I have a question about the following lemma:
Assume that the characteristic of $F$ is $p$ and $p>2$. Then $(t^m-1)/(t^n-1)$ is a square in $F[t, t^{-1}]$ if and only if $(\exists s \in \mathbb{Z}) m=np^s$.
($F[t,t^{-1}]$: the polynomials in $t$ and $t^{-1}$ with coefficients in the field $F$)
We assume that the characteristic of $F$ is $p>2$. Why can the characteristic not be $p=2$. Is it because then we would have $\frac{t^m-1}{t^n-1}=a^2=1$ ?
Can we modify a little the lemma so that it stands for characteristic $p=2$ ?
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EDIT1:
Can we say something about $p=2$ ?
$$(\exists s \in \mathbb{Z})m=2^sn \Leftrightarrow \dots$$
If $(\exists s \in \mathbb{Z})m=2^sn$ then we have that:
$$t^m=t^{2^sn}=\left (t^n\right )^{2^s} \Rightarrow t^m-1=\left (t^n\right )^{2^s}-1=\left (t^n-1\right )^{2^s}$$
Can we write with that a $\Leftrightarrow$ relation?
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EDIT2:
Using te lemma:
$t^n-1$ divides $t^m-1$ in $F[t, t^{-1}]$ ($F[t,t^{-1}]$: the polynomials in $t$ and $t^{-1}$ with coefficients in the field $F$) if and only if $n$ divides $m$ in $\mathbb{Z}$.
we have the following:
If $\exists s \in \mathbb{Z}$ so that $m=2^s n$, then $2 \mid m$ and $n \mid m$.
We have the following:
$$2 \mid m \Leftrightarrow t^2-1 \mid t^m-1$$
and $$n \mid m \Leftrightarrow t^n-1 \mid t^m-1$$
Is this an "iff" statement?
If $F$ is the $2$-element field, one has $\frac{t^2-1}{t-1}=t + 1$, but $1+t$ is not a square in $F[t, t^{-1}]$.