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Makhalan Duff
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Let $E$ be ana 1-dimensional integral scheme (if you need to assume more, do so). Let $\hat{E}$ be $Spec$ of the completion of a stalk of $E$ at some closed point (think for example $E=Spec(\mathbb{C}[t])$ and $\hat{E}=Spec(\mathbb{C}[[t]])$).

Say we are given a branched Galois cover of, say, $\mathbb{P}^1_{\hat{E}}$ with branch divisor $\hat{D}$ (the divisor will not be a prime divisor - meaning it's made up of several irreducibles). Say we are also given a divisor of $E$, $D$, such that $D \times_E \hat{E}=\hat{D}$. It's not obvious (and probably not true) that this cover extends to a cover of $\mathbb{P}^1_E$ with branch divisor $D$ (by extends I mean that I require the new cover to be Galois as well with the same group).

Is there a way to measure the obstruction? Can I calculate something to check if I may really extend the cover?

Assume whatever you need to in order to avoid wild ramification since I'm not interested in that. And if you can solve a particular case of this, assume whatever you want.

Let $E$ be an 1-dimensional integral scheme (if you need to assume more, do so). Let $\hat{E}$ be $Spec$ of the completion of a stalk of $E$ at some closed point (think for example $E=Spec(\mathbb{C}[t])$ and $\hat{E}=Spec(\mathbb{C}[[t]])$).

Say we are given a branched Galois cover of, say, $\mathbb{P}^1_{\hat{E}}$ with branch divisor $\hat{D}$ (the divisor will not be a prime divisor - meaning it's made up of several irreducibles). Say we are also given a divisor of $E$, $D$, such that $D \times_E \hat{E}=\hat{D}$. It's not obvious (and probably not true) that this cover extends to a cover of $\mathbb{P}^1_E$ with branch divisor $D$ (by extends I mean that I require the new cover to be Galois as well with the same group).

Is there a way to measure the obstruction? Can I calculate something to check if may really extend the cover?

Assume whatever you need to in order to avoid wild ramification since I'm not interested in that. And if you can solve a particular case of this, assume whatever you want.

Let $E$ be a 1-dimensional integral scheme (if you need to assume more, do so). Let $\hat{E}$ be $Spec$ of the completion of a stalk of $E$ at some closed point (think for example $E=Spec(\mathbb{C}[t])$ and $\hat{E}=Spec(\mathbb{C}[[t]])$).

Say we are given a branched Galois cover of, say, $\mathbb{P}^1_{\hat{E}}$ with branch divisor $\hat{D}$ (the divisor will not be a prime divisor - meaning it's made up of several irreducibles). Say we are also given a divisor of $E$, $D$, such that $D \times_E \hat{E}=\hat{D}$. It's not obvious (and probably not true) that this cover extends to a cover of $\mathbb{P}^1_E$ with branch divisor $D$ (by extends I mean that I require the new cover to be Galois as well with the same group).

Is there a way to measure the obstruction? Can I calculate something to check if I may really extend the cover?

Assume whatever you need to in order to avoid wild ramification since I'm not interested in that. And if you can solve a particular case of this, assume whatever you want.

Source Link
Makhalan Duff
  • 5.9k
  • 43
  • 80

Measuring the obstruction of extending a cover

Let $E$ be an 1-dimensional integral scheme (if you need to assume more, do so). Let $\hat{E}$ be $Spec$ of the completion of a stalk of $E$ at some closed point (think for example $E=Spec(\mathbb{C}[t])$ and $\hat{E}=Spec(\mathbb{C}[[t]])$).

Say we are given a branched Galois cover of, say, $\mathbb{P}^1_{\hat{E}}$ with branch divisor $\hat{D}$ (the divisor will not be a prime divisor - meaning it's made up of several irreducibles). Say we are also given a divisor of $E$, $D$, such that $D \times_E \hat{E}=\hat{D}$. It's not obvious (and probably not true) that this cover extends to a cover of $\mathbb{P}^1_E$ with branch divisor $D$ (by extends I mean that I require the new cover to be Galois as well with the same group).

Is there a way to measure the obstruction? Can I calculate something to check if may really extend the cover?

Assume whatever you need to in order to avoid wild ramification since I'm not interested in that. And if you can solve a particular case of this, assume whatever you want.