# Intersection of Cyclotomic Fields

How would I prove that $\mathbb{Q_m} \cap \mathbb{Q_n} = \mathbb{Q_{(m, n)}}$ (here $\mathbb{Q_n}$ denotes the $n$th 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 $\text{Gal}(ML/L) \cong \text{Gal}(M/M \cap L)$, but does there exist a solution that doesn't require such a theorem?

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It's proved as Theorem 4.25 at math.ku.dk/~olsson/manus/alg3-2009/ek4-2009.pdf --- but the proof does seem to rely on that theorem from Galois Theory (although only one field is required to be normal, not both). –  Gerry Myerson Jan 31 '13 at 2:41
I think all that is needed is to show that $\varphi(\gcd(m,n)) = \dfrac{\varphi(m)\varphi(n)}{\varphi(\text{lcm}(m,n))}$. For another proof see p. 5 of math.uconn.edu/~kconrad/blurbs/galoistheory/cyclotomic.pdf –  David Wheeler Jan 31 '13 at 2:44
It's also done at Example 3.4 of math.uconn.edu/~kconrad/blurbs/galoistheory/cyclotomic.pdf but again I think that Galois Theory result may be lurking in the background. Another proof at Proposition 3.6 of math.uconn.edu/~kconrad/math5230f12/weston.pdf –  Gerry Myerson Jan 31 '13 at 2:54
I don't think there is going to be an easy proof (unlike the "opposite" result ${\mathbf Q}_m{\mathbf Q}_n = {\mathbf Q}_{[m,n]}$) and addressed this already in a comment to the question at math.stackexchange.com/questions/93691/…, which this question is nearly a duplicate of. –  KCd Jan 31 '13 at 3:39

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.

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let $K = \mathbb{Q}(\zeta_m) \cap \mathbb{Q}(\zeta_n)$. Clearly $K \supset \mathbb{Q}(\zeta_{(m,n)})$; we want to show the reverse inclusion.

Observe that $\frac{\varphi([m,n])}{\varphi(n)} = [\mathbb{Q}(\zeta_{[m,n]}: \mathbb{Q}(\zeta_n)] = [\mathbb{Q}(\zeta_m):K]$ and $\varphi(m) = [\mathbb{Q}(\zeta_m):\mathbb{Q}] = [\mathbb{Q}(\zeta_m): K] [K:\mathbb{Q}]$. Therefore,

$$\frac{\varphi(m)}{[K:\mathbb{Q}]} = \frac{\varphi([n,m])}{\varphi(n)}.$$

We claim that

$$\varphi((n,m))\varphi([n,m]) = \varphi(n)\varphi(m)$$.

If $\prod p_i^{e_i}$ is the prime factorization of $n$ and $\prod q_i^{f_i}$ is the prime factorization of $m$, then the right hand side is $nm$ times $\prod \left( 1- \frac{1}{p_i} \right) \prod \left( 1-\frac{1}{q_i} \right))$. Since $nm = [n,m] (n,m)$, the left hand side is $nm$ times $\prod (1-\frac{1}{p_{i_j}})$ where the product runs over the primes dividing $n$ and $m$ (for the LCM) and the primes dividing both $n$ and $m$ again (for the gcd). IThese are clearly equal. So $[K:\mathbb{Q}] = \varphi((n,m)) = [\mathbb{Q}(\zeta_{(m,n)}):\mathbb{Q}]$, which implies that $K = \mathbb{Q}(\zeta_{(n,m)})$.

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How do you get $[{\bf Q}(\zeta_{[m,n]}):{\bf Q}(\zeta_n)]=[{\bf Q}(\zeta_m):K]$? I think that's where the proof I found calls in the Galois Theory bit that OP wants to avoid. –  Gerry Myerson Jan 31 '13 at 2:51
I'm not sure how much it is reasonable to avoid all field theory. I can show that the inequality < of that equation in about one line (with about two more lines, you could prove the theorem.) The inequality <, along with the arithmetic identity, is also enough to prove the result. –  Tony Jan 31 '13 at 3:24
@Tony Hmm I somewhat concur with Gerry that when you assume that equality you are already assuming the result from field theory that the OP is trying to avoid.... –  user38268 Jan 31 '13 at 11:50
As I suggested above, it suffices to note the inequality $\leq$, which follows from the fact that this is an injection of Galois groups, which is justified by the line: "Since the fixed field of of $G$ is $L$, the fixed field of its image in $\Gal(M/K)$ is $M \cap L$." If even this level of field theory is not allowed... –  Tony Jan 31 '13 at 17:23
At the moment I can't give a proof of the general case but only of the specific case that $m = p^r$ and $n = q^s$ hence $(m,n) = 1$. Let us write $L = \Bbb{Q}(\zeta_n)$ and $L' = \Bbb{Q}(\zeta_m)$. We want to show that $$L \cap L' = \Bbb{Q}(\zeta_{(m,n)}) = \Bbb{Q}.$$
Firstly it is clear that $L \cap L' \supseteq \Bbb{Q}$. Now take the same prime $p \in \Bbb{Z}$. Then the prime $p$ is totally ramified in $\mathcal{O}_L'$ because $$p\mathcal{O}_{L'} = (1- \zeta_m)^{\varphi(m)}.$$
Thus by using results from ramification theory we get that $p$ is also totally ramified in $\mathcal{O}_{L \cap L'}$. On the other hand because the discriminant of $L$ is a power of $q$, $p$ must be unramified in $\mathcal{O}_L$ and hence unramified in $\mathcal{O}_{L \cap L'}$. We conclude that $$L \cap L' = \Bbb{Q}.$$