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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
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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

3 Answers 3

up vote 2 down vote accepted

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 $$L \cap L' = \Bbb{Q}.$$

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I don't think the use of ramification from algebraic number theory is in the spirit of the question, which is seeking simple methods of explaining this intersection. Certainly the viewpoint of ramification is a nice way to understand the result, but it is not really a simple explanation at the level of field theory, say. –  KCd Jan 31 '13 at 4:10
    
@KCd I agree with you that my answer is too highfalutin for this question. Though because the OP asked for an "alternative proof" of this fact, I thought of presenting one in such a specific case. –  user38268 Jan 31 '13 at 4:13

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