Confusion about sum of ramification indices

Question

In algebraic number theory, I'm well aware of the following formula: Given the "$AKLB$ setup" and a prime $\mathfrak{p}$ of $A$, then \begin{equation}\sum_{\mathfrak{P}|\mathfrak{p}} e_{\mathfrak{P}|\mathfrak{p}}f_{\mathfrak{P}|\mathfrak{p}} = |L:K|. \tag{1}\label{ram rel}\end{equation} Now I'm working on Hartshorne IV.2.2(a) which asks to show that on a curve $X$ of genus 2, the canonical divisor determines a morphism $f:X\to\mathbb P^1$ of degree 2 which is ramified at 6 points, each of index 2. Indeed, the Riemann-Hurwitz theorem implies $$2 = 2g_X - 2 = n(2g_{\mathbb P^1}-2) + \deg R = -4 + \deg R$$ where $R$ is the ramification divisor, so $\deg R = 6$. However, this seems to contradict the ramification index-residue class degree formula. Indeed, the morphism of sheaves $f^\#:\mathcal{O}_{\mathbb P^1}\to f_*\mathcal{O}_{X}$ induces a ring map $f^\#(\mathbb A^1):k[t]\to B$ with $f^\#_\xi:k(t)\to K(X)$ having degree 2 ($\xi$ the generic point). So in this "$AKLB$ setup", $A = k[t]$, $K = k(t)$, and $L/K$ has degree 2. Now if $Q\in\mathbb P^1$ is a branch point, there being 6 ramified points in $X$ seems to imply that $$\sum_{P\to Q}e_P = 12$$ while $$|K(X):k(\mathbb P^1)| = 2$$ Where have I gone wrong?

Answer 1

While I was typing this I figured out what was wrong. A ramification (branch?) point $Q\in\mathbb P^1$ corresponds to a choice of ramified prime ideal $\mathfrak p$. Since $f$ has degree 2, viewing $\mathfrak p$ as a divisor, we have $f^*\mathfrak{p} = \mathfrak{P}^e$, and $$e=e\deg\mathfrak{P}=\deg \mathfrak{P}^e =\deg f^*\mathfrak p = (\deg f)\deg(\mathfrak{p}) = \deg f,$$ so each ramified $Q\in \mathbb P^1$ has ramification index 2. So there are 6 corresponding $Q\in\mathbb P^1$ to each of the 6 ramified $P\in X$ (this follows at once from the definition of ramification since we're in degree 2). Not sure what I was thinking. Anyhow here's the answer in case somebody finds this useful.