Can you help me to understand a proof about equivalency of polynomials? The lemma says that if m and n are natural numbers and m not divisible by n and $\zeta_m^j \equiv \zeta_m^k \pmod n$, then $\zeta_m^j = \zeta_m^k$
$\zeta_m$ is here a primitive mth root of unity of the complex numbers so $e^{\frac{2 * \pi * i}{m}}$.
Now the proof is the following: We multiply the conguence by $\zeta_m^{-k}$ and for this reason may assume:
$\zeta_m^j \equiv 1 \pmod n$ and want to prove then that $\zeta_m^j = 1$ this part is no problem... Now:
Because $\prod\limits_{l=1}^{m-1}(x-\zeta_m^l)=(x^m-1)/(x-1)$ we get that  $\prod\limits_{l=1}^{m-1}(1-\zeta_m^l)=m$. 
Thus no factor in this last product is zero modulo n, which proves the result.
Now the first product equivalency $\prod\limits_{l=1}^{m-1}(x-\zeta_m^l)=(x^m-1)/(x-1)$ is something I don't get...
But $(x^m-1)/(x-1)$ looks a lot like the solution of $\sum\limits_{l=0}^{m-1}x^l$ (geometric series) and if I insert x=1 in it I indeed get the solution m;
So what I don't understand in particular:


*

*The equivalency $\prod\limits_{l=1}^{m-1}(x-\zeta_m^l)=(x^m-1)/(x-1)$... I can clearly see that the zeros of the function are identical but this isn't enough to prove that those polynomials are identical??

*Can I say $(x^m-1)/(x-1)=\prod\limits_{l=1}^{m-1}x^l$ and then insert x=1.... Because the formula $(x^m-1)/(x-1)$ is just for those $x \neq 1$... And then I insert x=1.... This is confusing

*"Thus no factor in this last product is zero modulo n, which proves the result." Can someone explain that sentence more detailed?
 A: Let me address your numbered list of questions:


*

*In general, because of the fundamental theorem of algebra, the zeroes of a polynomial (together with multiplicities of those zeroes) do indeed determine that polynomial up to scalar multiplication. To deal with the "up to scalar multiplication" part in your case, we note that both expressions are monic polynomials (i.e. coefficient of top-degree term is $1$) - hence they are equal polynomials.

*I think you wrote the product symbol ($\Pi$) when you meant the sum symbol ($\Sigma$). In any case, yes, you can write $(x^m - 1)/(x-1) = \sum_{l=1}^{m-1}x^l$ and then insert $x = 1$. This is exactly what the author of the proof intended you to do.

*For this conclusion, the author is referring to the equation $\prod_{l=1}^{m-1}(1-\zeta_m^l) = m$ and essentially saying that since $m$ is not divisible by $n$ (as was assumed at the beginning) it is therefore impossible that any of the factors $(1-\zeta_m^l)$ are divisible by $n$ on the left side of the equation. (Remember "divisible by $n$" means the same as "$\equiv 0$ mod $n$"). This then completes the proof because $(1- \zeta_m^l)$ not divisible by $n$ means $1-\zeta_m^l \not\equiv 0$ mod $n$ which means $1 \not\equiv \zeta_m^l$ mod $n$ and thus $\zeta_m^j \not\equiv \zeta_m^{l+j}$ mod $n$ for any $1 \leq j < m$.

