7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Have you studied Galois theory? $\endgroup$ – CJD Jun 19 '17 at 2:10
  • $\begingroup$ Yes ! I think this problem is easy indeed, but I'm a bit confused. $\endgroup$ – jnaf Jun 19 '17 at 2:11
  • 3
    $\begingroup$ 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 '17 at 2:13
  • 3
    $\begingroup$ 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 '17 at 2:13
  • 3
    $\begingroup$ Niki, this was the argument I was thinking of: math.stackexchange.com/questions/460397/… $\endgroup$ – CJD Jun 19 '17 at 2:20
10
$\begingroup$

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 $p\geq 3$

$\endgroup$
  • $\begingroup$ You are using that normal subgroups correspond to normal extensions ? $\endgroup$ – reuns Jun 19 '17 at 3:53
  • $\begingroup$ @user1952009 yes. definitely. And that every subgroup of an abelian group is normal. $\endgroup$ – sharding4 Jun 19 '17 at 3:55
  • $\begingroup$ A very nice and elegant argument! $\endgroup$ – pisco Jun 19 '17 at 4:16
3
$\begingroup$

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 $\sqrt[p]{k} \notin \mathbb{Q}(\zeta_n)$.

$\endgroup$
  • $\begingroup$ Why does the lemma say that? $\endgroup$ – Mariano Suárez-Álvarez Jun 19 '17 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 '17 at 5:50
  • $\begingroup$ Yup, I know: I was suggesting you edit that into the answer :-) $\endgroup$ – Mariano Suárez-Álvarez Jun 19 '17 at 5:58
  • $\begingroup$ @MarianoSuárez-Álvarez Thanks for suggestion :), I updated my answer. $\endgroup$ – pisco Jun 19 '17 at 6:16
1
$\begingroup$
  • $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \sim \mathbb{Z}_n^\times$ 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 $\mathbb{Q}(\zeta_n)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.