Let $k$ be a finite field of order $q$ in characteristic $p$, let $n$ be a positive integer not divisible by $p$, and let $K$ be the splitting field of $X^n-1$ over $k$. Prove that $[K:k]$ equals the smallest positive integer $d$ such that $n\mid q^d-1$.

My approach: It's easy to show that if $k$ is a finite field of order $q$ with $\text{char} \ k=p$ then $k=\mathbb{F}_{q}$ with $q=p^m$ with $m\geq 1$. If $K$ is the splitting field of $X^n-1$ over $k$ then $K=\mathbb{F}_{q^r}$ for some suitable $r\geq 1$.

Let $\alpha \in K=\mathbb{F}_{q^r}$ be a root of $X^n-1$ then $\alpha^n-1=0$ and $\alpha\neq 0$.

Thus, $\alpha^n=1$ and $\alpha \in \mathbb{F}^{\times}_{q^r}$ where $\mathbb{F}^{\times}_{q^r}$ is the cyclic group of order $q^r-1$.

But I have some difficulties:

1) From $\alpha^n=1$ it follows that $o(\alpha)\mid n$. How to show that $o(\alpha)=n$? I was trying to show it using that $(n,p)=1$ but I failed.

Remark: Because if we can show that $o(\alpha)=n$ then the desired result follows immediately from cyclicity of multiplicative group of field $\mathbb{F}_{q^r}$.

I have spent one day in order to come up with this approach. So please do not duplicate this question and can anyone show how to answer my question, please.

Detailed explanation would be great!


It is not true that $o(\alpha)=n$ for every root $\alpha\in K$ of $X^n-1$; after all $1\in K$ is also a root of $X^n-1$. But because $K$ is the splitting field of $X^n-1$ there exists a root $\alpha\in K$ of $X^n-1$ with $o(\alpha)=n$:

Because $\gcd(n,p)=1$ the polynomial $X^n-1$ is separable over $k$, meaning that it has precisely $n$ roots in the splitting field $K$. These roots form a subgroup of $K^{\times}$ of order $n$, and because $K^{\times}$ is cyclic so is this subgroup. Then for any generator $\alpha$ of this subgroup you have $o(\alpha)=n$.

  • $\begingroup$ "But because $K$ is the splitting field of $X^n-1$ there exists $\alpha\in K$ with $o(\alpha)=n$." Here $\alpha$ is a root of $X^n-1$, right? $\endgroup$ – ZFR Mar 2 at 17:21
  • $\begingroup$ Yes indeed. I have edited my answer to use less machinery and be more explicit. $\endgroup$ – Servaes Mar 2 at 17:22
  • $\begingroup$ I think that I've got you: 1) Since $f=X^n-1$ and $f'=nX^{n-1}$ then $(f,f')=1$. Here we are using that $(n,p)=1$ because in this case $f'$ is not zero. If the gcd $\neq 1$ then some $X-\beta$ divides $f$ and $f'$ and hence $\beta^n=1$ and $n\beta^{n-1}=0$ and combining these two results we get: $n=0$ which is contradiction because $n\geq 1$. Thus the polynomial $f(X)=X^n-1$ is separable over $k$, i.e. in $K$ it has $n$ distinct roots $\alpha_1,\dots,\alpha_n$ and we can check that they form subgroup in $K^{\times}$ $\endgroup$ – ZFR Mar 2 at 17:40
  • $\begingroup$ and being subgroup of cyclic is itself cyclic. Thus we can take an element of order $n$ and proceed above as I said in the topic. Right? $\endgroup$ – ZFR Mar 2 at 17:40
  • $\begingroup$ Yes, that's right! $\endgroup$ – Servaes Mar 2 at 18:07

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