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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 $Q(\omega)$ 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 $Q(\omega)$.

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.

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marked as duplicate by Kerry, Gerry Myerson, Zhen Lin, Chris Eagle, t.b. Aug 17 '12 at 11:41

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A good keyword to google for is "Gauss sums". –  Mariano Suárez-Alvarez Feb 19 '12 at 1:38
    
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 1:40
    
@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 1:45
    
@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 1:46
    
@MarianoSuárez-Alvarez: I am very surprised by the wikipedia article... –  Kerry Feb 19 '12 at 1:51
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1 Answer 1

up vote 3 down vote accepted

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?

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:Thanks.very grateful. –  Kerry Feb 19 '12 at 2:04
    
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 2:05
    
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 2:09
    
@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 2:13
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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 6:17
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