MHB The equation has exactly m different solutions

[mathmari]

Hey!

Let $n=p^rm$, where $p$ is a prime, $m \in \mathbb{N}, r \geq 0$ an integer and $(p,m)=1$.
I have to show that the equation $x^n=1$ has exactly $m$ different roots in the algebraic closure $\overline{\mathbb{Z}}_p$ of $\mathbb{Z}_p$.

I have done the following:

In $\mathbb{Z}_p$ it stands that $x^p=x$.

So, we have that
$$x^{p^r}=(x^{p})^{p^r-1}=x^{p^r-1}=(x^p)^{p^r-2}=x^{p^r-2}= \dots =x^p=x$$

That means that $x^n=1 \Rightarrow x^{p^rm}=1 \Rightarrow (x^{p^r})^m=1 \Rightarrow x^m=1$

Is this correct?? Do we conclude from that, that the equation $x^n=1$ has exactly $m$ different roots in the algebraic closure $\overline{\mathbb{Z}}_p$ of $\mathbb{Z}_p$ ?? (Wondering)

 

[johng1]

You're on the right track, but all you've shown is for $x\in \mathbb{Z}_p$, $x^m=1$.
So let S be the collection of all $p^rm$ roots of unity in the algebraic closure. Clearly S is finite (it contains at most $p^rm$ elements). Then let $F$ be the field obtained by adjoining S to $\mathbb{Z}_p$. $F$ is then a finite extension of the base field and so is finite with a power of $p$ elements, say $p^s$. Assume $x\in F$ satisfies $x^{p^rm}=1$. Let $t$ be the multiplicative order of $x$. Then $t$ divides the order of the multiplicative group of $F$, namely $p^s-1$. So $t$ divides the gcd of $(p^rm,p^s-1)$, a divisor of m since $p$ is prime $p^s-1$. Thus $x^m=1$. Conversely, it is clear that $x^m=1$ implies $x^{p^rm}=1$. Since $m$ is prime to $p$, the derivative of $x^m-1$ is not 0 and so has no multiple roots. Since we started in the algebraic closure, there are exactly $m$ roots of this equation.