Here we give the promised proof of the ABC-conjecture for function fields.

As always, $k$ is a perfect (e.g. finite) field and $K=k(X)$ is the function field of a smooth projective curve $X$ defined over $k$. We take elements $u,v \in K^*$ satisfying $u+v=1$ and consider the cover $u : X \mapsto \mathbb{P}^1_k$ corresponding to the embedding $k(u) \hookrightarrow K$. We want to determine the (schematic) zero- and pole-divisors of $u$ and $v$ and call them $A=div_0(u), B=div_0(v)$ and $C=div_{\infty}(u)=div_{\infty}(v)$.

Let $R$ be the integral closure of $k[u]$ in $K$, then in $R$ we can write the ideals $u$ and $(v)=(1-u)$ as products of prime-ideals (which correspond to schematic points of $X$)

$(u) = P_1^{e_u(P_1)} \cdots P_r^{e_u(P_r)}$

$(v) = Q_1^{e_u(Q_1)} \cdots Q_s^{e_u(Q_s)}$

and so $A = \sum_i e_u(P_i) [P_i]$ and $B = \sum_j e_u(Q_j) [Q_j]$. If $S$ is the integral closure of $k[\frac{1}{u}]$ in $K$, then we have in $S$ a decomposition

$(\frac{1}{u}) = R_1^{e_u(R_1)} \cdots R_t^{e_u(R_t)}$

and therefore $C = \sum_l e_u(R_l)[R_l]$. We already know that $deg(A)=deg(B)=deg(C)=n=[K : k(u)]$.

**Case 1 :** Let us assume that the field extension $K/k(u)$ is separable. Then, by the Riemann-Hurwitz formula (or rather, the scheme-version of it) we get the inequality (use that the genus of $\mathbb{P}^1_k$ is zero) :

$2 g_K – 2 \geq -2n + \sum^{scheme}_{P \in C} (e_u(P)-1) deg(P)$

Because for all points $e_u(P)-1 \geq 0$, the inequality only becomes better if we restrict the sum to a subset of points, say to the support of $A+B+C$ (that is to ${ P_1,\cdots,P_r,Q_1,\cdots,Q_s,R_1,\cdots,R_t }$). Then we get

$2 g_K -2 \geq -2n + \sum_{P \in Supp(A+B+C)} e_u(P)deg(P) – \sum_{P \in Supp(A+B+C)}deg(P)$

$~\qquad = -2n+3n-\sum_{P \in Supp(A+B+C)} deg(P)$

which gives us the required form of the ABC-conjecture for curves

$n=deg(u)=deg_s(u) \leq 2g_K – 2 + \sum_{P \in Supp(A+B+C)} deg(P)$

**Case 2 :** If $K/k(u)$ is not separable, take a maximal separable subfield $k(u) \subset M \subset K$, then by definition of $deg_s(u)$ and case 1 we have

$deg_s(u) \leq 2g_M – 2 + \sum_{P’ \in Supp(A’+B’+C’)} deg(P’)$

where $A’$ (resp. $B’$,$C’$) are the schematic fibers of the cover $Y \mapsto \mathbb{P}^1_k$ over the $k$-rational points $0$ (resp. $1$, $\infty$) and where $Y$ is the curve with function field $M$. We are done if we can show that $g_K=g_M$ and that in the cover $X \mapsto Y$ there is a unique point $P$ lying over each point $P’$ with $deg(P)=deg(P’)$.

As $K/M$ is purely inseparable, we have a tower of subfields

$M=M_0 \subset M_1 \subset \cdots \subset M_z=K$

such that $M_i / M_{i-1}$ is purely inseparable of degree $p$ for all $i$. That is, raising to the $p$-th power gives a field-isomorphism $M_i \simeq M_{i-1}$. The genus is a field-invariant, so $g_{M_i}=g_{M_{i-1}}$ and there is a bijection between the dvr’s in $M_i$ and $M_{i-1}$. That is, a bijection between points $P_i \leftrightarrow P_{i-1}$ of the corresponding curves $Y_i \mapsto Y_{i-1}$. Finally, because $t_{P_i}^p = t_{P_{i-1}}$ it follows that $deg(P_i)=deg(P_{i-1})$, and we are done by induction on $i$.