Timeline for Solving "a, b, a+b have given divisors" problem
Current License: CC BY-SA 2.5
5 events
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| Oct 23, 2009 at 22:52 | comment | added | Ilya Nikokoshev | I was wrong about counting, btw. It's about proving things mostly. Your answer is very relevant. | |
| Oct 23, 2009 at 22:50 | comment | added | Ilya Nikokoshev | In other words, x should be an element of projective space minus three points over every prime p for primes not in S. This can be combined together to say that x is a section of a map (P^1-(0, 1, infty)) over (Spec ZZ - S). | |
| Oct 23, 2009 at 22:49 | comment | added | Ilya Nikokoshev | Now I want to solve an equation x + y = 1 where x and y have no primes other than the ones in S. In other words, they have only zeroes or poles on S when you consider them as rational functions. Ok, so let's say I have an x with this property. Then y is uniquely defined and I need it to have no poles or zeroes outside of S. This means x must have no poles or ones outside of S as well. So, our problem is reformulated as finding x such that at every prime p the value of x is neither 0 nor 1 nor infinity. | |
| Oct 23, 2009 at 22:46 | comment | added | Ilya Nikokoshev |
I can't resist explaining the connection to schemes. So, first let's consider the scheme Spec ZZ (if you don't know what this is, then thin about it as just the collection of primes numbers). Now what's an integer? Simple -- it's a function on this thing. Ok, what's a rational? Also easy -- it's a rational function, that is something with poles.
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| Oct 23, 2009 at 22:31 | history | answered | Alison Miller | CC BY-SA 2.5 |