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I am asked to prove that "$\sqrt[3]{2}\not\in\mathbb{Q}(\alpha_n)$ for all $n$ where $\alpha_n$ is the primitive nth root of unity"

I have attempted to use contradiction with the tower theorem. I got stuck at $[\mathbb{Q}(\alpha_n):\mathbb{Q}(\sqrt[3]{2})]=\frac{\phi(n)}{3}$ where $\phi(n)$ is the Euler totient function. Can someone give me some sort of hint?

Thanks in advance!

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1 Answer 1

up vote 2 down vote accepted

There's an answer using Galois Theory:

Since ${\bf Q}(\alpha_n)$ is normal over the rationals, if $\root3\of2$ is in it, so are all its conjugates. That makes the Galois group of the splitting field of $x^3-2$ a quotient group of the Galois group of ${\bf Q}(\alpha_n)$. But the Galois group of the splitting field of $x^3-2$ is the nonabelian group $S_3$, the Galois group of ${\bf Q}(\alpha_n)$ is abelian, contradiction.

Maybe someone will come up with a proof that uses less machinery.

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Quotient group? Do you mean that is it a subgroup of the Galois group? True by the fundamental theorem? –  user44322 Nov 22 '12 at 6:13
    
If $K$ is an extension of $F$, and $E$ is an intermediate field, normal over $F$, then the group of $E$ over $F$ is not a subgroup of the group of $K$ over $F$, it is a quotient. The group of $K$ over $E$ is a normal subgroup of the group of $K$ over $F$, and $G(E/F)$ is the quotient of $G(K/F)$ by $G(K/E)$. –  Gerry Myerson Nov 22 '12 at 6:38
    
"That makes the Galois group of the splitting field of $x^3−2$ a quotient group of the Galois group of $\mathbb{Q}(\alpha_n)$." Do you mean that the Galois group of $x^3−2$ is isomorphic to a quotient group of the Galois group of $\mathbb{Q}(\alpha_n)$? –  user44322 Nov 22 '12 at 7:49
    
Not isomorphic, equal --- on the (false) assumption that $\root3\of2$ is in ${\bf Q}(\alpha_n)$. –  Gerry Myerson Nov 22 '12 at 11:42
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