# Intersection of Cyclotomic Fields

How would I prove that $$Qm∩Qn=Q(m,n)\mathbb{Q}_m \cap \mathbb{Q}_n = \mathbb{Q}_{(m, n)}$$ (here $$Qn\mathbb{Q}_n$$ denotes the $$nn$$th cyclotomic field)?

I already know of a solution involving the fact that given two normal extension fields $$M,LM, L$$ of some field $$KK$$ contained in some common extension, then $$Gal(ML/L)≅Gal(M/M∩L)\text{Gal}(ML/L) \cong \text{Gal}(M/M \cap L)$$, 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
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