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I tried writing the full motivation for this question, but it turned out to be too long a detour for a question that is really quite specific. So I will only sketch the motivation this time.

Sketch of motivation

I'm taking a $G$-Galois cover of $\mathbb{P}^1_{\overline{\mathbb{C}(t)}}$ which is defined (not including the group action) over $\mathbb{C}(t)$. Then I descend to $\mathbb{C}[t]$ and observe the branching behavior. The following specific example came up from a very explicit computation of a $D_8$-cover, $X$ over $\mathbb{P}^1_{\overline{\mathbb{C}(t)}}$, where the interesting part (the part in the question) is the $C_4$-cover $X \rightarrow X/C_4$. In the question, I'm actually looking formally about $t=0$.

While this is the motivation, the question will require none of this.


Let $K=\mathbb{C}((t))(y)$, and $L=Quot(\mathbb{C}((t))[y,z]/z^4-(y-\sqrt{1-t})^2(y+\sqrt{1-t})^2(\sqrt{2(2-t)}-y)(\sqrt{2(2-t)}+y)^3)$. Let's look at the surface $R:=\mathbb{P}^1_{\mathbb{C}[[t]]}$ (with parameter $y$, so its function field is $K$), and the branched covering of it: $S:=$ the normalization of $R$ in $L$ (obviously with function field $L$). My question is: what is the ramification index of the vertical divisor, $t$, of $R$, in the branched covering $S \rightarrow R$?

It seems to me, through roundabout ways, that the ramification index should equal $2$, but surely there should be a systematic way of doing this that isn't completely terrible.

Do you have any tricks up your sleeves for computing vertical ramification in situations like this?


Fix a choice for $\sqrt{1-t}$ and $\sqrt{2(2-t)}$ in $\mathbb{C}[[t]]$ throughout.

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I think you left out an exponent of 3 on $sqrt(2(2-t))-y$. –  KConrad Aug 7 '10 at 19:47
No, it's supposed to be like that. –  H. Hasson Aug 7 '10 at 19:49
$\sqrt{1-t}$ doesn't really appear, so no need to fix choice. I haven't finished chapter 2 of Hartshorne, so I hope this isn't too wrong: the vertical divisor lives on the affine $y=1$, and using $\sqrt{2(2-t)} = 2-t/2+O(t^2)$, we get $z^4-t^2(27+O(t))$, showing that the ramification index is indeed 2. –  Dror Speiser Aug 7 '10 at 21:06
But $y=\sqrt{1-t}$ is a horizontal ramification. –  Qing Liu Aug 7 '10 at 22:00
Ey, look who it is! How are you? If I understand your comment correctly, you're plugging in y=1 because you're saying the ramification above y=1 t=0 is the same as the ramification along t=0. This isn't true, though - the ramification along t=0 means generically (a horizontal branch divisor intersects t=0 at y=1). But it is a good idea to plug in a y not equal to one of 1,-1,2,-2. One should also be careful. The equation I gave is fine for defining the birational class, but it won't give a regular C((t))-curve, let alone define the normalization as a C[[t]]-curve. –  H. Hasson Aug 7 '10 at 22:01
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1 Answer

up vote 3 down vote accepted

I think there is no vertical ramification. Denote by $D={\mathbb C}[[t]]$ and $K_0$ its fraction field. Your curve is defined over $K_0$ by an affine equation $z^n+f(y)=0$ with $f(Y)\in D[Y]$ monic. Let $U$ be the affine scheme associated to $D[y,z, {{1}\over{f(y)}}]/(z^n-f(y))$. It is clearly finite and étale over the open subset $V:={\mathrm{Spec}}(D[y, {{1}\over{f(y)}}])$ of $R$, so $U=S\times_R V$. On the special fibers we have $U_s=S_s\times_{R_s} V_s$. As $V_s$ contains the generic point of $R_s$, we see that $S_s$ has a unique generic point, so $U_s$ is dense in $S_s$. But $U_s$ is étale over $V_s$, so there is no vertical ramification. What is needed here is only $D$ is a DVR and $n$ is coprime with the residue characteristic of $D$.

In fact, in your example, the zeros of $f(Y)$ remain pairwise distinct mod $t$, this implies that the function field of $U_s$ has the same genus as $L$. It is then easy to see that the whole curve $S_s$ is smooth.

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Maybe this is a trivial matter, but I don't see why D[y,z, 1/f(y)]/(z^n-f(y)) would be integrally closed. –  H. Hasson Aug 7 '10 at 22:36
It is smooth over $D$ (either directly using Jacobian criterion or because it is étale over $V$ which is smooth over $D$). As $D$ is regular, $U$ is regular, hence integrally closed. –  Qing Liu Aug 8 '10 at 10:39
...and I just found the problem in my roundabout way. Thanks! –  H. Hasson Aug 8 '10 at 16:01
On a different level, for any Dedekind domain $R$, with 2 and $\epsilon$ units, $R[\sqrt{\epsilon}]$ is integrally closed. This is clear from the minimal polynomial of $a+b\sqrt{\epsilon}$. Applying this twice we get the specific example. –  Dror Speiser Aug 8 '10 at 17:43
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