Let $p$ be a prime number, $n\in\mathbb{N}$, with $p \nmid n$. Let $K$ be the splitting field of $x^n-1$ over $\mathbb{F}_p$. Show that the dimension of $K$, $[K:\mathbb{F}_p] = d$, is the smallest positive number such that $p^d = 1 \mod n$.
First I found that since $[K:\mathbb{F}_p] = d$, we have that $x^n-1$ has $d$ roots that are added to $\mathbb{F}_p$, so that $K=\mathbb{F}_p(\alpha_1, \alpha_2 \dots \alpha_d)$ for $\alpha_i$ the roots of the polynomial. I had no idea how to continue on this matter. I tried the Frobenius function, but as the binomials in $(x^n-1)^p$ are all divisible by $p$, we get that $(x^n-1)^p=(x^n)^p+(-1)^p$, and since $x^p=x \mod p$ and $p$ is prime and therefore odd, we have that $(x^n-1)^p = x^n-1$ which leaves nothing changed.
Furthermore, I know that in $K[x]$, we have $x^n-1 = (x-\alpha_1 )(x-\alpha_2)\dots(x-\alpha_d)$. The problem is that most of the examples and theorems I can use are of the cases where we have the fields $\mathbb{F}_q$ where $q=p^k$ for some prime $p$ and a natural number $k$. In this case, in particular, $p\nmid n$, so I'm not sure how to continue.
Edit: the linked post was very helpful, but did not have the exact same details as my problem. I'm not sure, but the link in the post seems like a generalization. I have tried to come up with a proof myself, using the details in the linked post. I'm still not sure of some steps, and in particular how to show that such $d$ is the smallest such that $p^d=1 \mod n$
Proof:
We have that $[K:\mathbb{F}_p]=d$ so that $\#K=p^d$ so $K=\mathbb{F}_q$ with $q=p^d$. Now: $[x^n-1]'=nx^{n-1}=0 \mod n$ so this polynomial does not contain double roots; hence there are $n$ distinct roots $\alpha_i\in K$. All of these roots are invertible, so we can check that the multiplicative group of roots $L$ in $\mathbb{F}_q^*$ is a subgroup. By what we had from above it follows also that $\#L=n$, and by Lagrange's theorem, we have $\#L\mid\#\mathbb{F}_q^*$ so $n\mid q-1$. We can write that for some integer $k$ we have $q-1=p^d-1=kn \implies p^d=1+kn=1\mod n$.
I'm unsure about the first step, to say that $\#K=p^d$ and not a general $p^r$ for some $r\in\mathbb{N}$. I also haven't shown how this $d$ is the smallest number for which the equality holds. In the linked post they used elements of $L$ and their orders to prove for instance that there is a generating root $\alpha\in L$ with $ord(\alpha) = n$ and $L=<a>$. In this case I didn't know how to use that, as it would just impose an extra step which felt unnecessary since we already had that $\#L=n$
Best Answer
I do not understand your argument for $x^n-1$ not having double roots; you say this is because the derivative is congruent to $0$ modulo $n$, but I would say it is because the derivative is not $0$ modulo $p$, because $p\nmid n$.
Also, the structure of your argument is a bit unclear to me. At the very beginning, what is $d$? Is it the $\Bbb{F}_p$-dimension of the splitting field $K$, or is it the smallest positive integer such that $n\mid p^d-1$?
I would start off by saying that $[K:\Bbb{F}_p]=r$ for some positive integer $r$, and then proving that $r$ is the smallest positive integer such that $n\mid p^r-1$. Your proof already shows that $n\mid p^r-1$, but not yet that $r$ is minimal. To prove this, note that if $n\mid p^s-1$ for some positive integer $s$, then $\Bbb{F}_{p^s}^{\times}$ contains a cyclic subgroup of order $n$, and hence the polynomial $X^n-1$ splits in the field $\Bbb{F}_{p^s}$, so $r\mid s$. I leave the details to you.