Homomorphism from integral module generated by roots of unity to cyclic group? Let $S$ be the set of all roots of unity. Consider the $\mathbb{Z}$-module,  $\mathbb{Z} S$, as an additive abelian group (that is, $\mathbb{Z} S$ is the subset of complex numbers that can be expressed as a finite integral combination of roots of unity). The question is: Given an integer $k > 2$, is there is a group homomorphism $\phi: \mathbb{Z} S \to \mathbb{Z}_k$ such that for any element $s \in S$, $s \not\in Ker(\phi)$? 
I suspect the answer is no for any such $k$, but it is not clear why.
Additionally, if $S$ is restricted to $2pq$-th roots of unity, I suspect that such a homomorphism does indeed exist. Is there a clear reason why?
 A: I believe that the answer to your question is actually yes.  I'll only sketch the idea, leaving the details to you.
First,  I will use slightly different notation.  By $\mathbb{Z}S$ I will mean the group ring over $S$ (so these are formal $\mathbb{Z}$-linear combinations of elements in the group $S$), and I'll let $R$ denote the subset of $\mathbb{C}$ generated by finite integral combinations of roots of unity.
Consider the "obvious" map $\psi:\mathbb{Z}S\to R$.  Its kernel consists of those formal sums which, when thought of as elements of $\mathbb{C}$ (rather than as formal sums), are zero.  Schoenberg in 1964 ("A note on the cyclotomic polynomials") apparently showed that elements of the kernel are linear combinations of the trivial relations
$$
1+\zeta_p+\zeta_p^2+\cdots + \zeta_p^{p-1}=0
$$
when $p$ is a prime, and their rotations.  (By a rotation, we mean multiplying each term of the relation by a fixed root of unity.)
Thus, your map $\phi$ should exist as long as you can send each root of unity to a non-zero element of $\mathbb{Z}_k$ and yet the trivial relations (and their rotations) are still zero.
So, consider the special case when $k=3$.  We can set $\phi(\zeta_1)=1$.  The trivial relation $1+\zeta_2=0$ forces $\phi(\zeta_2)=-1$.  The next trivial relation $1+\zeta_3+\zeta_3^2=0$ forces $\phi(\zeta_3)=\phi(\zeta_3^2)=1$.  The next trivial relation $1+\zeta_5+\zeta_5^2+\zeta_5^3+\zeta_5^4=0$ allows us to take $\phi(\zeta_5)=\phi(\zeta_5^2)=\phi(\zeta_5^3)=1$ and $\phi(\zeta_5^4)=2$ (among other options).  Continuing in this way, we can define $\phi(\zeta_p^n)$, and the trivial relations are forced to be zero.
However, we still need to guarantee that the rotations are zero, and extend this map to other roots of unity not indexed by primes.
First, one should extend $\phi$ to roots of unity indexed by square-free numbers.  For instance, what about $\zeta_6$?  It's value is forced (from the choices above) from the rotation $\zeta_3^2+\zeta_6=\zeta_3^2(1+\zeta_2)=0$.  I'll leave it to you to decide how to continue the process from here for other square-free numbers.
What about $\zeta_4$?  Here, it doesn't arise from a "square-free" rotation.  So we should be able to take $\phi(\zeta_4)=1$.  After that, there are many rotational relations involving $\zeta_4$, for instance $\zeta_4+\zeta_4\zeta_2=0$, $\zeta_4+\zeta_4\zeta_3+\zeta_4\zeta_3^2=0$, etc...  Just as above we can extend $\phi$ to have values on these other rotated roots of unity.
