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Topic: tough number theory prob. (Read 3027 times) 

Grimbal
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Re: tough number theory prob.
« Reply #25 on: Oct 18^{th}, 2008, 11:54pm » 
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The way I see it is that the correct notation is: a b (mod n) and it should be understood as a b (please understand as the equivalence modulo n) So, "(mod n)" is not part of the expression, but gives a context to it. The equivalence a b (mod n) means a = b + k·n for some k which implies rem(a,n) = rem(b,n) This has nothing to do with the binary operator 'mod' that you can see in some computer languages. In mathematics you use " ... (mod n)" equivalence and express the remainder with a rem(x,n) function, while in some computer languages, or calculators, you use "mod" as an operator to compute a remainder.


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Michael Dagg
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Re: tough number theory prob.
« Reply #26 on: Nov 7^{th}, 2008, 12:23pm » 
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Nice discussion and as pointed out, this is an interesting problem as well as ones like it. It is crucial in many problems to know the group of units in an algebraic number field. (For example, when "ideals", then viewed as "ideal numbers" were first invented by Kummer, it was in the context of proving Fermat's Last Theorem. He got as far as calculating that some important ideal would be principal, but then had to wrestle with the possible ways that the generator could be presented; the options differ from one another by multiplication by a unit. So it was very important to know the group of units in those algebraic number fields.) A summary of what happens is that the group of units in the ring of integers in an algebraic number field F is a finitelygenerated abelian group, (that's not obvious!) hence (easy) it's isomorphic to a finite group T times a free abelian group Z^r . The finite subgroup T is (easy) the group of roots of unity that lie in the field, and that' not usually hard to determine; at worst you have to try factoring a finite list of cyclotomic polynomials like x^n  1 over the field. The torsionfree part can be shown to be of rank s1+s2 where s1 is the number of embeddings of the field into the real numbers, and s2 is the number of complexconjugate pairs of nonreal embeddings. (These numbers make s1 + 2 s2 = [ F : Q ] .) The calculation of this value of r is a little bit clever but not hard to follow. The hardest part of all is to actually compute a set of r generators of this group; typically one can find a set of r independent units, i.e. one can find a group of units which is of finite index in the full group of units, but then you have to decide whether for example any simple combination of the units you know is the square of another unit you didn't previously know about. I like the book Number Theory by Borevich and Shafarevich (despite its numerous misprints and lousy index) because it treats all these topics. Definitely one to add to your bookshelf.

« Last Edit: Nov 7^{th}, 2008, 12:24pm by Michael Dagg » 
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Regards, Michael Dagg



