# cardinality of the fibre of a constantly branched, finite morphism over the branch locus

Let $\pi:Y\to X$ be a Galois cover, i.e. a finite morphism of nonsingular varieties over an algebraically closed field $\Bbbk$ such that $K(X)\hookrightarrow K(Y)$ is Galois. Let $H\subset X$ be the branch locus of $\pi$. We assume that each component of $H$ is nonsingular1. I seek your help in proving the following statement:

Proposition Let $P\in X$ be a regular point of $H$ and let $\eta$ be the generic point of the component $Z$ of $H$ which contains $P$. Assume that every point $\tau$ of codimension one with $\pi(\tau)=\eta$ has ramification index $n$. Then, $\left|\pi^{-1}(P)\right|=\deg(\pi)/n$.

Of course, if you can refer me to a text where this or a more general result is proven, that'd be great, since I couldn't find anything. Also, feel free to provide a proof which is different from my approach, but this is what I did so far:

1. Restrict to the case where $X=\mathrm{Spec}(A)$ and $Y=\mathrm{Spec}(B)$ are affine (duh).
2. Use the primitive element theorem and further localization to get $B=A[t]$ for some $t$ which is integral over $A$ and the minimal polynomial $F\in A[x]$ of $t$ has degree equal to $\deg(\pi)$.
3. Denote by $F_P$ the image of $F$ under $A[x]\twoheadrightarrow (A/\mathfrak{m}_P)[x]=\Bbbk[x]$, then for any $Q\in\pi^{-1}(P)$, we have $0=F(t)(Q)=F_P(t(Q))$. The cardinality of the fiber of $P$ is therefore equal to the number of distinct zeros of $F_P$.
4. I thought I could write $F=\prod_i (x-f_i)$ for certain $f_i\in B$ by possibly localizing further, since $K(Y)$ is normal over $K(X)$. These $f_i$ will all be distinct, but at $P$, exactly $n$ of them should have the same value. I don't know how to prove that, though.
5. I also thought about considering the discriminant of the $(k-1)$-st formal derivative of $F$ - this is a function on $Y$ which 'knows' where $F$ has $k$-uple zeros. I have even less of an idea how to relate that to ramification, though.

1 Of course, if you want to weaken these assumptions anywhere, be my guest.

• Since $\pi \colon Y \to X$ is a Galois cover, we have $X=Y/G$, where the finite group $G$ (with $|G|=\deg \pi$) is the Galois group. Take a generic point $\eta$ of a component of the branch locus with ramification index $n$. You are assuming that the points $\tau$ over $P$ have stabilizer isomorphic to $\mathbb{Z}/n \mathbb{Z}$. Since $|\pi^{-1}(\eta)|$ is the cardinality of the orbit of $\tau$, it must be equal to $|G|$ divided by the order of the stabilizer at $\tau$, and this gives precisely $\deg (\pi) / n$. – Francesco Polizzi Aug 5 '11 at 8:29
• By the way, since the stabilizer subgroup of different points in $\pi^{-1}(\eta)$ are conjugate in $G$, all points in $\pi^{-1}(\eta)$ have necessarily the same ramification index. Of course, this is false for non-Galois covers. – Francesco Polizzi Aug 5 '11 at 8:31
• That sounds interesting, I'd really like to read more about Galois covers - is there any Book which covers all the stuff you used there? – Jesko Hüttenhain Aug 5 '11 at 12:34
• You can read Chapter III, Section 3 "Group action on Riemnann surfaces" of Miranda's book "Algebraic curves and Riemann surfaces". It develops the theory only in dimension 1, but it is enough in order to understand how things work. – Francesco Polizzi Aug 5 '11 at 12:54
• @Francesco: The assumption is that the generic stabilizer along $Z$ has order $n$. Why can't the order jump at $P$? – Laurent Moret-Bailly Aug 7 '11 at 10:39