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I am trying to prove that $\Phi_m(x)$ is irreducible over $\Bbb Q(\zeta_n)$ if and only if $(m,n)\leq2$.

The left implication turns out to be somewhat easy since without loss of generality, $2\mid m$ and no higher power of $2$ divides $m$ and using the fact that $\Bbb Q(\zeta_{m/2})=\Bbb Q(\zeta_m)$ (this assumes the degrees aren't relatively prime to begin with, which I think is even easier).

My problem comes in doing the forward implication. I tried and tried without success, so I started trying to use the contrapositive, that is, if $(m,n)\geq3$, then $\Phi_m(x)$ is reducible over $\Bbb Q(\zeta_n)$. This hasn't proven fruitful yet, so if possible, will someone point me in the right direction?

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Are you wedded to this formulation? I’d rather have said that for odd $m$ and $n$, $\mathbb Q(\zeta_m)\cap\mathbb Q(\zeta_n)=\mathbb Q(\zeta_g)$ with $g=\gcd(m,n)$. And go from there to the case of even indices. But maybe how to do that is exactly the heart of your question? –  Lubin Aug 21 '13 at 16:29

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First show that $\Phi_m(x)$ is irreducible over $\mathbb{Q}(\zeta_n)$ if and only if $[\mathbb{Q}(\zeta_m,\zeta_n):\mathbb{Q}(\zeta_n)]=\varphi(m)$. Then compute $[\mathbb{Q}(\zeta_m,\zeta_n):\mathbb{Q}(\zeta_n)]$ and the result will fall out.

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Exactly what I had in mind, using the fact that the big field here is the compositum of the two smaller cyclotomic fields, and indeed equal to $\mathbb Q(\zeta_\ell)$, where $\ell$ is the lcm of $m$ and $n$. The only question is how to get a nice description involving the gcd of the indices, in light of the annoying fact that $\mathbb Q(\zeta_n)=\mathbb Q(\zeta_{2n})$ if $n$ is odd. –  Lubin Aug 21 '13 at 20:15
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@Lubin, why does that "annoying fact" matter? Irreducibility is equivalent to $\varphi(\ell)/\varphi(n)=\varphi(m)$, then use that $\varphi(k)=k\prod_{p\mid k}(1-1/p)$ to turn this into $1=g\prod_{p\mid g}(1-1/p)$ where $g=\text{gcd}(m,n)$. Then observe that this holds just when $g\le 2$. –  Michael Zieve Aug 21 '13 at 21:18
    
Alternately you could use the fact that $\varphi(\ell)/\varphi(n)=\varphi(m)/\varphi(g)$ to show that irreducibility is equivalent to $\varphi(g)=1$, and thus to $g\le 2$. –  Michael Zieve Aug 21 '13 at 21:29
    
Well, @Zieve, I just didn’t give it the requisite amount of thought. –  Lubin Aug 21 '13 at 23:36
    
I'm not sure, but doesn't this primarily focus on the part I already know? At any rate, this part is obvious. If $\Phi_m(x)$ is irreducible, we get an extension of degree $\varphi(m)$, and conversely we already know $\Phi_m(x)$ has degree $\varphi(m)$, so if we know the degree of the extension, it must be irreducible. (Sorry, I've been gone a while and just got on to check). –  Clayton Aug 21 '13 at 23:41

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