$\newcommand{\u}{\mathfrak{U}}$I have a question about the proof of Riemann-Roch theorem in Farkas&Kra Riemann surfaces.
Definition. For a divisor $\u$ on $M$, we set a $\Bbb C$-vector space $$L(\u) = \{f\in\mathcal{K}(M);(f)\geq \u\},$$ where $\mathcal{K}(M)$ denotes a field of meromorphic functions on $M$. $$r(\u) =\dim_{\Bbb C}L(\u).$$ Similarly, we define a $\Bbb C$-vector space $$\Omega(\u) = \{\omega;\omega\ \text{is a meromorphic differential with}\ (\omega)\geq\u\}.,$$ and $$i(\u) = \dim\Omega(\u).$$
Theorem (Riemann-Roch). Let $M$ be a compact Riemann surface of genus $g$ and $\u$ an integral divisor on $M$. Then $$r(\u^{-1}) = \deg\u-g+1+i(\u).$$ During the proof of this 'integral divisor' case, we get the Riemann inequality $$r(\u^{-1})\geq\deg\u-g+1.$$ Corollary of the theorem is that The Riemann-Roch theorem holds for the divisor $\u$ provided that either $(a)$ $\u$ is equivalent to an integral divisor, or $(b)$ $Z/\u$ is equivalent to an integral divisor for some canoncial divisor $Z$.
Now if $\u$ is a divisor on $M$ and neither $\u$ nor $Z/\u$ is equivalent to an integral divisor, then $i(\u) = 0 = r(\u^{-1})$. So to prove Riemann-Roch for the general case, need to show $$\deg\u = g-1$$ in this case.
Theorem. The Riemann-Roch theorem holds for every divisor on a compact surface $M$.
Proof. We write the divisor $\u$ as $$\u = \u_1/\u_2,$$ with $\u_j(j=1,2)$ integral and the pair relatively prime (no points in common). We now have $$\deg\u =\deg \u_1-\deg \u_2,$$ and by the Riemann inequality $$r(\u_1^{-1})\geq\deg\u_1-g+1 = \deg\u_2+\deg\u-g+1.$$ Assume that $$\deg\u\geq g.$$ We then have $$r(\u_1^{-1})\geq\deg\u_2+1.$$ Thus, we can find a function $0\neq f\in L(\u_1^{-1})$ that vanishes at each point in $\u_2$ (to the order specified by this divisor). (Vanishing at points of $\u_2$ imposes $\deg\u_2$ linear conditions on the vector space $L(\u_1^{-1})$. If $r(\u_1^{-1})$ is big enough, linear algebra provides the desired function.) Thus $f\in L(\u_2/\u_1) = L(\u^{-1})$, which contradicts the assumption that $r(\u^{-1}) =0$. Hence $$\deg u\leq g-1.$$ Since $0 = i(\u) = r(\u/Z)$, it follows that $$\deg(Z/\u)\leq g-1,$$ or $$\deg\u\geq g-1,$$ concluding the proof of the Riemann-Roch theorem.
I don't understand why $\deg(Z/\u)\leq g-1$ follows. Riemann inequality seemed to imply this, but we don't know whether $Z/\u$ is an integral divisor. Actually, the proof handles the case when neither $\u$ nor $Z/\u$ is equivalent to an integral divisor. Could you explain why this inequality holds?