Newsgroups: sci.math From: kwhyte@math.uchicago.edu (Kevin Whyte) Subject: Re: Mersenne Composites Date: Thu, 2 Apr 1992 20:08:48 GMT Since Mersenne primes seem popular right now, here's an amusing attempt to prove there are only finitely many. Define a double field to be a pair of fields F and F' together with and isomorphism between the mult. group of F and the additive group of F'. It is easy to show that the only finite double fields are those with |F|=2^n , |F'|=2^n-1 with 2^n-1 prime. Now, the compactness theorem in this case gives that if there are infinitely many finite double fields (i.e. infinitely mant Mersenne primes) then there must be an infinite double field. Infinite double fields are hard to come by, so proving there aren't any seemed plausible. This would show that there are only finitely many Mersenne primes. I'm not sure if the compactness theorem gives a countable double field or not (logic is nowhere near my subject), but I don't know of any countable double fields (but then, I haven't looked very hard). Can anyone come up with a similiar approach to Fermat primes? There one might actually hope to prove something. Finally, is it known that the sequence 2^2-1 =3 , 2^3-1=7, 2^7-1=127 , ... contains a composite number? Kevin kwhyte@math.uchicago.edu ============================================================================== Newsgroups: sci.math From: rusin@mp.cs.niu.edu (David Rusin) Subject: Re: Mersenne Composites Date: Fri, 3 Apr 1992 18:55:50 GMT In article <1992Apr2.200848.16009@midway.uchicago.edu> kwhyte@math.uchicago.edu (Kevin Whyte) writes: >Define a double field to be a pair of fields F and F' together >with and isomorphism between the mult. group of F and the additive >group of F'. [motivation deleted] >... Infinite double fields are hard >to come by, so proving there aren't any seemed plausible. I have this memory of having looked into the multiplicative groups of fields once and having found a paper (probably mid 20th century) that the only restriction on the multiplicative group of a field is that there be at most n solutions to x^n=1 (obviously). Thus for example, there would be a field with multiplicative group isomorphic to the additive reals. (It would be of characteristic 2). I may be forgetting some other condition or two but the long and short of it is that really there will be lots of double fields. (It would be a lot more fun to have "amicable pairs" of fields: each isomorphic to the other's multiplicative group.) dave rusin@math.niu.edu ============================================================================== From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin) Newsgroups: sci.math Subject: Re: Mersenne Composites Date: 6 Apr 92 19:29:13 GMT In <1992Apr3.185550.20987@mp.cs.niu.edu> rusin@mp.cs.niu.edu (David Rusin) writes: >I have this memory of having looked into the multiplicative groups of fields >once and having found a paper (probably mid 20th century) that the only >restriction on the multiplicative group of a field is that there be at most >n solutions to x^n=1 (obviously). Thus for example, there would be a field >with multiplicative group isomorphic to the additive reals. (It would >be of characteristic 2). >I may be forgetting some other condition or two but the long and short of >it is that really there will be lots of double fields. >(It would be a lot more fun to have "amicable pairs" of fields: each isomorphic >to the other's multiplicative group.) >dave rusin@math.niu.edu References would be appreciated. -- Arthur L. Rubin: a_rubin@dsg4.dse.beckman.com (work) Beckman Instruments/Brea 216-5888@mcimail.com 70707.453@compuserve.com arthur@pnet01.cts.com (personal) My opinions are my own, and do not represent those of my employer. Ich bin ein Virus. Mach' mit und kopiere mich in Deine .signature. ============================================================================== Newsgroups: sci.math From: rusin@mp.cs.niu.edu (David Rusin) Subject: Re: Mersenne Composites Date: Wed, 8 Apr 1992 14:55:15 GMT In article a_rubin@dsg4.dse.beckman.com (Arthur Rubin) writes: >In <1992Apr3.185550.20987@mp.cs.niu.edu> rusin@mp.cs.niu.edu (David Rusin) writes: > >> ... the only >>restriction on the multiplicative group of a field is that there be at most >>n solutions to x^n=1 (obviously). ... > >References would be appreciated. Everything I know about this question (not much) started with the relevant section of Fuchs, "Infinite Abelian Groups" (Vol II), p 312 et seq (especially the notes at the end of the chapter). Relying now not on my poor memory but on this source we see: Given any infinite abelian A with an element of order 2 and a torsion subgroup embeddable in Q/Z, there is a field K whose multiplicative group is isomorphic to A x F for some free abelian group F. On the other hand, there is no way to characterize exactly the groups A which occur as the multiplicative subgroup of some K (without the extra free factor, that is). This is a result from Logic. dave rusin@math.niu.edu