tag:blogger.com,1999:blog-5376628436133716219.post885074468465342933..comments2024-03-29T02:14:39.189-07:00Comments on Why I hate physics: The Seventh Roots of UnityMarty Greenhttp://www.blogger.com/profile/17624084719249673373noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-5376628436133716219.post-2105716626855286632014-01-02T01:04:12.768-08:002014-01-02T01:04:12.768-08:00C6 is not actually a Galois group of a cubic, as i...C6 is not actually a Galois group of a cubic, as it is a permutation on 6 objects and S3 is permutation on 3, implying C6 is not a subgroup of S3.<br /><br />You're right, it does not really shows what happens for S3, but it does ensures that no such swapping is allowed where the Galois group is C3, as there is no complex root for the later case. What is really superior understanding here is the rotational symmetry, as you have mentioned in the previous blogpost.<br /><br />I am eager to see your next blogpost on C3, hope you'll post it soon.Balarkanoreply@blogger.comtag:blogger.com,1999:blog-5376628436133716219.post-53259199872614325922014-01-01T11:09:51.583-08:002014-01-01T11:09:51.583-08:00Balarka, I have to observe that the essence of you...Balarka, I have to observe that the essence of your explanation is that the complex roots give you a two cycle, the overall order of the splitting field is six, and therefor there has to be a three-cycle in the Galois group. I'm not entirely sure why C6 is ruled out as a possiblity, but maybe that falls our of an examination of where the three roots a,b and c have to go.<br /><br />My real issue with this type of analysis is that while it may logically prove that S3 is the Galois group, it fails to show in a constructive way just why it is that you're entitled to swap a real root with a complex one. That's why I was so pleased to notice that for every such asymmetrical swap, there is a corresponding swap with the "opposite" asymmetry. If you know what I mean.<br /><br />Anyhow, I'm going to continue in my next post with a discussion of the C3 Galois group...Marty Greenhttps://www.blogger.com/profile/17624084719249673373noreply@blogger.comtag:blogger.com,1999:blog-5376628436133716219.post-51714508787444495002013-12-31T02:16:28.081-08:002013-12-31T02:16:28.081-08:00Oh, and as a note, the polynomial is supposed to b...Oh, and as a note, the polynomial is supposed to be irreducible over Q, which I forgot to add.Balarkanoreply@blogger.comtag:blogger.com,1999:blog-5376628436133716219.post-24547245563867952232013-12-31T02:13:03.072-08:002013-12-31T02:13:03.072-08:00Hello again,
I just wanted to encourage everyone ...Hello again,<br /><br />I just wanted to encourage everyone to think about the relation between finding a cubic with Galois group C3 and the plausibility of swapping a complex root with a real root, which lies significantly deep in the theory.<br /><br />Take any cubic with rational coefficients with only one real root in x. Let x = p + q be the real root [note how p and q are of degree 3 over Q]. Then the fact that (x + a + b)(x + w * a + w^2 * b)(x + w^2 * a + w * b) = x³ - 3abx + a^3 + b^3, imply every such polynomial has splitting field Q[2^(1/3), w)] where w is the cube root of unity.<br /><br />Let's compute the degree of extension by multiplicativity of [..]: [Q[2^(1/3), w] : Q] = [Q[2^(1/3)] : Q] * [Q[w] : Q] = 3 * 2 = 6.<br /><br />As Q[2^(1/3), w] is Galois over Q, which follows from the fact that it is the splitting field of a polynomial over Q, we have that |Aut(Q[2^(1/3), w]/Q)| = [Q[2^(1/3), w] : Q] = 6 by Artin's theorem but then Aut(...) is in turn Gal(...), so we have that Galois group of any polynomial over Q with only one real root has Galois group of degree 6, implying that the Galois group is not C3 [C3 is of degree 3].<br /><br />So the only other possibility is the discriminant is positive, i.e., the polynomial has three distinct real roots [as it is easy enough to exclude the possibility of having non-distinct roots].<br /><br />Thus, not only having to permute a real and a complex root of a cubic in the complex plane imply that the Galois group is not cyclic but replacing that with "having a complex root" does the same.<br /><br />As a remark, I'd like to note that it is a necessary condition [i.e., having no complex roots] for a cubic to have cyclic Galois group, but not sufficient enough. Looking for such polynomials are especially tough. Balarkanoreply@blogger.com