)p,
$$
where zeros have positive order and poles have negative order.
Now let $D=\sum_i a_ip_i$ have degree zero on the <projective line>. Choose a homogeneous linear form $L_i(X_0,X_1)$ whose zero is $p_i$. Since $\sum_i a_i=0$, the expression
$$
f=\prod_iL_i^{a_i}
$$
is homogeneous of degree zero and therefore defines a rational function on $\mathbb P^1$. Each $L_i$ has one simple zero, and the cancellation of total degree removes any common scaling ambiguity, so
$$
(f)=\sum_i a_ip_i=D.
$$
Thus every degree-zero divisor on $\mathbb P^1$ is principal. If $E$ and $E'$ have the same degree, then $E-E'$ is principal, whence
$$
\boxed{E\sim E'}.
$$
This is the <divisor class on the projective line>.
Write $p_\infty=[1:0]$. The coordinate $t=X_0/X_1$ has no critical zero or pole in the finite chart. Near infinity use $s=X_1/X_0=t^{-1}$; then
$$
dt=-s^{-2}ds.
$$
Hence the rational differential has one double pole at infinity and no other zero or pole:
$$
\boxed{(dt)=-2p_\infty},
$$
which represents the <canonical divisor of the projective line>.
If $D\sim mK_{\mathbb P^1}$, then $D\sim-2mp_\infty$, and <linear equivalence of divisors> gives $L(D)\cong L(-2mp_\infty)$. A function in the latter space has no finite poles, so it is a polynomial, and its degree is at most $-2m$. Therefore
$$
\boxed{\ell(D)=\max\{1-2m,0\}}.
$$
This calculation uses only the rational functions on $\mathbb P^1$, rather than the <Riemann-Roch theorem>.
Finally, suppose distinct $p,q\in C$ satisfy $p-q=(f)$. The function $f$ has one simple pole and therefore defines a degree-one <finite morphism> $f:C\to\mathbb P^1$. A degree-one finite morphism between smooth projective curves is an <isomorphism of algebraic varieties>[isomorphism]. It would follow that $C\cong\mathbb P^1$ and hence that $C$ has <geometric genus>[genus] zero, contrary to the hypothesis. Thus the <principal divisor with one simple zero and one simple pole> cannot occur on $C$:
$$
\boxed{p-q\text{ is not principal}.}
$$
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