I have found a fun problem concerning compass and straight edge constructions, which I am unable to solve completely at the moment.

The problem is the folowing: let $\zeta$ be a primitive elevent root of unity and consider the set $R = \{ z \in \mathbb{C} | z^{11}=1\}$. For every $S \subset R$ we define $z_{S} = \sum_{s\in S}s$. If our starting field is $\mathbb{Q}$, how many subsets $S$ are such that $z_{S}$ is constructible?

For example, if we take $S=\{1, \zeta, \zeta^3 \zeta^4,\zeta^5, \zeta^9 \}$. One can show that this leads to a constructible $z_{S}$. One does this by noting that the extension $\mathbb{Q} \subset \mathbb{Q}(\zeta)$ has a cyclic galois group of order 10 and obviously contains $R$. If a number in this extension is constructible it is necessarily contained in a subfield of degree a power of 2, considering our observation concerning our galois group, this can only be the case if the corresponding galois group is a power of two and because our extension has order ten, if this galois group has order two. This group is generated by the automorphism $\varphi$ which holds $\mathbb{Q}$ fixed and $\varphi(\zeta) = \zeta^9$. When applying some homomorphism properties and modulo $11$ calculations, we conclude that $\varphi(z_{S}) = z_{S}$ Therefore $z_{S}$ is contained in this extension of order two and is therefore constructible (I think, correct me if I'm wrong).

I tried generalizing this for other sets and I think I reduced it to the following combinatorial problem. Let $i \in \{0,1,2, \dots 10\}$. Doing the same calculations as when applying the homomorphism properties in the last example, we find that also $9i, 4i, 3i, 5i$ need to be contained in the sum expression of $z_{S}$. I suppose that choosing $i = 0$ bears no consequences,so we can restrict ourselves to counting the combinations of $i \in \{1, \dots 10 \}$ and multiply these by $2$. My first guess (or at least upper bound I suppose) is that choosing one such $i$ is equivalent to choosing $4$ of them, therefore there are at most $10 \choose 4$ $\cdot 2 = 520$. Any help would be greatly appreciated!


You are mostly on the right track, only the combinatorial part is off.

  1. The Galois group $G$ of $\Bbb{Q}(\zeta)\Bbb{Q}$ is cyclic of order $11-1=10$. Because $2$ is a primitive root modulo $11$, a generator for this cyclic group is the mapping $\sigma:\zeta\mapsto \zeta^2$.
  2. For $z_S$ to be constructible it is necessary that $n:=[\Bbb{Q}(z_S):\Bbb{Q}]$ is a power of two.
  3. Because $z_S\in\Bbb{Q}(\zeta)$ it follows that $n\mid 10$, and we are left with the alternatives $n=1$ and $n=2$.
  4. Because $G$ is cyclic the Galois correspondence tells us that there exists a single quadratic intermediate field, namely the fixed points of $\sigma^2:\zeta\mapsto \zeta^4$. All the elements of that quadratic field are constructible, so $z_S$ is constructible if and only if $\sigma^2(z_S)=z_S$.
  5. Because $11$ is a prime, the only $\Bbb{Q}$-linear relation among the elements of $R$ is the obvious $$1+\zeta+\zeta^2+\cdots+\zeta^{10}=0.$$
  6. If $J\subseteq \{0,1,\ldots,10\}$ and we denote $z_J=\sum_{j\in J}\zeta^j$, then $$\sigma^2(z_J)=\sum_{j\in J}\zeta^{4j}=z_{4J},$$ where we calculate $4j$ modulo $11$.
  7. Items 4 and 5 imply that $z_J$ is constructible if and only if $J=4J$ (you seem to have reached this point under your own steam as $\varphi:\zeta\mapsto \zeta^9$ and $\sigma^2$ generate the same subgroup of $G$).

But at this point something went wrong. You still seem to be clear on the idea that in the interesting cases $j\in J\implies 4j\in J$. To wrap up:

  • If $1\in J$, then also $4,5=4^2,9=4^3,3=4^4\in J$. Similarly if $4\in J$, then $4^2=5,4^3=9,4^4=3$ and $4^5=1$ must all be in $J$. The same with others. In other words, if $J$ contains any of $1,4,5,9,3$ it must contain all five of them.
  • Similarly, if $2\in J$, then $8=4\cdot2,$ $10=4^2\cdot2$, $7=4^3\cdot2$ and $6=4^4\cdot2$ must all be in $J$. Repeating the argument, either $2,8,10,7,6$ are all in, or none of them are.
  • This leaves us with the choices:
    • $J=\emptyset$,
    • $J=\{0\}$,
    • $J=\{1,4,5,9,3\}$,
    • $J=\{0,1,4,5,9,3\}$,
    • $J=\{2,8,10,7,6\}$,
    • $J=\{0,2,8,10,7,6\}$,
    • $J=\{1,4,5,9,3,2,8,10,7,6\}=\Bbb{Z}_{11}\setminus\{0\}$,
    • $J=\Bbb{Z}_{11}$.

So a total of $8$ alternatives. Basically you need to make three binary choices: i) either you include or don't include $0$, ii) either you include or don't include $1$ (and hence all of $1,4,5,9,3$), iii) either you include or don't include $2$ (and hence all of $2,8,10,7,6$). This makes $2^3=8$ the answer.

Although (see item 5 again) the empty set and all of $\Bbb{Z}_{11}$ give rise to the same sum $z_J=0$. The others pair up as negatives of each other due to item 5. Namely $$ z_S=-z_{R\setminus S} $$ for all $S\subseteq R$.

In a different language: you split the set $R$ into the orbits of the subgroup $\langle \phi\rangle=\langle\sigma^2\rangle\le G$, and use collections of orbits.


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