# Do there exist non-trivial integer coefficients that break linear independence of the roots of unity?

Let $$n$$ be a positive integer, $$k=0,\cdots,n-1$$, $$\omega_k=e^{\tfrac{2\pi i}{n}k}$$ be the roots of unity, $$c_k \in \mathcal{Z}$$ be integer coefficients, trivial and non-trivial be two subcategories for $$c_k$$ defined as follows: Trivial is when $$c_k$$ contains equal integers that are evenly spaced under modulo addition of a composite divisor of $$n$$, and non-trivial $$c_k$$ is when evenly spaced integers under modulo addition of a composite divisor of $$n$$ are not equal. An example of a trivial case for $$n=6$$ is $$(2)\omega_0+(2)\omega_1+(1)\omega_2+(2)\omega_3+(2)\omega_4+(1)\omega_5=0$$ because $$(2)\omega_0+(2)\omega_3=0$$, and $$(2)\omega_1+(2)\omega_4=0$$, and $$(1)\omega_2+(1)\omega_5=0$$. Then regarding the linear independence of $$\sum_k c_k \omega_k = 0$$ consider the following: Does their exist a non-trivial case?

Note: Evenly spaced under modulo addition of a composite divisor of $$n$$ means the following: Let $$m_p$$ be a composite divisors of $$n$$ such that $$n=m_0 m_1 \cdots$$, then the trivial case of $$c_k$$ satisfies $$c_k=c_j$$ for $$(k-j) \mod m_p =0$$.

• This is somewhat poorly worded. I think that what you want to ask is whether given integers $a_0,\ldots,a_{n-1}$ such that $\sum a_i\omega_i = 0$, must is follow that the nonzero $a_i$ are all equal. Correct? – Arturo Magidin Apr 16 at 18:56
• @ArturoMagidin The case when all $a_i$ are equal is not the most general case in the trivial case mentioned in the OP. This is because $n$ can be a composite number. – linuxfreebird Apr 16 at 20:13
• I'm trying to understand what you wrote; "that area evenly spaced" is a typo, but even taking that into account I'm not sure what you mean. I don't know what 'evenly spaced under modulo addition' means.If my attempt at understanding it is incorrect, fine, but just telling me I'm wrong doesn't clarify what you are trying to say. – Arturo Magidin Apr 16 at 20:34
• @ArturoMagidin. Thank you for pointing out my poor wording and the typo. I have corrected the typo and provide clarification as a note in the OP. – linuxfreebird Apr 16 at 22:00

## 1 Answer

I am not sure I understood your definition correctly but I believe the fact that the $$n$$th cyclotomic field extension of $$Q$$ has degree $$\phi(n)$$ says that there are no non trivial combinations.

• Thanks for the answer and the reference. May I ask: how can I improve the wording of my question to address your confusion? – linuxfreebird Apr 16 at 18:52
• @linuxfreebird sure, use more maths in your definitions and less words. This is not a general rule but it would help in this case. Could you make sense of my answer? Do you think it answers your original question? – Bananach Apr 17 at 5:49