# Splitting field of $x^{n}-1$ over $\mathbb{Q}$

From I.N.Herstein's Topics in Algebra. Chap 5 Sec 5.3 Page 227 Problem 8

• Problem 8: If $n>1$ prove that the splitting field of $x^{n}-1$ over the field of rational numbers is of degree $\Phi(n)$ where $\Phi$ is the Euler $\Phi$-function. ( This is a well known theorem. I know of no easy solution, so don't be disappointed if you fail to get it. If you get an easy proof, I would like to see it.)

First, I would like to see a proof of this result. Next, I think I have seen this proof in Dummit and Foote's Abstract Algebra book, but not sure. Anyway, next question is: Has an easy solution been found to this problem? If not, I would like to know what efforts have been taken to make the proof more simple. And why does Herstein think an easy solution can exist.

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It's in section 13.6 in Dummit & Foote. –  Bruno Stonek Feb 27 '11 at 19:12
@Bruno: Yeah i got it. –  anonymous Feb 28 '11 at 4:50

I think the difficulty is proving that the $n$th cyclotomic polynomial is irreducible. Wikipedia says it's a non-trivial result. This gives a factorization of $x^n-1$ as the product of all cyclotomic polynomials $\Phi_d$ for $d$ dividing $n$.
Over $\mathbb{Q}$ every splitting field is a Galois extension. let $w$ be a primitive n-th root of unity. It is relatively easier to prove $Gal(\mathbb{Q}(w)/\mathbb{Q})$ is isomorphic to $U(Z/nZ)$ whose order is $\phi(n)$. For a galois extension $K$ of $F$, $[K:F]=Gal(K/F)]$. Is it not easier than the traditional answer? Or are both of these the same?
Not sure about this. To prove $Gal(\mathbb{Q}(w)/\mathbb{Q}) \cong U(\mathbb{Z}/n\mathbb{Z})$ don't you end up needing to prove $\Phi(n)$ is irreducible anyway? Without this result I think the best you can say is that $Gal(\mathbb{Q}(w)/\mathbb{Q}) \subset U(\mathbb{Z}/n\mathbb{Z})$. –  Ben Blum-Smith Sep 13 '11 at 16:13
let $G=Gal(\mathbb{Q}(w)/\mathbb{Q})$ and $\sigma$ is in $G$. now $\sigma$ is determined by $\sigma(w)$. And $\sigma(w)=w^{i}$ where i is allowed only to be relatively prime to $n$. Hence There are only $\phi(n)$ such $\sigma$ possible in $G$. Is there already a nice bijection which is also a homomorphism? –  Dinesh Sep 13 '11 at 18:11
@Ben Blum-Smith The homomorphism is $\sigma$ which sends $w$ to $w^i$ is sent to to the residue class of $i$. –  Dinesh Sep 13 '11 at 20:47
@Dinesh, but the difficulty is in showing that for every $i$, the map $w \mapsto w^i$ is an element of the Galois group. Of course the other direction is easy, i.e. that every element of the Galois group must be of this form. –  Bruno Joyal Sep 16 '11 at 19:53
@Bruno agreed, in fact the claim that $w\mapsto w^i$ extends to an element of the Galois group for all $i$ relatively prime to $n$ is equivalent to the claim that $\Phi(n)$ is irreducible. (If $\Phi(n)$ factors nontrivially, then the various $w^i$'s with $i$ relatively prime to $n$ would sort into more than one orbit under action of the Galois group, and conversely.) –  Ben Blum-Smith Jan 10 '12 at 1:57