tag:blogger.com,1999:blog-5376628436133716219.post5527440666082641590..comments2024-03-18T23:22:43.330-07:00Comments on Why I hate physics: Cubic Equations With Galois Group C3Marty Greenhttp://www.blogger.com/profile/17624084719249673373noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-5376628436133716219.post-49163160929421216652014-01-03T23:49:19.184-08:002014-01-03T23:49:19.184-08:00Yes, that does. I half assumed it does not, but st...Yes, that does. I half assumed it does not, but still went ahead to take a look at because even if the sextic wasn't reducible, it would have been nice to look at it's Galois group.Balarkanoreply@blogger.comtag:blogger.com,1999:blog-5376628436133716219.post-15895385140341561642014-01-03T14:35:48.944-08:002014-01-03T14:35:48.944-08:00So the sextic factors for x^2-4?? Well, that chang...So the sextic factors for x^2-4?? Well, that changes everything! Maybe we do have a machine for generating cyclic cubics after all. By the way, you have a sign error on the second factor...it shoould be plus 3 for the x term, not minus 3. No, I didn't multiply it out...I figured it out a different way, quite a funny way actually, which I'm going to explain in my next post...<br /><br />Marty Greenhttps://www.blogger.com/profile/17624084719249673373noreply@blogger.comtag:blogger.com,1999:blog-5376628436133716219.post-42522257475466940662014-01-03T13:33:43.622-08:002014-01-03T13:33:43.622-08:00Ingenious! You excluded the factor of the sextic t...Ingenious! You excluded the factor of the sextic that contains the well-known cyclic cubic and get another; very nice, quite nice indeed.<br /><br />The fact that a sextic gives rise to two cubics satisfying the same cyclic relationship, I think this has something to do with the fact that the cyclic relations are all quadratic. Saying this, I don't think it is hard afterwards to prove why such phenomena occurs.<br /><br />Now, about A^2 - 4 = B, etc., I get the octic x^8 - 16x^6 + 88x^4 - 192x^2 - x + 140 and after the elimination of trivial quadratic <br />this reduces to a sextic that gives rise to two cubics x^3 - x^2 - 6x + 7 and x^3 + 2x^2 - 3x - 5, both having the alternating group of order 3, as desired. I can't take all credits for this apparently tedious factorization, btw : GP did it all. [:-p]Balarkanoreply@blogger.com