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WLOG I'm going to use $\zeta = \sqrt[12]1$ in this example. I'm trying to find some way to work with sums of distinct powers of $\zeta$, specifically so that I can analyze sums that are equal to zero (such as $1 + \zeta + \zeta^2 + \zeta^5 + \zeta^6 + \zeta^8 + \zeta^9$) (EDIT: I want the coefficients in the sum to be either $0$ or $1$, so each distinct power is only included once at most). I initially tried to work in the quotient ring $\mathbb F_2[x] / ( x^{12} + 1)$, but it quickly became clear that while I expected $P=1 + x+ \cdots + x^{11}$ to be equal to $0$ (since $Px = P$), it was not. I'm not sure what the flaw in my logic here was (though I suspect it has something to do with $1$ being a root of $x^{12}+1$), but moving to $\mathbb F_2[x]/P$ obviously fixed that issue. My problem is that I can't find a sense in which the polynomial (for example) $1 + x + x^2 + x^5 + x^6 + x^8 + x^9 = 0$, even though that's what I should expect to see if $x = \zeta$. Where am I going wrong?

Erez Israeli Miller
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Over $\mathbb{F}_2[x]$ we have $x^{12} - 1 = (x^3 - 1)^4$ so the quotient ring $\mathbb{F}_2[x]/(x^{12} - 1)$ is less well-behaved than you might have initially guessed and in particular is not a field or even an integral domain ($x^3 - 1$ is nilpotent); this means it is not the subfield of $\overline{\mathbb{F}_2}$ obtained by adjoining a $12^{th}$ root of unity, which may have been what you were actually going for. The problem persists for $\frac{x^{12} - 1}{x - 1} = (x - 1)^3 (x^2 + x + 1)^4$.

Linear combinations of powers of $\zeta_{12}$ equal to zero with integer or rational coefficients can be completely understood as follows. The minimal polynomial of $\zeta_{12}$ over $\mathbb{Z}$ is the cyclotomic polynomial

$$\Phi_{12}(x) = x^4 - x^2 + 1$$

and every linear relation between the powers of $\zeta_{12}$ is a consequence of the relation $\Phi_{12}(\zeta_{12}) = 0$ in the sense that it corresponds to a polynomial relation $p(\zeta_{12}) = 0$ and $\Phi_{12}(x) \mid p(x)$. In your example, WolframAlpha gives

$$p(x) = (x^4 - x^2 + 1)(x^5 + x^4 + x^3 + 2x^2 + x + 1).$$

Qiaochu Yuan
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  • Thanks! This makes a lot of sense, but I specifically want to analyze sums where each power is either included once or not at all so the coefficients are either $0$ or $1$ (I probably should have clarified this in my post, and I'll edit it now). Does this mean I need $x^4 + x^2 + 1$? Will $\mathbb F_2[x] / (x^4 + x^2 + 1)$ allow me to analyze those types of sums, and will in general $\mathbb F_2[x] / \Phi_n(x)$ work if I take all coefficients of $\Phi_n$ to be $\mod 2$? – Erez Israeli Miller Jan 07 '21 at 03:07
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    @Erez: I don't think working over $\mathbb{F}2$ does what you want. Over $\mathbb{F}_2$ the cyclotomic polynomial further factors as $x^4 + x^2 + 1 = (x^2 + x + 1)^2$. Basically the deal is that $\overline{\mathbb{F}_2}$ does not actually contain any primitive $12^{th}$ roots of unity, and in fact doesn't contain any primitive $n^{th}$ roots of unity for any even $n$. So we're going all the way down to the third roots. But over $\mathbb{C}$ we have $\zeta{12}^2 + \zeta_{12} + 1 = \frac{\zeta_4 - 1}{\zeta_{12} - 1} \neq 0$. – Qiaochu Yuan Jan 07 '21 at 03:12
  • Ah, that makes sense. I'm marking your answer as correct, since you've basically explained my entire question is impossible :) – Erez Israeli Miller Jan 07 '21 at 03:15
  • Do you have any ideas as to how I might approach analyzing such zero-valued sums (with coefficients of either 0 or 1) in that case? – Erez Israeli Miller Jan 07 '21 at 03:16
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    @Erez: sums of distinct powers of a root of unity equal to $0$ can be understood geometrically as tracing out an equilateral polygon whose angles are suitable multiples of a fixed angle. The sum you wrote down corresponds to some heptagon. So at least there is some geometry available. What do you want this for? – Qiaochu Yuan Jan 07 '21 at 03:17
  • I'm not sure I understand what you mean by that equilateral polygon (since I don't think the distance from $1$ to $\zeta$ is the same as from $\zeta^6$ to $\zeta^8$, can you elaborate on that? And this is mostly out of curiosity: I was playing with the balanced centrifuge problem and was attempting to get some idea of the structure of the solution (but clearly messed up). I just found some stuff online right now relating to what you were saying with the cyclotomics; if you're interested it's here: https://mattbaker.blog/2018/06/25/the-balanced-centrifuge-problem/ – Erez Israeli Miller Jan 07 '21 at 03:24
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    @Erez: I mean that if $\sum \zeta^{e_i} = 0$ then consider the polygon with vertices $0, \zeta^{e_1}, \zeta^{e_1} + \zeta^{e_2}, \dots$ which goes all the way back to the origin. – Qiaochu Yuan Jan 07 '21 at 03:26
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    @Erez: the "Mann's theorem" link in that blog post takes this point of view: https://www.cambridge.org/core/journals/mathematika/article/abs/on-linear-relations-between-roots-of-unity/346C7FE8D77A7D987662F7310265271D – Qiaochu Yuan Jan 07 '21 at 03:32
  • Awesome, thanks! I think that covers everything! – Erez Israeli Miller Jan 07 '21 at 03:32
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$\zeta_{12}\,$ has minimal poly $\, \Phi_{12}(x) = x^4-x^2+1\mid x^6+1\,$ (by $(x^6)^2 = 1,\ x^6\neq 1\Rightarrow x^6= -1)\,$ so by the method of simpler multiples, to reduce $\!\bmod \Phi\,$ we can first reduce $\!\bmod x^6+1,\,$ where we can apply the simple rewrite rules $\,\color{#c00}{x^6 \equiv -1},\,$ so $\,\color{#0a0}{x^8 \equiv -x^2},\,$ and $\,\color{#90f}{x^9 \equiv -x^3},\,$ e.g. in your case

$$\begin{align}&1+x+x^2+x^5+\color{#c00}{x^6}+\color{#0a0}{x^8}+\color{#90f}{x^9}\\[.1em] \equiv\ & \color{#c00}{1}+x+\color{#0a0}{x^2}+x^5\color{#c00}{-1}\,\color{#0a0}{-\,x^2}\,\color{#90f}{-x^3}\\[.1em] \equiv\ &\ \ \ \ \ \ \ x\ \ \ \,+\ \ \ \ \ x^5\ \ \color{#90f}{-\ \ x^3}\\[.1em] \equiv\ &\ \ x\ (1\ \ \ \ +\,\ \ \ \ x^4\ \ -\ \ x^2)\\[.1em] \equiv\ &\ \ x\,\Phi_{12}\equiv\, 0 \end{align}\qquad\qquad$$

Bill Dubuque
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