From: Robin Chapman Subject: Re: Infinite division ring that is not a field? Date: Sat, 26 Aug 2000 10:07:06 GMT Newsgroups: sci.math Summary: [missing] In article <8o80dq$bce$1@front7.grolier.fr>, "Panh" wrote: > > Pertti Lounesto a écrit dans le message : > 39A63433.9C673F65@pp.htv.fi... > > Robin Chapman wrote: > > > > > For a more exciting example consider the ring of 3 by 3 matrices > > > generated over Q by > > > (cos 2pi/7 0 0 ) > > > ( 0 cos 4pi/7 0 ) > > > ( 0 0 cos 6pi/7) > > > and > > > (0 1 0) > > > (0 0 1) > > > (2 0 0). > > > > Could you prove that this is a division ring? Could you > > prove that it is not a field? Is this also an algebra ove Q? > > If so, is this an infinite-dimensional algebra over Q? I didn't respond to this before because PL is in my killfile. The answers are (i) yes (unless I have made a mistake, but if I have all anyone need do to confirm it is to produce a singular nonzero matrix in the ring), can you? (ii) well, do the two matrices I wrote down commute? (iii) How can a ring generated over Q by a couple of matrices fail to be a Q-algebra? (iv) Is it not obvious that this is a subring of M_3(K) where K is a cubic extension of Q? Another exercise for the reader: prove that this ring is not isomorphic to its opposite. > Can you explain why one element is invertible? I can explain why the *two* elements I wrote down are invertible: their determinants are 1/8 and 2 respectively :-) -- Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html "`The twenty-first century didn't begin until a minute past midnight January first 2001.'" John Brunner, _Stand on Zanzibar_ (1968) Sent via Deja.com http://www.deja.com/ Before you buy. ============================================================================== From: Robin Chapman Subject: Re: Infinite division ring that is not a field? Date: Sun, 27 Aug 2000 15:34:16 GMT Newsgroups: sci.math In article <8o8k0a$l90$1@vega.worldonline.fr>, "Panh" wrote: > > For a more exciting example consider the ring of 3 by 3 matrices > > generated over Q by > > (cos 2pi/7 0 0 ) > > ( 0 cos 4pi/7 0 ) > > ( 0 0 cos 6pi/7) > > and > > (0 1 0) > > (0 0 1) > > (2 0 0). > > > > > > > I can explain why the *two* elements I wrote down are invertible: > > their determinants are 1/8 and 2 respectively :-) > Can you explain why all element is invertible? > Excuse my bad english. I'm sure Pertti will tell you, since this is an example of a standard construction very similar to those of his beloved Clifford algebras. So I'll just make a few remarks about the structure of this ring. Let A and B denote my two matrices and let r_j denote cos 2pi j/7. Note that r_1, r_2 and r_3 are algebraic numbers, the roots of X^3 + X^2/2 - X/2 - 1/8 = 0, and so are all conjugate. It follows that the ring generated over Q by A is isomorphic to the field K = Q(r_1). Note that this field is a Galois extension of Q; it's the real part of the 7-th cyclotomic field. It follows that K has automorphism group of order 3 generated by the map sigma with sigma(r_1) = r_2, sigma(r_2) = r_3 and sigma(r_3) = r_1. We also see that BAB^{-1} = (r_2 0 0 ) ( 0 r_3 0 ) ( 0 0 r_1) and B^3 = 2I. Thus the ring in question is isomorphic to the Q-algebra generated by K and an element b with the relations that b^3 = 2 and bxb^{-1} = sigma(x) for x in K. Let me draw a parallel with the familiar construction of the quaternions. This can be described as the R-algebra generated by C and an element j with the relations j^2 = -1 and jzj^{-1} = complex conjugate of z for z in C. I'm sure Pertti will enlighten us about the name of this standard construction, and why it does produce a division ring in my example (as well as in this quaternion construction). -- Robin Chapman, http://www.maths.ex.ac.uk/~rjc/rjc.html "`The twenty-first century didn't begin until a minute past midnight January first 2001.'" John Brunner, _Stand on Zanzibar_ (1968) Sent via Deja.com http://www.deja.com/ Before you buy. ============================================================================== From: baez@galaxy.ucr.edu (John Baez) Subject: Re: Infinite division ring that is not a field? Date: 29 Aug 2000 01:11:38 GMT Newsgroups: sci.math In article <8obcdo$gaq$1@nnrp1.deja.com>, Robin Chapman wrote: >Let me draw a parallel with the familiar construction of the >quaternions. This can be described as the R-algebra generated by >C and an element j with the relations j^2 = -1 and jzj^{-1} = complex >conjugate of z for z in C. > >I'm sure Pertti will enlighten us about the name of this standard >construction, and why it does produce a division ring in my example >(as well as in this quaternion construction). I'm not Pertti, but I assume you're trying to get him to say "Cayley-Dickson construction". Nice example!