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I have the following problem

Let $p\geq3$ a prime. Show that $\mathbb{Q}(\sqrt[p]{p})$ is not contained in any cyclotomic extension.

I don't know how to start the problem. Any hint or help will be appreciated !

Thanks in advance.

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  • $\begingroup$ Have you studied Galois theory? $\endgroup$
    – CJD
    Jun 19, 2017 at 2:10
  • $\begingroup$ Yes ! I think this problem is easy indeed, but I'm a bit confused. $\endgroup$
    – jnaf
    Jun 19, 2017 at 2:11
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    $\begingroup$ I'm not sure how big of a hint this will be, but the cyclotomic extensions are Galois and have abelian Galois group... $\endgroup$
    – CJD
    Jun 19, 2017 at 2:13
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    $\begingroup$ What's the minimum polynomial of $\sqrt[p]{p}$? Why is it irreducible? Cyclotomic extensions are normal, so the polynomial would have to split. Compare degrees. $\endgroup$
    – sharding4
    Jun 19, 2017 at 2:13
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    $\begingroup$ Niki, this was the argument I was thinking of: math.stackexchange.com/questions/460397/… $\endgroup$
    – CJD
    Jun 19, 2017 at 2:20

3 Answers 3

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Note for $n\geq 3,\, \Bbb{Q}[\zeta_n]$ is complex and for any $n$ is normal with abelian galois group. Suppose $\sqrt[p]{p} \in \Bbb{Q}[\zeta_n]$. Since $\sqrt[p]{p}$ is real, it is contained in the fixed field of complex conjugation, call it $K$. As $Gal(\Bbb{Q}[\zeta_n])$ is abelian, $K$ is galois hence must be normal. But if $p\geq 3$, $K$ doesn't contain the roots of $x^p-p$ conjugate to $\sqrt[p]{p}$, namely $\zeta_p\sqrt[p]{p},\, \zeta_p^2\sqrt[p]{p},\dots$ since the roots are complex, so it can't be normal. Hence $\sqrt[p]{p} \not \in \Bbb{Q}[\zeta_n]$ for any $p\geq 3$

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  • $\begingroup$ You are using that normal subgroups correspond to normal extensions ? $\endgroup$
    – reuns
    Jun 19, 2017 at 3:53
  • $\begingroup$ @user1952009 yes. definitely. And that every subgroup of an abelian group is normal. $\endgroup$
    – sharding4
    Jun 19, 2017 at 3:55
  • $\begingroup$ A very nice and elegant argument! $\endgroup$
    – pisco
    Jun 19, 2017 at 4:16
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We use a lemma:

Let $p$ be a prime number, $k\in \mathbb{Q}$, if $x^p-k$ has no rational root, then $x^p-k$ is irreducible over $\mathbb{Q}[x]$

Assume $\sqrt[p]{k} \in \mathbb{Q}(\zeta_n)$. Since the extension $\mathbb{Q}(\zeta_n)$ is normal over $\mathbb{Q}$ and $\sqrt[p]{k}$ is a root of the polynomial $x^p-k$, the lemma says the polynomial $x^p-k$ splits completely in $\mathbb{Q}(\zeta_n)$, hence $\zeta_p \in \mathbb{Q}(\zeta_n)$. Consider the chain of extensions:

$$ F:=\mathbb{Q}\quad \subset \quad L:=\mathbb{Q}(\zeta_p, \sqrt[p]{k}) \quad \subset \quad K:=\mathbb{Q}(\zeta_n)$$

Both extensions $K/F$ and $L/F$ are Galois, the Galois group for $K/F$ is abelian of order $\varphi(n)$, while the Galois group for $L/F$ has order $p(p-1)$, it is a group which is not abelian when $p\geq 3$, (more specifically, it is the general affine group over $\mathbb{F}_p$), a contradiction, hence $\sqrt[p]{k} \notin \mathbb{Q}(\zeta_n)$.

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  • $\begingroup$ Why does the lemma say that? $\endgroup$ Jun 19, 2017 at 5:33
  • $\begingroup$ @MarianoSuárez-Álvarez If an irreducible polynomial has a root in a normal extension, then it has all its roots in this normal extension. "Irreducible" is indispensable here. Since we assume $\sqrt[p]{k}$ is in this normal extension, $x^p-k$ splits completely there $\endgroup$
    – pisco
    Jun 19, 2017 at 5:50
  • $\begingroup$ Yup, I know: I was suggesting you edit that into the answer :-) $\endgroup$ Jun 19, 2017 at 5:58
  • $\begingroup$ @MarianoSuárez-Álvarez Thanks for suggestion :), I updated my answer. $\endgroup$
    – pisco
    Jun 19, 2017 at 6:16
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  • $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \sim \mathbb{Z}_n^\times$ is an abelian (and Galois) extension. Thus for any field $F \subseteq \mathbb{Q}(\zeta_n)$, $Gal(F/\mathbb{Q})$ is a subgroup of $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ and $ F/\mathbb{Q}$ is an abelian extension.

  • Let $K = \mathbb{Q}(\sqrt[p]{p},\zeta_p)$.
    $[\mathbb{Q}(\zeta_p):\mathbb{Q}]= p-1$ and $[\mathbb{Q}(\sqrt[p]{p}):\mathbb{Q}]= p$ so $[K:\mathbb{Q}] = p(p-1)$ and its Galois group $Gal(K/\mathbb{Q})$ has elements of the form $$\sigma_{a,b}(\zeta_p^l \sqrt[p]{p}) = \sigma_{a,b}(\zeta_p^l)\sigma_{a,b}( \sqrt[p]{p})=\zeta_p^{al}\zeta_p^b \sqrt[p]{p}, \qquad a \in (\mathbb{Z}/p\mathbb{Z})^\times,b \in \mathbb{Z}/p\mathbb{Z}$$

    and hence for $p \ge 3$ : $$\sigma_{2,1}( \sigma_{1,2}(\zeta_p^l \sqrt[p]{p}))=\zeta_p^{2l+4+1} \sqrt[p]{p} \ne \sigma_{1,2}(\sigma_{2,1}( \zeta_p^l \sqrt[p]{p}))=\zeta_p^{2l+3} \sqrt[p]{p}$$ Therefore $Gal(K/\mathbb{Q})$ is not an abelian group so neither $K$ nor $\mathbb{Q}(\sqrt[p]{p})$ is contained in $\mathbb{Q}(\zeta_n)$.

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