From pmontgom@cwi.nl Fri Jan 8 14:46:00 CST 1999 Article: 228395 of sci.math Newsgroups: sci.math Path: news.math.niu.edu!husk.cso.niu.edu!vixen.cso.uiuc.edu!news-peer1.sprintlink.net!news.sprintlink.net!news.maxwell.syr.edu!cpk-news-hub1.bbnplanet.com!news.gtei.net!remarQ-easT!supernews.com!newsfeed.wirehub.nl!sun4nl!cwi.nl!pmontgom From: pmontgom@cwi.nl (Peter L. Montgomery) Subject: Re: Roots of the fifth degree Message-ID: Sender: news@cwi.nl (The Daily Dross) NNTP-Posting-Host: bark.cwi.nl Organization: CWI, Amsterdam References: <01be38f7$734a9da0$0100a8c0@mgreen> Date: Thu, 7 Jan 1999 08:27:37 GMT Lines: 67 Xref: news.math.niu.edu sci.math:228395 In article <01be38f7$734a9da0$0100a8c0@mgreen> "Martin Green" writes: >Suppose the roots of a third degree equation are a,b, and c. > >I can find combinations of these terms that are roots of a second-degree >equation: they are > >a^2b + b^2c + c^2a , ab^2 + bc^2 + ca^2 >This helps me solve the third degree equation. >Suppose the roots of a fourth degree equation are a,b,c, and d. >I can find combinations of these terms that are roots of a third-degree >equation: they are >ab+cd, ac+bd, ad+bc >This helps me solve the fourth-degree equation. >Suppose the roots of a fifth degree equation are a,b,c,d, and e. >I have looked for combinations of these roots that are solutions of a >sixth-degree equation, because I know that the "resolvent" is of sixth >degree. I can't find any such combinations. Does anyone know what they are? Note how you did degree-4. One order-8 = 24/3 subgroup of S4 (S4 = symmetric group on four elements) has identity (1234) (1432) (13)(24) (12)(34) (14)(23) (13) (24) Your ac + bd is invariant under any of these permutations. For example, (1234) sends a->b->c->d->a, whence ac + bd mape to bd + ca = ac + bd. One order-20 = 120/6 subgroup of S5 has the arithmetic progressions modulo 5: those satisfy tau(i) -= c1*i + c2 (mod 5) for some c1, c2 and all i. For example tau(i) == 3i + 1 (mod 5) sends 1->4->3->5->1 while leaving 2 fixed. One such progression sends i -> 4 - i, whence a<->c and d<->e with b fixed. Start with a term which is symmetric under this transformation, such as a b^2 c or b*(a - c)*(d - e) Add the images of these under other group members, getting (for example) a^2 (be + cd) + b^2 (ac + de) + c^2 (ae + bd) + d^2 (ab + ce) + e^2 (ad + bc) as one of the six roots. Apply a random permutation in S5 to this to get another root of your degree-6 polynomial, until you have all six roots. -- Peter-Lawrence.Montgomery@cwi.nl San Rafael, California The bridge to the 21st century is being designed for the private autombile. The bridge to the 22nd century will forbid private automobiles. From testware@pangea.ca Fri Jan 8 14:46:04 CST 1999 Article: 228428 of sci.math Path: news.math.niu.edu!husk.cso.niu.edu!vixen.cso.uiuc.edu!howland.erols.net!sunqbc.risq.qc.ca!cyclone.mbnet.mb.ca!pumpkin.pangea.ca!news.pangea.ca!not-for-mail From: "Martin Green" Newsgroups: sci.math Subject: Re: Roots of the fifth degree Date: 7 Jan 1999 15:39:20 GMT Organization: Pangea.CA, Inc. Lines: 66 Message-ID: <01be3a53$cec89120$0100a8c0@mgreen> References: <01be38f7$734a9da0$0100a8c0@mgreen> NNTP-Posting-Host: dock08-00-32.ner.pangea.ca Mime-Version: 1.0 Content-Type: text/plain; charset=ISO-8859-1 Content-Transfer-Encoding: 7bit X-Trace: pumpkin.pangea.ca 915723560 23318 207.161.114.144 (7 Jan 1999 15:39:20 GMT) X-Complaints-To: abuse@pangea.ca NNTP-Posting-Date: 7 Jan 1999 15:39:20 GMT X-Newsreader: Microsoft Internet News 4.70.1155 Xref: news.math.niu.edu sci.math:228428 > > Note how you did degree-4. One order-8 = 24/3 subgroup of S4 > (S4 = symmetric group on four elements) has > > identity > (1234) > (1432) > (13)(24) > (12)(34) > (14)(23) > (13) > (24) > > Thank you for the insight...I had not previously recognized the role of the above subgroup. HOWEVER.... I took 3rd year Algebra 25 years ago so I vaguely know that there's something significant about NORMAL subgroups...as I look at the above group, I think it's not a normal subgroup. My reasoning is that it has isomorphic conjugates...for example, you've generated it starting with (identity, (1234)...) etc. whereas one could generate a similar subgroup starting with (identity, (1243)...) etc. Or maybe I'm wrong, and this IS the "normal subgroup" referred to in Galois theory where they speak of a "chain of normal subgroups...". Regards, Marty Green > Your ac + bd is invariant under any of these permutations. > For example, (1234) sends a->b->c->d->a, whence > ac + bd mape to bd + ca = ac + bd. > > One order-20 = 120/6 subgroup of S5 has the arithmetic > progressions modulo 5: those satisfy tau(i) -= c1*i + c2 (mod 5) > for some c1, c2 and all i. For example tau(i) == 3i + 1 (mod 5) > sends 1->4->3->5->1 while leaving 2 fixed. > > One such progression sends i -> 4 - i, > whence a<->c and d<->e with b fixed. Start with a term > which is symmetric under this transformation, such as > > a b^2 c or b*(a - c)*(d - e) > > Add the images of these under other group members, getting (for example) > > a^2 (be + cd) + b^2 (ac + de) + c^2 (ae + bd) > + d^2 (ab + ce) + e^2 (ad + bc) > > as one of the six roots. Apply a random permutation in S5 to this to get > another root of your degree-6 polynomial, until you have all six roots. > > > -- > Peter-Lawrence.Montgomery@cwi.nl San Rafael, California > The bridge to the 21st century is being designed for the private autombile. > The bridge to the 22nd century will forbid private automobiles. >