How would I prove that Qm∩Qn=Q(m,n) (here Qn denotes the nth cyclotomic field)?

I already know of a solution involving the fact that given two normal extension fields M,L of some field K contained in some common extension, then Gal(ML/L)≅Gal(M/M∩L), but does there exist a solution that doesn’t require such a theorem?

**Answer**

I included this exercise on an article I wrote (which I presume is where the question poser got this problem from if he’s the same bzprules on Art of Problem Solving) for a reason: nothing advanced is needed. Just some tricky ideas.

Denote \displaystyle \omega_n = \text{exp} \left ( \frac{2 \pi i}{n} \right). Denote \ell = \text{lcm}[m,n] and d = \gcd(m,n).

Let F = \mathbb{Q}[\omega_n] \cap \mathbb{Q}[\omega_m]. Consider the automorphisms over \mathbb{Q}[\omega_\ell] which fix \mathbb{Q}[\omega_n]. It is clear they are defined by f_k : \omega_{\ell} \to \omega_{\ell}^k where \gcd(k,\ell) = 1 and k \equiv 1 \pmod{n}. Now for the tricky part of the proof : note that these are also automorphisms for \mathbb{Q}[\omega_m] which fix F! It is clear that they fix F due to k \equiv 1 \pmod{n} (or more simply, its a subfield of \mathbb{Q}[\omega_n]). To see that they also are isomorphisms on \mathbb{Q}[\omega_m], just note its effectively exponentiating \omega_m by k for \gcd(k,m) = 1 so of course it works. It follows \frac{\phi(\ell)}{\phi(n)} = \frac{\phi(m)}{\phi(d)} \le [\mathbb{Q}[\omega_m]:F]

due to the fact there are at least \displaystyle \frac{\phi(\ell)}{\phi(n)} automorphisms. Thus [F : \mathbb{Q}] \le \phi(d) \implies F = \mathbb{Q}[\omega_d] because that [\mathbb{Q}[\omega_d] : \mathbb{Q}] = \phi(d) and \mathbb{Q}[\omega_d] \subset F.

The motivation behind this proof is that we need to bound [F : \mathbb{Q}] from above, so it is natural to consider [K : F] for some field K. As cyclotomic fields have nice Galois groups, we can often bound the dimension by finding automorphisms.

**Attribution***Source : Link , Question Author : bzprules , Answer Author : dinoboy*