Recently I've been using something similar to Mordell-Weil sieve to study certain Diophantine equations. This has been pretty successful, which is wonderful. And I've been wondering, if these are successful since I am lucky, I am not unlucky, or because this is just going to work. Here is a bit more clear explanation (I'm going to focus in 1-d version).
We can actually abstract the sieving problem I'm facing with in the following way. Say for any prime $p$, we have a set $S_p \subset \ZZ$, and we have that $\bigcup S_p = \ZZ$. Can we find a finite subset of primes so that $\bigcup S_p = \ZZ$? In the particular case that I'm faced with the sets $S_p = \cup (a+m_p \ZZ)$ with $m_p>0$ (in fact, $m_p=\#E(\FF_p)$). Will there always be a finite subcover in here? This, of course, will be true if $\ZZ$ is compact under a certain topology for integers (I think it is called profinite topology). I'm guessing that $\ZZ$ is not compact under profinite topology though (otherwise, that would make life a lot nicer). However, I am still unsure on weather or not in my particular case, these particular open coverings of $\ZZ$ have a finite subcover or not.
We can actually abstract the sieving problem I'm facing with in the following way. Say for any prime $p$, we have a set $S_p \subset \ZZ$, and we have that $\bigcup S_p = \ZZ$. Can we find a finite subset of primes so that $\bigcup S_p = \ZZ$? In the particular case that I'm faced with the sets $S_p = \cup (a+m_p \ZZ)$ with $m_p>0$ (in fact, $m_p=\#E(\FF_p)$). Will there always be a finite subcover in here? This, of course, will be true if $\ZZ$ is compact under a certain topology for integers (I think it is called profinite topology). I'm guessing that $\ZZ$ is not compact under profinite topology though (otherwise, that would make life a lot nicer). However, I am still unsure on weather or not in my particular case, these particular open coverings of $\ZZ$ have a finite subcover or not.
For the record, $\mathbb Z$ is *not* compact under the full profinite topology. An example can be devised by looking at $U_n=2^{n-1}+2^n \mathbb Z$ for $n>1$, and $U_0=3 \mathbb Z$. The union of $U_n$ for $n>1$ covers all non-zero integers, and $U_0$ takes care of $0$. (This example came up while discussing this problem with Mike Bennett and Patrick Ingram.)
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