Why is it true that if $p\equiv1\pmod{4}$ then $\sqrt{p}\in\mathbb{Q}(\zeta_p)$?
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Since $p$ is prime, $2$ divides the degree of the cyclic extension $\Bbb Q(\zeta_p) / \Bbb Q$, which has then a unique quadratic subextension $K/\Bbb Q$. Since only $p$ ramifies in $K$, and has odd discriminant, one finds $K = \Bbb Q(\sqrt p)$ (the possibility $K = \Bbb Q(\sqrt{-p})$ is excluded). – Watson Nov 06 '18 at 22:14
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@Watson: Why does only $p$ ramifies in $K$? Why does $K$ have odd discriminant? – Nov 16 '18 at 09:50
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- if $p$ ramifies in $K$, then it ramifies in $L =\Bbb Q(\zeta_p)$, but it is a general fact that the only ramified prime in $L$ is $p$. 2) If $K$ had even discriminant, then $2$ would be ramified in $K$, but we've just seen that the only ramified prime in $K$ is $p$, which is odd.
– Watson Nov 16 '18 at 09:55 -
@Watson: 1) shows that the only prime possibly ramified in $K$ is $p$, but why is it impossible that it is not ramified? – Nov 16 '18 at 11:14
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Dear @sai, it is a theorem of Minkowski that any number field of degree $>1$ over $\Bbb Q$ has at least one ramified prime. – Watson Nov 16 '18 at 11:19
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@Watson: How come $\mathbb{Q}(\sqrt{-p})$ is excluded? – Nov 16 '18 at 11:30
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Because $-p \equiv 3 \pmod 4$, so $\Bbb Q(\sqrt{-p})$ has even discriminant $-4p$. – Watson Nov 16 '18 at 12:44
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@Watson: Thank you very much for your explanations, now I understand logically. But how come just one sentence in what I was reading amount to this much? And how does one come up with this proof? – Nov 17 '18 at 02:17
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For $p \equiv 1 \bmod 4$ then $(-1)^{(p-1)/2} = 1$ so there is $c^2 \equiv -1 \bmod p$ and $(a,b) \to (a+cb,a-cb)$ is bijective $Z/(p)\times Z/(p) \to Z/(p)\times Z/(p)$ therefore $$(\sum_{a \bmod p} e^{2i \pi a^2/p})^2= \sum_{a,b \bmod p} e^{2i \pi (a^2+b^2)/p}=\sum_{a,b \bmod p}e^{2i \pi (a+cb)(a-cb)/p}=\sum_{u,v \bmod p}e^{2i \pi uv/p}\\= p+\sum_{u\bmod p,u \ne 0} \sum_{v \bmod p}e^{2i \pi uv/p}= p$$
The class field theory approach is to find the conductor of $\mathbb{Q}(\sqrt{p})/\mathbb{Q}$ is $(p)$, not sure how without using $\sum_{a \bmod p} e^{2i \pi a^2/p} = \pm \sqrt{p}$ (for $p\equiv 3 \bmod 4$ the conductor is $(4p)$, that is the Artin map is obtained from Dirichlet characters $\bmod 4p$)
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Thank you very much for the answer. I think I can understand, at least logically, the part before "The class field theory approach ..." How did you come up with this answer? – Oct 30 '18 at 07:16
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1@sai: 1) Look up Gauss sums. They are kinda "famous". 2) The Galois group acts by $\zeta_p\mapsto \zeta_p^a$, $\gcd(a,p)=1$, and is known to produce a cyclic group. So the sole quadratic subfield must be fixed by the squares of all the automorphisms. IOW automorphisms of the form $\zeta_p\mapsto \zeta_p^{a^2}$. Therefore the sum $\sum_a\zeta_p^{a^2}$ is in the fixed field. – Jyrki Lahtonen Oct 30 '18 at 15:19