20200411, 15:00  #1 
Jul 2014
2^{2}×3×37 Posts 
need guidance with some maths
Hi, I've been trying to understand some maths in one of my text books for a long time now. I've written out all the relevant material and then explained why I can't understand it. If anyone can point out the right way to understand it, as usual I'd be most appreciative.

20200412, 09:44  #2 
Dec 2012
The Netherlands
7^{2}×29 Posts 
I'm having trouble understanding what you have written.
You start with q, which I assume is a fixed prime number, and define zeta as a primitive qth root of unity in the complex numbers. You then choose a polynomial F and suppose that \(F(\zeta^m)=F(\zeta)\) whenever \(v(m)\equiv 0\pmod{e}\). Presumably this e is not the constant 2.718... that you used earlier! Is e a fixed integer here? In explaining what v(n) means, you then refer to both e and f  what is f? 
20200413, 11:47  #3 
Jul 2014
674_{8} Posts 

20200413, 13:54  #4 
Dec 2012
The Netherlands
58D_{16} Posts 
Let's take your example of q=7. If
\[f_0\zeta^0+f_1\zeta^1+f_2\zeta^2+f_3\zeta^3+f_4\zeta^4+f_5\zeta^5= g_0\zeta^0+g_1\zeta^1+g_2\zeta^2+g_3\zeta^3+g_4\zeta^4+g_5\zeta^5\] where the \(f_i\) and \(g_i\) are polynomials with integer (or rational) coefficients then \(f_0=g_0\), \(f_1=g_1\),...,\(f_6=g_6\). So it's not a problem if the polynomial F has a constant term. Replacing \(\zeta\) with \(\zeta^m\) permutes the coefficients of \(\zeta^1\) up to \(\zeta^6\) and leaves \(\zeta^0\) unaltered. Last fiddled with by Nick on 20200413 at 20:19 Reason: Corrected typo 
20200413, 15:44  #5 
Jul 2014
1BC_{16} Posts 
Thanks very much. That's just what I needed. I can get on now with your help

20200413, 16:11  #6 
Jul 2014
2^{2}×3×37 Posts 
Except I still don't know how it can be
that if \[ F(x) = \prod_{R}(x  \zeta^{R}) \] has a constant term, then \[ F(\zeta) = A_{0}(x)\eta_{0} + A_{1}(x)\eta_{1} \] as it seems to me there can't be a constant term on the right hand side of the latter. Last fiddled with by wildrabbitt on 20200413 at 16:12 
20200413, 16:59  #7 
Dec 2012
The Netherlands
7^{2}·29 Posts 
On your 1st sheet, you have
\[ F(\zeta)=\sum_{r=1}^{q1}A_r\zeta^r\] so there is no term with \(\zeta^0\) (or \(\zeta^q\) which is also 1). On your 2nd sheet, you apply the idea to a polynomial which does have a term with \(\zeta^0\). Is the book you are using a well known one? If one of has it, perhaps we can help you pinpoint where the misunderstanding arises. 
20200413, 18:49  #8 
Jul 2014
2^{2}×3×37 Posts 
Multiplicative Number Theory by Harold Davenport published by Springer (part of a series  Graduate Texts in Mathematics.
Thanks. 
20200413, 20:29  #9 
Dec 2012
The Netherlands
7^{2}·29 Posts 
OK, I think I understand the confusion.
Let's use your example of q=7 with \(\zeta=e^{2\pi i/7}\). Then we have \[ \zeta^0=1,\zeta^1=\zeta,\zeta^2,\zeta^3,\zeta^4,\zeta^5,\zeta^6\] all distinct and \(\zeta^7=1\) again. We also have the equation \[ \zeta^6+\zeta^5+\zeta^4+\zeta^3+\zeta^2+\zeta+1=0\] so it follows that \[ 1=\zeta^6+\zeta^5+\zeta^4+\zeta^3+\zeta^2+\zeta \] Thus you can use \(\zeta^0\) to \(\zeta^5\) inclusive OR \(\zeta^1\) to \(\zeta^6\) inclusive and still equate coefficients (they are linearly independent). I hope this helps! 
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