1
$\begingroup$

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 $\#L=n$

$\endgroup$
  • 3
    $\begingroup$ Possible duplicate of Degree of splitting field of $X^n-1$ over some finite field $\endgroup$ – Servaes Mar 3 at 16:21
  • $\begingroup$ @Servaes thank you for the reference, I hadn't found a similar question myself. I have used it to find my own solution to my problem. If I would post an edit, answer, or comment with a short proof, would you mind taking a look if it's correct? As there are some details in the link you posted that are slightly different/more general than in my case. I don't want to skip over important details in my proof $\endgroup$ – Marc Mar 3 at 17:41
  • $\begingroup$ Sure, I can take a look in a bit if you post an edit. $\endgroup$ – Servaes Mar 3 at 17:50
  • $\begingroup$ @Servaes I updated the post with an edit. Thanks again for taking your time to help me out! $\endgroup$ – Marc Mar 3 at 20:10
  • 1
    $\begingroup$ Your argument is good, but not quite complete, see my answer. And my pleasure! $\endgroup$ – Servaes Mar 3 at 21:04
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I think I can go from where you left off in your last argument! Thanks a bunch; as to clarify what I meant in my own edit: I'm actually doing what you are doing in at the beginning too. Instead of $r$ I just call it $d$. The reason I thought I might've needed to use two different integers is because that's what they did in the other post as well. However, maybe that was necessary for the generalization where the ground field was not $\mathbb{F}_p$ but rather a field of order $q$, a power of $p$. $\endgroup$ – Marc Mar 3 at 21:10
  • $\begingroup$ Actually, I can use your results but I can't quite figure out the following step: $\mathbb{F}_{p^s}^*$ contains a cyclic subgroup of order $n$, and hence the polynomial $X^n-1$ splits in this field. Why is this the case? How can we connect an arbitrary cyclic subgroup of order $n$ to the splitability of the polynomial? Also, am I correct to conclude that since $K$ is the splitting field of $X^n-1$, it is the minimal field for which the polynomial splits, and thus any other field in which the polynomial splits is larger than $K$? $\endgroup$ – Marc Mar 4 at 13:14
  • $\begingroup$ If $\Bbb{F}_p^{\times}$ contains a subgroup of order $n$, then its elements are $n$ distinct roots of $X^n-1$, hence the polynomial splits; the fact that the subgroup is cyclic is not even needed. $\endgroup$ – Servaes Mar 4 at 15:20
  • $\begingroup$ As for your second question; yes, because $K$ is the splitting field of $X^n-1$ over $\Bbb{F}_p$, by definition every field extension of $\Bbb{F}_p$ in which $X^n-1$ splits contains a subfield isomorphic to $K$. $\endgroup$ – Servaes Mar 4 at 15:28
  • $\begingroup$ But why can't there be a different subgroup of $n$ elements that does not contain all the distinct roots? If we do not yet know if $K\subset \mathbb{F}_{p^s}$ how do we know the latter contains any roots at all? $\endgroup$ – Marc Mar 4 at 18:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.