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Unique quad. subfield of $\mathbb{Q}(\zeta_p)$ is $\mathbb{Q}(\sqrt{p})$ if $p \equiv 1$ $(4)$, is $\mathbb{Q}(\sqrt{-p})$ if $p \equiv 3$ $(4)$

This question may be quite naive. In here $\omega=e^{\frac{2\pi i}{p}}$ and $\mathbb{Q}(\omega)$ is the cyclotomic field. I believe this should be an easy execrise for Galois theory, but I could not prove it using standard tools available (norm and trace, discriminants) as I have not used Galois theory for years.

After some algebraic manipulation I even constructed a ${\bf{wrong}}$ proof as this:

All elements of $\mathbb{Q}(\omega)$ can be written in form $\sum^{i=p-1}_{i=0} a_{i}w^{i}$. For $\sqrt{p}=\sum^{i=p-1}_{i=0} a_{i}w^{i}$, this would imply $p=\sum a_{i}^{2}w^{2i}+2\sum_{i\not=j}a_{i}a_{j}w^{i+j}$. This implies all the powers remaining in the reducted form must be equal to $p$. Hence the non-constant terms must vanish, implying for all $w^{k},k>0$ its coefficient must be 0. But adding equations of the type $a_{i}^{2}+2\sum a_{j}a_{k}=0$ and $a_{0}^{2}+2\sum_{s+t=p}a_{s}a_{t}=p$ would imply $(\sum a_{i})^{2}=p$. This is impossible since $\sum a_{i}\in \mathbb{Q}$. Thus $\sqrt{p}$ is not in $\mathbb{Q}(\omega)$.

I think I am definitely on the wrong track somehow. After struggling with this HW problem for more than 4 hours I decided to give up. If someone could give some directions(instead of solving the problem) I would be grateful.

Batrachotoxin
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Kerry
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    A good keyword to google for is "Gauss sums". – Mariano Suárez-Álvarez Feb 19 '12 at 01:38
  • What do you have at your disposal? In addition to Gauss sums, as Mariano mentioned, there are also a few ways of proving this using discriminants and possibly ramification degrees. [I also don't think it's an easy problem, but I'd love to be proven wrong.] – Dylan Moreland Feb 19 '12 at 01:40
  • @Dylan Moreland: I will search about Gauss sums. The author gives me the hint to use discriminant for $disc(\mathbb{Q}(\omega))=p^{p-2}$. But I do not see how I can use this result explicitly in here. For my knowledge in number theory, I know elementary level arguments(A course in arithemetic) but not to the level of algebraic number theory. – Kerry Feb 19 '12 at 01:45
  • @ChangweiZhou Ah, I see now that you mentioned discriminants in your question. That's a very good hint. I can try to say something about it. – Dylan Moreland Feb 19 '12 at 01:46
  • @MarianoSuárez-Alvarez: I am very surprised by the wikipedia article... – Kerry Feb 19 '12 at 01:51
  • $\operatorname{disc},\mathbb{Q}(\omega)=\color{Blue}{(-1)^{(p-1)/2}}p^{p-2} $, doesn't it? I don't know the answer, but I see a $p\equiv 1;(4)$ in there... – anon Feb 19 '12 at 01:51
  • @anon: Yes, this is the hint the author give to me, but I do not know how it could help. I am too naive in number theory. – Kerry Feb 19 '12 at 01:58
  • By the way (sorry if you already realized this), your "wrong proof" is wrong because in the reduced form for elements of $\mathbb{Q}(\omega)$, the highest power is $p-2$, not $p-1$. You have to account for the relation $\sum_{i=0}^{p-1} \omega^i = 0$. – Ted Feb 19 '12 at 02:19
  • @Ted: I realized this after doing a few computation with Gauss sums, that $Q(\omega)$ is a degree $p-2$ extension of $Q$. I was obvious confused. – Kerry Feb 19 '12 at 02:34
  • No, it is a degree $p-1$ extension. The set ${1, \omega, \omega^2, \ldots, \omega^{p-2}}$, which has $p-1$ elements, is a basis. – Ted Feb 19 '12 at 03:01
  • @Ted: I see. Thanks for pointing out. I guess I was twice confused. – Kerry Feb 19 '12 at 03:17

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Let $K = \mathbf Q(\omega)$, and let $f$ be the minimal polynomial of $\omega$. For the purpose of calculation, the formula \[ d = (-1)^{p(p - 1)/2}N_{K/\mathbf Q}(f'(\omega)) \] for the discriminant is probably best. You've found that this is equal to $p^{p - 2}$. But we have another expression for $d$, namely \[ \prod_{0 \leq i < j \leq p} (\omega^i - \omega^j)^2. \] See Proposition 2.6 of Milne's notes for proofs. Thus, $d$ has a square root in $K$. Why does this help?

Dylan Moreland
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  • The original version of this said maybe too much, in view of your last paragraph. I hope I didn't ruin the fun! – Dylan Moreland Feb 19 '12 at 02:05
  • No, I am still trying to prove this by evaluating Gauss sums. So no fun is ruined at this stage. – Kerry Feb 19 '12 at 02:09
  • @ChangweiZhou Cool. I do think that Serre shows that a certain Gauss sum squares to $p$ within the first few pages of A Course in Arithmetic, if that's a book you like. – Dylan Moreland Feb 19 '12 at 02:13
  • Yes, I read that in college freshmen year and can hardly recall the contents, which is why I am very surprised to see the wikipedia article. I do not have the book with me so I prefer to work out everything myself.Thank you. – Kerry Feb 19 '12 at 02:31
  • I can show the Guass sum's square must be $p$, but I am missing a factor of $i$. – Kerry Feb 19 '12 at 03:18
  • I finally reached the stage to follow your proof. I found I am having trouble to evaluate the $p-2$th root of $\prod (w^{i}-w^{j})$. It seems quite complicated to me. Is there any simple way to avoid this issue? – Kerry Feb 19 '12 at 05:18
  • For Gauss sums, I try to follow the direction in Exercise 17, but I could not prove how $\tau$ would be real is $-1$ is a square mod $p$, and imaginary otherwise. – Kerry Feb 19 '12 at 05:43
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    @ChangweiZhou: You don't need to take the $p-2$th root of $\prod(\omega^i - \omega^j)$. You can just divide it by $p^{(p-3)/2}$ to get what you want. – Jonas Kibelbek Feb 19 '12 at 06:17
  • @JonasKibelbek: Thanks. I understand. – Kerry Feb 19 '12 at 22:15
  • @DylanMoreland: forget to note I was following the directions in this sheet:http://www.dpmms.cam.ac.uk/study/II/Galois/2009-2010/ex2.pdf – Kerry Feb 19 '12 at 22:41