tag:blogger.com,1999:blog-9004452969563620551.post4175474098940705361..comments2019-01-05T13:47:05.461+05:30Comments on Paramanand's Math Notes: Gauss and Regular Polygons: Gaussian Periods Contd.testernoreply@blogger.comBlogger7125tag:blogger.com,1999:blog-9004452969563620551.post-53048360739126388552014-10-31T11:18:14.137+05:302014-10-31T11:18:14.137+05:30@Gerald Schmidt
There is a typo in the subgroups y...@Gerald Schmidt<br />There is a typo in the subgroups you have mentioned. The partitioning is with $\{3, 10, 5, 11, 14, 7, 12, 6\}$ and $\{1, 9, 13, 15, 16, 8, 4, 2\}$. In terms of Modern Algebra, the second set containing $1$ is a subgroup and the other one is a coset. Together these cosets form the quotient group of order $2$. Thus the subgroup $\{1, 9, 13, 15, 16, 8, 4, 2\}$ is normal subgroups of the full group $\{1, \ldots, 16\}$. The same technique (based on a primitive root of $p$) would work for any Fermat prime $p$ apart from $17$.Paramanand Singhhttps://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-79446734500370239422014-10-30T22:09:32.550+05:302014-10-30T22:09:32.550+05:30Dear Paramanand
I was reading your contributions w...Dear Paramanand<br />I was reading your contributions with delight. I totally agree that the theory of Gaussian periods has been forgotten at the expense of more modern Theories. At university students are confronted with abstract terms about wonderful objects that they cannot see with their beautiful 'colors' or cannot feel with their amazing 'shape'.<br />I use the heptadecagon as a lighthouse for all my Algebra-Studies, because it's viewpoint is so enlightening.<br /><br />I have still not fully understood what Gauss intuitively saw when he separated the 2 subgroups of order 8.<br />All I know is this:<br />The first sub-group {3,10,5,11,14,7,2,16} are all the primitive roots of 17<br />The second sub-group {1,9,13,15,16,8,4,2} are all the quadratic residues of 17<br /><br />2^p-1 with q = (p-1)/2 can be factorized into 2^q+1 × 2^q-1 <br />Is this a natural way to divide these 2 periods?<br />Many thanks in advance for further clarification<br />Gerald Schmidthttps://www.youtube.com/user/Gerald56noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-19709806755086756902014-07-12T15:47:21.783+05:302014-07-12T15:47:21.783+05:30@Kazbich-the-bandit,
By definition $eta$, a period...@Kazbich-the-bandit,<br />By definition $eta$, a period of $f$ terms always looks like $$\eta = \zeta_{r} + \zeta_{e + r} + \zeta_{2e + r} + \cdots + \zeta_{e(f - 1) + r}$$ and clearly if we apply $\sigma^{e}$ to $\eta$ each term above is replaced by next term and last term is replaced by first term thereby the sum $\eta$ remains fixed. Hence by definition of $\eta$ it belongs to the field $K_{f}$.Paramanand Singhhttps://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-19478328711671052712014-07-11T11:43:31.652+05:302014-07-11T11:43:31.652+05:30In the proof of the first property it is claimed t...In the proof of the first property it is claimed that eta, a period of f terms is a member of Kf, the fixed field of sigma repeated e times. So that sigma^e(eta) = eta, right? As I understand it, the operation should send the terms zeta(0) to zeta(e), zeta(1) to zeta(e+1),...,zeta(f-1) to zeta(e+(f-1)). I am to understand that the operation leaves eta fixed because Kf is of dimension e - so that (zeta(0),...,zeta(f-1)) denotes the same vector as (zeta(e),...,zeta(e+f-1))?Kazbich-the-bandithttps://www.blogger.com/profile/18247629194967955510noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-25693274196320868922014-07-10T09:34:36.346+05:302014-07-10T09:34:36.346+05:30@Kazbich-the-bandit,
First a small typo in your co...@Kazbich-the-bandit,<br />First a small typo in your comment. Symbol $\eta$ in $\mathbb{Q}(\eta_{6})$ is called "eta" and not "nu" and "nu" is denoted by $\nu$. Next your interpretation that "a period of two terms is a root of a cubic with coefficients that are elements of $\mathbb{Q}(\eta_{6})$" is correct. The proof of this is given in the post. See the result "Every element in $K_{f}$ is a root of a polynomial ..." and its proof thereafter. Let me know if you have any specific trouble in understanding this proof. You have to focus on the part which deals with automorphisms $\sigma$ and its powers.Paramanand Singhhttps://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-11398932076002357922014-07-10T03:02:52.255+05:302014-07-10T03:02:52.255+05:30I am seeking to grasp the theorem on an intuitive ...I am seeking to grasp the theorem on an intuitive level, so please bear with me. Firstly, the second property of Gaussian periods, with particular application to p = 13, states that a period of two terms is a root of cubic equation (that is obvious) with coefficients that are rational expressions in a period of 6 terms. Would I be correct if I paraphrase the last statement as saying that the each coefficient of the cubic equation can be written as expression of 6 terms, ie as an element of Q(nu6)? <br />Secondly, this particular property must be a direct consequence of the fact that, as in the example, 2 divides 6, which in turn divides 12. However, I fail to see how it follows. I mean, why is it true?<br />Kazbich-the-bandithttps://www.blogger.com/profile/18247629194967955510noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-35417398388008693882013-04-09T09:31:35.995+05:302013-04-09T09:31:35.995+05:30Truly no matter if someone doesn’t know then its u...Truly no matter if someone doesn’t know then its up to other users that they will help, so here it happens.Eduardonoreply@blogger.com