Divisors and Riemann-Roch on the Riemann sphere and on a complex torus

Each factor $z-a_j$ has divisor $[a_j]-[\mathrm{\infty }]$ by the coordinates in [F3], so [F2] gives $(u)=-D+s[\mathrm{\infty }]$. For any nonzero $f$, $(f\mathrm{/}u)+s[\mathrm{\infty }]=(f)+D$, so division by $u$ identifies $L(D)$ with $L(s[\mathrm{\infty }])$. A rational function with no finite poles is a polynomial: in a reduced quotient of polynomials, a nonconstant denominator would have a finite root by [F4], hence a pole, contrary to the divisor bound. A polynomial of degree $d$ has pole order $d$ at infinity in the coordinate $w=1\mathrm{/}z$, so for $s\ge 0$ this space consists precisely of the polynomials of degree at most $s$. Its basis $1,z,\dots ,z^s$ yields the displayed basis of $L(D)$ and its divisor formula. If $s<0$, [F2] gives $L(D)=0$.

By [F8], different local lifts of a torus point differ by a lattice translation, whose derivative is $1$, so their differentials $dz$ agree; by [F5] they define a nowhere-vanishing holomorphic differential with divisor $0$. Thus [F6] identifies the canonical bundle with ${\mathcal{O}}_X(0)$. By [F7], $\mathrm{\ell }(0)=1$ and $i(0)=\mathrm{\ell }(K)=1$; applying Riemann–Roch at $0$ gives $0=1-g$, hence $g=1$. For $n\ge 1$, the divisor $K-n[o]=-n[o]$ has negative degree, so [F2] gives $\mathrm{\ell }(K-n[o])=0$. Riemann–Roch and duality [F7] now give $\mathrm{\ell }(n[o])=n$ and $L(0)=\mathbb{C}$.

On the finite chart $dz$ has no zeros or poles; at infinity $dz=-w^{-2}dw$, so [F5] gives $(dz)=-2[\mathrm{\infty }]=K$. Step 1.1 applied to $K$ gives $\mathrm{\ell }(K)=0$, and [F6] identifies this zero space with the holomorphic differentials. Applied to $K-D$, the same calculation gives $\mathrm{\ell }(K-D)=\max (0,-s-1)$. If $s\ge -1$, subtracting it from $\mathrm{\ell }(D)$ gives $s+1$; if $s\le -2$, the difference is $0-(-s-1)=s+1$. This is Riemann–Roch for the genus-zero sphere, and [F7] identifies the second term with $i(D)$.

By [F9], ${\mathrm{\wp }}^j({\mathrm{\wp }}^{\mathrm{\prime }}{)}^{\epsilon }$ descends to the torus, has no poles away from $o$, and has pole order exactly $2j+3\epsilon$ at $o$, with nonzero leading coefficient $(-2{)}^{\epsilon }$. The chosen representatives have distinct orders $2,3,\dots ,n$, so each belongs to $L(n[o])$; together with $1$ they are linearly independent, because in a nontrivial linear combination the term of largest pole order has a principal coefficient that no other term can cancel. There are $n$ such functions for $n\ge 1$ (only $1$ when $n=1$), so step 1.2 makes them a basis. This proves the gap and the first nonconstant-function claims.