Unicity of decomposition for the monoid generated by roots of the unity.

Question

This question is linked to Minimal number of generators of the monoid generated by roots of the unity.

The inspiration of this question comes from a proof I made in which I had to decompose relative coefficients into sums of $p$-th roots of the unity.

Let us set $M_n$ the monoïd generated by all $n$-th roots of the unity, i.e. all elements $z \in \mathbb{C}$ of the form $$z = \sum_{\lambda \in U_n} k_\lambda \cdot \lambda$$ where $k_\lambda \in \mathbb{N}$.

My question is the following, it is easy to prove that $\forall z \in M_n$ there exists $k_\lambda \in \mathbb{N}, \, \lambda \in U_n$ such that $\exists \lambda_0 \in U_n: k_{\lambda_0} = 0$ and $$z = \sum_{\lambda \in U_n} k_\lambda \cdot \lambda.$$ but Is this decomposition unique at least for $n$ without any perfect square divider or prime $n$?

• My motivation for the additional condition is that $\sum_{\lambda \in U_n} \lambda = 0$ so I need at least to ensure that there is at least one zero coefficient.
• For the second part we know that $\sum_{\lambda \in U_n^\times} \lambda = \mu(n)$ where $U_n^\times$ is the set of primitive $n$-th roots of the unity and $\mu$ is the möbius function.

This gives us:

• If $n$ has at least three different primes $p,q,r$ in its decomposition, since $U_{pqr}^\times \subset U_n$, we need to add extra conditions since $\sum_{\lambda \in U_{pqr}^\times} \lambda = -1$ thus we won't have unicity without changes of conditions.
• If $n$ has at least two different primes $p,q$ in its decomposition, since $U_{pq}^\times \subset U_n$, we need to remove $1$ as an index for the coefficients since $\sum_{\lambda \in U_{pq}^\times} \lambda = 1$.
• If $a^2 \mid n$ then we must ensure that at least one of the coefficients indexed by primitive roots is zero since $\sum_{\lambda \in U_{n}^\times} \lambda = 0$

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There are probably additional conditions but my intuition is the following:

• I think it is at least true for prime $n$
• It is likely to be true if we revise conditions and remove $1$ when $n$ is the product of two primes.
• It is probably much more complicated when $n$ is the product of at least three primes.
• Lastly if some $a^2 \mid n$ it will be even more complicated and I have no clue.

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Useful Information

For $n \leq 6$ and $n \neq 5$ $Z[e^{\frac{2i\pi}{n}}]$ is euclidian.

Update

The case is very complicated for non-prime $n$ since we need to consider all relations of the type: $$ \sum_{\lambda \in U_m^\times} \zeta_n^k\lambda = \mu(m) \zeta_n^k$$ for $m \mid n$ and $0\le k <n$

My guess would be that all the relations are generated by the $$ \sum_{\lambda \in U_n^\times} \zeta_n^k\lambda = \mu(n) \zeta_n^k$$ for $0\le k <n$.

Three interesting cases:

