I want to prove the following lemma:
$t^n-1$ divides $t^m-1$ in $F[t, t^{-1}]$ if and only if $n$ divides $m$ in $\mathbb{Z}$.
I have done the following:
$\Leftarrow $ :
$n\mid m \Rightarrow m=kn, k \in \mathbb{Z}$
$t^m-1=t^{kn}-1=(t^n)^k-1=(t^n-1)(t^{n(k-1)}+\dots +1)$
So, $t^n-1\mid t^m-1$.
Is this correct?
How could we show the other direction?
If $t^n-1\mid t^m-1$ there is $f(t)\in F[t,t^{-1}]$ such that $(t^n-1)f(t)=t^m-1$. Moreover, there is $p\ge 0$ such that $t^pf(t)\in F[t]$. Set $g(t)=t^pf(t)$. We then have $(t^n-1)g(t)=t^p(t^m-1)$. Without loss of generality we may assume $m,n\ge 0$. Then $t^n-1\mid t^p(t^m-1)$ in $F[t]$, and since $\gcd(t^n-1,t^p)=1$ we have $t^n-1\mid t^m-1$ in $F[t]$. Now write $m=nq+r$ with $0\le r<n$. Then $ t^m-1=t^r(t^{nq}-1)+t^r-1$, so $t^n-1\mid t^r-1$ which implies $r=0$, that is, $n\mid m$.
Your proof is correct. To prove the other direction suppose $(t^n-1)P(t) = t^m-1$ (think about why $P(t) \in F[t,t^{-1}]$ has to be a polynomial in $t$). The quick way now would be to look at this equation modulo $t^n$ and deduce that $P(t)\equiv 1$ modulo $t^n$ and hence $P$ is a polynomial in $t^n$.
If you don't feel comfortable with this sort of argument write $P(t) = \sum_{k=0}^{m-n} a_k t^k$. By multiplying out we know that $a_{m-n}=1$ and $a_0=1$. We have $$ (t^n-1)P(t) = \sum_{k=0}^{m}b_kt^k =t^m-1, $$ where $$b_k= \begin{cases} a_{k-n} &\text{if }k>m-n,\\ a_{k-n}-a_k &\text{if } n\leq k\leq m-n,\\ -a_k &\text{if } k<n \end{cases} $$ From that you can deduce $a_0 = 1$ and then by induction that $a_k=0$ if $n\nmid k$ and $a_k=1$ if $n\mid k$ and $k\leq m-n$. Since the coefficient of $t^{m-n}$ is $1$ we must have $n\mid m-n$ which implies $n\mid m$.
Let's do the difficult direction.
Let $\zeta\in\mathbb C$ be a primitive $n$th root of unity.
If $f=t^n-1$ divides $g=t^m-1$, then $\zeta$ must be a root of $g$ because it is a root of $f$. From this and the basic property of primitive roots we see at once that $n\mid m$.
NB. Suppose that $F$ is any field and that $f$ divides $g$ in $F[t]$. If $\bar F$ is the algebraic closure of $F$, then $f$ divides $g$ in $\bar F[t]$, so we may just as well suppose that $F$ is algebraically closed. If $n=p^an'$ with $(n',p)=1$ and $\alpha\geq0$ and $f_0=t^{n'}-1$, then $f_0$ divides $f$ so it divides $g$. We may thus suppose also that $(n,p)=1$. In that case, there is a primitive $n$th root of unity $\zeta\in F$ and we do exactly the same as before.
