How can we show the other direction?

[mathmari]

Hello!

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 $ :

$n\mid m \Rightarrow n=km, k \in \mathbb{Z}$

$t^n-1=t^{km}-1=(t^m)^k-1=(t^m-1)(t^{m(k-1)}+\dots +1)$

So, $t^n-1\mid t^m-1$.

Is this correct?

How could we show the other direction?

 

[Euge]

Hi mathmari,

Your work is correct so far. To prove the converse, use the division algorithm to express $m = nq + r$, where $q$ and $r$ are integers with $0 \le r < n$. Since $t^n-1\, |\, t^m - 1$, then $t^m \equiv 1\pmod{t^n - 1}$. Also $t^n \equiv 1\pmod{t^n -1}$ (as $t^n - 1\, |\, t^n - 1$). So $t^m \equiv t^{nq + r}\pmod{t^n - 1}$ $\implies$ $1 \equiv t^r \pmod{t^n - 1}$. Therefore $t^n - 1\, |\, t^r - 1$. If $r \neq 0$, then the latter condition implies $n \le r$, contradicting the inequality $r < n$. So $r = 0$, which gives $m = nq$. Consequently, $n\, |\, m$.

 

[mathmari]

Your work is correct so far. To prove the converse, use the division algorithm to express $m = nq + r$, where $q$ and $r$ are integers with $0 \le r < n$. Since $t^n-1\, |\, t^m - 1$, then $t^m \equiv 1\pmod{t^n - 1}$. Also $t^n \equiv 1\pmod{t^n -1}$ (as $t^n - 1\, |\, t^n - 1$). So $t^m \equiv t^{nq + r}\pmod{t^n - 1}$ $\implies$ $1 \equiv t^r \pmod{t^n - 1}$. Therefore $t^n - 1\, |\, t^r - 1$. If $r \neq 0$, then the latter condition implies $n \le r$, contradicting the inequality $r < n$. So $r = 0$, which gives $m = nq$. Consequently, $n\, |\, m$.

I see... Thank you very much! (Mmm)

 

[Opalg]

Hello!

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 $ :

$n\mid m \Rightarrow n=km, k \in \mathbb{Z}$

$t^n-1=t^{km}-1=(t^m)^k-1=(t^m-1)(t^{m(k-1)}+\dots +1)$

So, $t^n-1\mid t^m-1$.

Is this correct?

That argument is basically correct. But notice that you have $m$ and $n$ the wrong way round. If $n$ divides $m$ then $m=nk$, not $n=mk$.

 

[mathmari]

That argument is basically correct. But notice that you have $m$ and $n$ the wrong way round. If $n$ divides $m$ then $m=nk$, not $n=mk$.

Oh, you're right! Thank you! (Mmm)