• $n=4 = 2^2$ we just need the generators $1,i,-1,-i$ and the relations $k_1 k_{-1} = 0$ and $k_i k_{-i} =0$.
• $n=6 = 2\cdot 3$ the monoid is generated by $\zeta_6,-1, \zeta_6^5$ and it is the smallest number of generators and we need the relation $k_{\zeta_6} k_{-1} k_{\zeta_6^5} = 0$.
  Indeed, the half lines $\zeta_6 \mathbb{R}_+, \, -\mathbb{R}_+, \, \zeta_6^5 \mathbb{R}_+$ trisect the complex plane, so if $$k_{\zeta_6}\zeta_6 - k_{-1} +k_{\zeta_6^5}\zeta_6^5 = k_{\zeta_6}'\zeta_6 - k_{-1}' +k_{\zeta_6^5}'\zeta_6^5,$$ necessarily since they both belong to one sector, there exists $\lambda \in \{\zeta_6,-1,\zeta_6^5\}$ such that $k_\lambda = 0 = k_\lambda'$.
  This gives us a real linear combination of two non-linear vectors which means we have $k_\lambda = k_\lambda' \, \forall \lambda \in \{\zeta_6,-1,\zeta_6^5\}$.
  In fact we could have directly seen that $M_6 = M_3$.
• $n=12 = 2^2 \cdot 3$ in this case, we denote that $V= \mathbb{Q} \otimes_{\mathbb{Z}} M_{12}$ (yes $M_n$ is always a group) is infact decomposable as $$V = \mathbb{Q} \cdot U_6 \oplus \mathbb{Q} \cdot U_{12}^\times$$ Since $U_4^\times \subset M_{12}^\times $ it suffices to check their real parts. Since they are both abelian groups, this mean we can at least reason on each part of the decomposition: $$M_{12} = M_6 \oplus M_{12}^\times$$ For the first one we already know the conditions and for the second we need all $4$ generators, with the relations $k_{\zeta_{12}} k_{\zeta_{12}^7} = 0$ and $k_{\zeta_{12}^5} k_{\zeta_{12}^{11}} = 0$.
  So we have the generators $\{\zeta_{12},\, \zeta_{12}^2, \, \zeta_{12}^5, \, \zeta_{12}^6 , \, \zeta_{12}^7, \, \zeta_{12}^{10}, \zeta_{12}^{11}\}$ with the relations $$k_{\zeta_{12}^{2}} k_{\zeta_{12}^{6}} k_{\zeta_{12}^{10}} = k_{\zeta_{12}} k_{\zeta_{12}^{7}} = k_{\zeta_{12}^{5}} k_{\zeta_{12}^{11}} = 0.$$ The main mechanic probably has to deal with the square-free part of $n$ apart from the rest.

Now as hinted before, we can also see that $M_{2p} = M_p$ for $p \neq 2$ prime since $M_p$ is a group. So it only needs $p$ generators with the same relations. By going even further, if $n\neq 1$ is odd, $M_n = M_{2n}$.

Answer 1

First note that $U_n=\{\zeta_n^m: m=0,\ldots,{n-1}\}$, where $\zeta_n$ is any primitive $n$-th root of unity, and so $$M_n=\left\{\sum_{i=1}^nk_i\zeta_n^i:\ k_0,\ldots,k_{n-1}\in\Bbb{N}\right\}.$$ Your conjecture is not true for all integers $n>1$. For $n=6$ you have $$\zeta_6^0=\zeta_6^1+\zeta_6^5.$$

On the other hand, it is true if $n$ is prime. Let $z\in M_n$ and suppose $k_0,\ldots,k_{n-1},k_0',\ldots,k_{n-1}'\in\Bbb{N}$ are such that $\prod k_i=\prod k_i'=0$ and $$\sum_{i=0}^{n-1}k_i\zeta_n^i=z=\sum_{i=1}^nk_i'\zeta_n^i.$$ Then it follows that $\zeta_n$ is a zero of the polynomial $$\sum_{i=0}^{n-1}(k_i-k_i')X^i=0,$$ and hence that the $n$-th cyclotomic polynomial divides the polynomial $$P:=\sum_{i=0}^{n-1}(k_i-k_i')X^i.$$ This is a polynomial of degree $n-1$. Of course if $n$ is prime then $$\Phi_n=\sum_{i=0}^{n-1}X^i,$$ from which it follows that $k_i-k_i'=c$ for all $i$. Because $k_j=0$ for some $j$ we see that $-k_j'=c$ and so $c\leq0$. Because $k_{j'}'=0$ for some $j'$ we see that $k_{j'}=c$ and so $c\geq0$. We conclude that $c=0$ and so $k_i=k_i'$ for all $i$, meaning that the representation is unique.

Answer 2

This monoid is contained in the abelian group underlying the ring of integers $\mathbb{Z}[\zeta_n]$ of the cyclotomic field $\mathbb{Q}[\zeta_n]$. The field has $\mathbb{Q}$-dimension $\phi(n)$ (where $\phi$ is Euler's totient function), and the former is a free $\mathbb{Z}$-module, and $\mathbb{Z}[\zeta]\otimes \mathbb{Q}\cong \mathbb{Q}[\zeta]$, so the rank (minimal number of generators as a group) is also equal to $\phi(n)<n$ (Euler's totient function, the number of primitive roots of unity). Therefore, there are many linear relations

$$\sum_{i=0}^n n_i\zeta_n^i=0$$

with $n_i\in \mathbb{Z}$ among your roots of unity. So, if you use all roots of unity, then the representation will not be unique.

This also means that the representation is unique if you use only primitive roots of unity for $n$ squarefree, as these are a $\mathbb{Z}$-basis for the ring of integers.