MHB Is the Lemma True for All Fields?

[mathmari]

Hey!

I want to show 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 $F[t, t^{-1}]$ ($F[t,t^{-1}]$: the polynomials in $t$ and $t^{-1}$ with coefficients in the field $F$) if and only if $(\exists s \in \mathbb{Z}) m=np^s$. I have done the following:

$\Leftarrow $ :

$$\frac{t^m-1}{t^n-1} \overset{ m=np^s }{ = } \frac{t^{np^s}-1}{t^n-1}=\frac{(t^n)^{p^s}-1}{t^n-1}=\frac{(t^n-1)((t^n)^{p^s-1}+\dots +1)}{t^n-1}=(t^n)^{p^s-1}+\dots +1$$

Is this correct so far?

How can we continue to conclude that $(t^m-1)/(t^n-1)$ is a square ? $\Rightarrow $ :

$$\frac{t^m-1}{t^n-1}=a^2 \Rightarrow t^m-1=a^2(t^n-1)$$

How can we continue?

 

[PaulRS]

Hi mathmari! Yes it is correct, but that expansion is not that useful because we seek a multiplicative property. Also, the fact that $t^{-1}$ can be used won't ever be of great help (try to justify why), and thus we shall assume that $a$ is a polynomial in $t$.

Converse.

Try a few small cases like $n=s=1$ with $p=3$ and $p=5$. You should see a little pattern emerge: $$ \frac{t^m - 1}{t^n - 1} = \left(t^n - 1\right)^{p^s - 1}\,,$$
and this would do it, since $p$ is odd.

To prove it I recommend you recall a very particular property that is true finite fields and that is in general also true for fields of finite characteristic (in this case you are working inside of $F(t)$ which has characteristic $p>2$), namely that $(\alpha+\beta)^p = \alpha^p+\beta^p$ for all $\alpha,\beta$ in the field.

Direct.

If $m = p\, m^\prime$ for some $m^\prime$ we have $$t^m - 1 = \left(t^{m^\prime} - 1\right)^p\,.$$ Keep applying until we get $m = p^l \, \tilde{m}$ where $p\not| \tilde{m}$ and we have $$ t^m - 1 =\left(t^{\tilde{m}} - 1\right)^{p^l}\,.$$ Here observe that $t^{\tilde{m}} - 1$ is free of squares (why?), that is, it has no repeated roots over any extension of $F$.

Now, from
$$a^2 \,\left(t^n - 1\right) = \left(t^{\tilde{m}} - 1\right)^{p^l}\,,$$
we do the same on the LHS getting $\tilde{n}$ such that $p \not| \tilde{n}$ and $n = p^{r}\, \tilde{n}$. Then
$$a^2 \,\left(t^{\tilde{n}} - 1\right)^{p^{r}} = \left(t^{\tilde{m}} - 1\right)^{p^l}\,.$$
Here we claim $\tilde{n}=\tilde{m}$, which would prove the result, since we must have $r \leq t$ by comparing degrees (and your $s$ would be $l-r$).

To prove it, we show that the roots (over some extension field of $F$) of $t^{\tilde{m}}-1$ and $t^{\tilde{n}}-1$ coincide, and since both have simple roots we get the equality of the polynomials and thus of $\tilde{n}$ and $\tilde{m}$. If $t^{\tilde{m}}-1$ had a root (in some extension field of $F$) that is not present in $t^{\tilde{n}}-1$, then it can only appear on the LHS of $a^2 \,\left(t^{\tilde{n}} - 1\right)^{p^{r}} = \left(t^{\tilde{m}} - 1\right)^{p^l}$ in $a^2$... with even multiplicity, while it appears on the RHS with odd multiplicity (an absurd). Then, of course all roots of $t^{\tilde{n}}-1$ must be roots of $t^{\tilde{m}}-1$ due to $a^2 \,\left(t^{\tilde{n}} - 1\right)^{p^{r}} = \left(t^{\tilde{m}} - 1\right)^{p^l}$.

