Equivalent Splitting Fields for two polynomials I'm working on the following question: 

Let $n$ and $m$ be relative prime, positive integers. Show that the splitting field in $\mathbb{C}$ of $x^{nm}-1$ over $\mathbb{Q}$ is the same as the splitting field in $\mathbb{C}$ of $(x^n - 1)(x^m-1)$ over $\mathbb{Q}$. 

I think I have a general idea about what to do. The zeros of these polynomials are the $mn$ roots of unity and the $n$ roots of unity and $m$ roots of unity. Since $n,m$ are relatively prime, I want to say these are the same, but I am unsure how to prove that. 
Then, in terms of the splitting fields, is it enough to say the splitting fields will be the same because the roots we must attach to $\mathbb{Q}$ are the same? 
I hope this outline makes sense, but I would like help filling in the details. 
 A: More precisely, you show this : any field containing all the roots of $x^{mn} - 1$ also contains all the roots of $(x^m - 1)(x^n - 1)$, and vice-versa. The roots of each polynomial are different, but it so happens that they are related to each other in terms of field operations : this will be sufficient to show that any field containing one set of roots must contain the other set of roots, and vice-versa.
For this, you realize that every root of $x^{mn} -1$ is a power of $\xi_{mn}$ ,with $\xi_{mn}$ denoting a primitive $mn$th root of unity. Similarly, any root of $(x^m-1)(x^n-1)$ is either a power of $\xi_n$ or a power of $\xi_m$.
Therefore, all you need to show is that $\mathbb Q[\xi_{mn}] = \mathbb Q[\xi_m,\xi_n]$. One side is obvious : $\xi_m,\xi_n$ are powers of $\xi_{mn}$. The other way must use the fact that $m,n$ are coprime. For example, take $\xi_n = e^{\frac{2 \pi i}{n}} , \xi_{m} = e^{\frac{2 \pi i}{m}}$ and see if you can get $\xi_{nm} = e^{2 \pi i \over nm}$ as a product of powers of $\xi_n,\xi_m$ so it will belong in any field containing these two. Hint : Use Bezout's lemma, and divide by $mn$.
