The Riemann inequality is not an equality for special divisors

The zero divisor is special. By [F3], $i(0)=g\ge 1$, so it is nonzero; hence [F1] says that $D=0$ is special. By [F2], $l(0)=1$. [F1, F2, F3] 1.2 (Smoothness of the Fermat quartic.) Let $k$ be algebraically closed of characteristic zero, set $F=x_0^4+x_1^4+x_2^4$, and let $X=V_{+}(F)\subseteq {\mathbb{P}}_k^2$. In the chart $x_i\mathrm{\ne }0$, set $x_i=1$ and write the other coordinates as $u,v$; the equation is $f=1+u^4+v^4$. The opens $D(u)$ and $D(v)$ cover this affine hypersurface, since no prime containing both $u$ and $v$ can contain $f=1+u^4+v^4$. On $D(u)$, the derivative $\mathrm{\partial }f\mathrm{/}\mathrm{\partial }u=4u^3$ is a unit; on $D(v)$, $\mathrm{\partial }f\mathrm{/}\mathrm{\partial }v=4v^3$ is a unit. Each is therefore a standard smooth presentation by [F5], so the three projective charts show that $X$ is smooth over $k$. The scheme is nonempty: choose $a\in k$ with $a^4=-1$; then $[1:a:0]\in X$. [F5, given] 1.3 (Scheme dimension.) In each of the three standard charts, identified by projective hypersurface affine pieces, the coordinate ring is $k[u,v]\mathrm{/}(f)$ with $f=1+u^4+v^4$, a nonzero nonunit. The polynomial ring $k[u,v]$ is Noetherian. Every minimal prime $\mathfrak{q}$ over $(f)$ is nonzero and has height one by [F6] and the principal ideal theorem. The dimension formula gives ${\mathrm{trdeg}}_k\mathrm{Frac}(k[u,v]\mathrm{/}\mathfrak{q})=1$, and affine-domain dimension equals this transcendence degree. Thus every irreducible component in every nonempty chart has dimension one. Since the standard charts cover $X$, it is pure of dimension one; the same calculation at a closed point gives local dimension one. The scheme $X$ is proper because it is the closed subscheme $V_{+}(F)\hookrightarrow {\mathbf{P}}_k^2$: projective space is proper over $k$, and the closed immersion and composite are proper by [F12]. [F6, F12, given] 2.1 The inequality at $D=0$ is strict. By [F1] and [F4], $l(0)-i(0)={\mathrm{deg}}_k(0)+1-g=1-g\mathrm{.}$ Using step 1.1 gives $l(0)=1>1-g={\mathrm{deg}}_k(0)+1-g$ because $g\ge 1$, and the excess is $1-(1-g)=g=i(0)\mathrm{.}$ [F1, F4, step 1.1] 2.2 For any divisor $D$, rearranging [F1] gives $l(D)={\mathrm{deg}}_k(D)+1-g+i(D)\mathrm{.}$ Since $i(D)\ge 0$, the Riemann inequality is strict exactly when $i(D)>0$, which is exactly when $D$ is special; equality holds exactly for nonspecial divisors. This proves the general claim independently of the concrete example. [F1, step 1.1] 2.3 (Integrality.) We show that $F$ is square-free and irreducible. Suppose an irreducible homogeneous factor $G$ occurs at least twice. Choose a line $\mathrm{\ell }=0$ not containing $G$; [F7] gives a closed point $p\in V_{+}(G,\mathrm{\ell })$. On a chart through $p$, the actual equation $f$ of $X$ lies in the square of the maximal ideal, so its Jacobian row is zero. The local ring has dimension one by step 1.3, and [F5] says it is not regular, contradicting smoothness from step 1.2 and [F8]. Hence $F$ is square-free. If the square-free $F$ were reducible, choose a nonconstant irreducible factor $G$ and let $H$ be the product of the remaining factors. Then $G,H$ are coprime and have positive degree. By [F7], they meet at a closed point $p$. There $f=gh$ lies in the square of the maximal ideal, so the Jacobian row again vanishes. The local ring has dimension one, contradicting regularity exactly as above. Thus $F$ is irreducible. Since $k[x_0,x_1,x_2]$ is a UFD by [F11], $(F)$ is prime, and $X$ is integral. As $k$ is algebraically closed, this proves geometric integrality. [F5, F6, F7, F8, step 1.2, step 1.3] 3.1 (Genus three.) Steps 1.2, 1.3, and 2.3 show that $X$ is a smooth proper geometrically integral plane curve cut out by a homogeneous quartic. The arithmetic-genus theorem in [F9] gives $p_a(X)=\frac{(4-1)(4-2)}{2}=3$, and the genus definition in [F9] identifies $g(X)=p_a(X)$ for this smooth curve. Hence $g(X)=3$, as needed for the concrete instance in the Counterexample section. [F9, step 1.2, step 1.3, step 2.3] 4.1 (Conclusion and choice accounting.) Steps 1.1 and 2.1 show that the zero divisor on any curve of genus at least one gives a strict Riemann inequality with excess exactly $i(0)$. Step 2.2 proves the stated criterion for all divisors. Steps 1.2, 1.3, 2.3, and 3.1 give the promised characteristic-zero genus-three example, where the inequality is $1>-2$. The Axiom of Choice [F10] is inherited through the Riemann--Roch, affine-dimension, Jacobian, and Bezout suppliers; no additional choice is made.