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If you don't specify more about the structure of the field $K$, then we can't say much about the structure of the group $E(K)$. There are special cases (described in the Wikipedia article): If $K$ i …

This revision (June 2015) is mostly to say that I've verified the answer by ell. I found the Proceedings volume in question in the U. Tokyo library. It is basically a bound volume of photocopies of …

You haven't really explained your notation, but it seems that you are covering the log curve $X$ with two neighborhoods $U$ and $V$, where $U$ contains a node, and $V$ contains a log point. I think t …

You won't get $A$ as a quotient of a vector space in general without some kind of strange transcendentality. For example, if $V$ is defined over $\mathbb{F}_p$, then any $S$-valued point of $V$ is $p$ …

The standard definition of "unramified" applies to the function field case, so any finite branched cover of the projective line yields a Galois representation that is ramified in finitely many places. …

Yes. See Kedlaya's Fourier transforms and p-adic Weil II. This is a proof using Berthelot's rigid cohomology.

There is another type of uniformization introduced in Mochizuki's book Foundations of $p$-adic Teichmüller theory. It uses curves equipped with nilpotent indigenous bundles. I don't see what local c …

One reason why one needs a large coefficient group like $\overline{\mathbb{Q}_\ell}$ instead of $\mathbb{Q}$ is that a reasonable cohomology theory should be a functor. Sometimes varieties over finit …

The answer is yes (assuming you are not demanding that the map be unramified away from $x$ and $y$). Choose any Belyi map $f: X \to \mathbb{P}^1$. The points $f(x)$ and $f(y)$ are defined over some …

In general, maps over $W$ from $W$ to $X \times W$ are the same as maps from $W$ to $X$, by the universal property of products. Here, $W = \operatorname{Spec} \mathbb{C}$. I think one possible reaso …

The set of points should be given by the second choice, i.e., the set of $\operatorname{Spec} \mathbb{C}$-points, over $\operatorname{Spec} \mathbb{C}$. However, there is an additional step you need …

I think the confusion concerns the definition of free action on a scheme. We do not demand that free actions act freely on the underlying topological space. Instead, we ask that for every scheme $S$ …

Here's a quick and dirty version of George Lowther's calculation that I learned from Bjorn Poonen. It is presented in a bit more generality in the next-to-last slide of this talk, so it is in a sense …

(More editing for cleanliness) The statement is false. I learned of this example from "James" at this blog post. If you take a nodal cubic curve (notably quasiprojective), there is a flat, unramifi …

Kedlaya gave a talk in August in which he mentioned some work of Daniel Caro on finiteness for rigid cohomology with coefficients (some of which is on the ArXiv). On the same page, you can find notes …