 

[mathmari]

we shall assume that $a$ is a polynomial in $t$.

Why do we assume that $a$ is a polynomial in $t$ and not a polynomial in $t$ and $t^{-1}$ ?

Direct.

If $m = p\, m^\prime$ for some $m^\prime$ we have $$t^m - 1 = \left(t^{m^\prime} - 1\right)^p\,.$$ Keep applying until we get $m = p^l \, \tilde{m}$ where $p\not| \tilde{m}$ and we have $$ t^m - 1 =\left(t^{\tilde{m}} - 1\right)^{p^l}\,.$$ Here observe that $t^{\tilde{m}} - 1$ is free of squares (why?), that is, it has no repeated roots over any extension of $F$.

Why is $t^{\tilde{m}} - 1$ free of squares?

To prove it, we show that the roots (over some extension field of $F$) of $t^{\tilde{m}}-1$ and $t^{\tilde{n}}-1$ coincide, and since both have simple roots we get the equality of the polynomials and thus of $\tilde{n}$ and $\tilde{m}$.

When we have shown that the set of roots of $t^{\tilde{m}}-1$ coincide with the set of roots of $t^{\tilde{n}}-1$, why do we conclude that $\tilde{m}=\tilde{n}$ ? Having shown that $\tilde{m}=\tilde{n}$ we have the following:

$$m=\tilde{m}p^l \ \ , \ \ n=\tilde{n}p^r \Rightarrow n=\tilde{m}p^r \Rightarrow \tilde{m}=np^{-r} \\ \Rightarrow m=np^{-r}p^l \Rightarrow m=np^{l-r}$$

So, $\exists s=l-r \in \mathbb{Z}$ such that $m=np^s$.

Is this correct?

 

[mathmari]

I thought about it again... Can we say the following?

Let $a=\frac{b(t)}{t^k}$, where $b(t)\in F[t]$ such that $\frac{t^m-1}{t^n-1}=a^2 \Rightarrow \frac{t^m-1}{t^n-1}=\frac{b^2(t)}{t^{2k}} \Rightarrow t^{2k}(t^m-1)=b^2(t)(t^n-1)$.

Let $m=\tilde{m} p^l $, where $p \nmid \tilde{m}$. Then $t^m-1=t^{\tilde{m}p^l}-1=(t^{\tilde{m}}-1)^{p^l} \Rightarrow t^{2k}(t^{\tilde{m}}-1)^{p^l}=b^2(t)(t^n-1) \ \ (*)$.

Let $n=\tilde{n} p^r$, where $p \nmid \tilde{n}$. Then $t^n-1=t^{\tilde{n}p^r}-1=(t^{\tilde{n}}-1)^{p^r}$.

Then $(*) \Rightarrow t^{2k}(t^{\tilde{m}}-1)^{p^l}=b^2(t)(t^{\tilde{n}}-1)^{p^r} \ \ (**)$

We assume that $t^{\tilde{m}}-1$ has a nonzero root, let $u$, in an extension of $F$ that is not a root of $t^{\tilde{n}}-1$.

Then $0=u^{2k}(u^{\tilde{m}}-1)^{p^l}=b^2(u)(u^{\tilde{n}}-1)^{p^r} \Rightarrow b^2(u)=0$.
So, $u$ is a root of $b^2$ with even multiplicity.
In the left side of the equation $(**)$, $u$ is a root of $(t^{\tilde{m}}-1)^{p^l}$, so it is a root of odd multiplicity, since $p^l$ is odd.

Is this correct?

So, $t^{m}-1$ and $t^{n}-1$ have the same roots.
Since the number of roots is equal to the degree of an equation, we have that $\tilde{n}=\tilde{m}$.

Is this right?

 

[mathmari]

I have a question about the lemma above...

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 $\frac{t^m-1}{t^n-1}=a^2=1$ ?