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 2020-04-11, 15:00 #1 wildrabbitt   Jul 2014 3·149 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.
 2020-04-12, 09:44 #2 Nick     Dec 2012 The Netherlands 7×239 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?
 2020-04-13, 11:47 #3 wildrabbitt   Jul 2014 1BF16 Posts Thanks for your interest. I wrote out what I understand of it.
 2020-04-13, 13:54 #4 Nick     Dec 2012 The Netherlands 7·239 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 2020-04-13 at 20:19 Reason: Corrected typo
 2020-04-13, 15:44 #5 wildrabbitt   Jul 2014 3×149 Posts Thanks very much. That's just what I needed. I can get on now with your help
 2020-04-13, 16:11 #6 wildrabbitt   Jul 2014 3·149 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 2020-04-13 at 16:12
 2020-04-13, 16:59 #7 Nick     Dec 2012 The Netherlands 68916 Posts On your 1st sheet, you have $F(\zeta)=\sum_{r=1}^{q-1}A_r\zeta^r$ 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.
 2020-04-13, 18:49 #8 wildrabbitt   Jul 2014 6778 Posts Multiplicative Number Theory by Harold Davenport published by Springer (part of a series - Graduate Texts in Mathematics. Thanks.
 2020-04-13, 20:29 #9 Nick     Dec 2012 The Netherlands 167310 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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