Minimal number of generators of the monoid generated by roots of the unity.

Question

This question is linked to Unicity of decomposition for the monoid generated by roots of the unity.

We set for $n \in \mathbb{N}^*$ $M_n$ the monoid generated by $n$-th roots of the unity and $M_n^\times$ the monoid generated by primitive $n$-th roots of the unity. They are respectfully $$M_n = \left\{ \sum_{\lambda \in U_n} k_\lambda \lambda, \, k_\lambda \in \mathbb{N} \right\}\\ M_n^\times = \left\{ \sum_{\lambda \in U_n^\times} k_\lambda \lambda, \, k_\lambda \in \mathbb{N} \right\}$$ Where $U_n$ is the set of $n$-th roots of the unity and $U_n^\times$ the set of primitive $n$-th roots of the unity.

My question is what is the minimal number of generators for these monoïds?

I already know that for $n = p_1 \cdots p_m$ where the $p_i$ are distinct primes and $m$ is even, we can remove $1$ from the list of the generators in the non-primitive case since $$\sum_{\lambda \in U_n^\times} \lambda = \mu(n) = 1$$ with $\mu$ the Möbius function.

And for $M_n$ where $n$ has at least two distinct primes in its decomposition, we need to remove $1$ as well.

The cases $n=1,2,3,4$ are easy to prove by reasoning geometrically:

• for $n=1$ it is direct since $M_1 = M_1^\times = \mathbb{N}$.
• for $n=2$ it is direct since $M_2 = \mathbb{Z}$ and $M_2^\times = -\mathbb{N}$
• for $n= 3$, if we keep only two generators we only have points contained in a half-plane which is absurd so three is the minimum for $M_3$, and since $M_3^\times = \mathbb{N}j + \mathbb{N} j^2$ is not contained in a line, we need these two generators.
• for $n = 4$ which is our first square example, we can easily see that we need all four of them.
  Indeed $M_4 = \mathbb Z [i]$ which elements are uniquely determined by their real and imaginary parts, so as an abelian group it needs two non-zero generators which are $1$ and $i$ (or their opposite). And since it only has a monoid structure we need their opposites.
  Now $M_4^\times = \mathbb{Z}i$ which obviously needs its two generators $i$ and $-i$.

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My thoughts on this:

• It must be true for prime $n$.
• It might be true when $n$ is a product of two primes and we remove $1$.
• I have no clue if $n-1$ is the right number for $M_n$ when $n$ is a product of an even number of distinct primes.
• It is more complicated when some $a^2 \mid n$

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Useful Information

For $n \leq 6$ and $n \neq 5$ $Z[e^{\frac{2i\pi}{n}}]$ is euclidian. It could be used to deal with these cases.

Answer 1

I have edited as I got a stronger result:

Note that $-1\in M_n$ for all $n$, since

$$\zeta_n^{n-1}+\zeta_n^{n-2}+...+1=0$$ $$\implies \zeta_n^{n-1}+\zeta_n^{n-2}+...+\zeta_n=-1\in M_n$$

Then multiplying this by $\zeta_n^k$, we find $-\zeta_n^k\in M_n$ for all $k$. Thus, $M_n$ is actually a group, and it is just equal to $\mathbb{Z}[\zeta_n]$ for all $n$. We know that the number of elements needed to generate it as a group is $\phi(n)$, but the number of generators needed as a monoid may be larger a priori. The minimum number of generators of $M_n$ as a monoid is certainly at most $2\phi(n)$, since we can take all primitive $n$th roots and their negatives. It is also at most $n$ as all roots of unity would give a generating set too. Thus, the number is between $\phi(n)$ and $\min(n,2\phi(n))$.