I am trying to solve the following exercice:
Let $ω = e^{2πi/ p}$ with $p$ an odd prime. Show that $\Bbb Q[ω]$ contains $\sqrt p$ if $p ≡ 1(mod 4)$, and $\sqrt {− p}$ if $p ≡ −1 (mod 4)$.
Express $\sqrt {−3}$ and $\sqrt 5$ as polynomials $P(ω)$.
I need help for the second part.
We know that for a cyclotomic extension, that $disc(ω)=\pm p^{p-2}$. Which means that $\frac{disc(ω)}{p^{p-3}}=\pm p$
So for $p=3$ it is straightforward: $disc(ω)=-3 \Rightarrow P(ω)=\sqrt{disc(ω)} = \pm (ω-ω^2)$.
However for $p=5$, we have $\frac{disc(ω)}{5^{5-3}}=5$ so we can't get $P(ω)$ directly from discriminant.
However, after some calculations, I find $disc(ω)=5$ so $P(ω)=\sqrt{disc(ω)}$ indeed. How to explain the missing denominator $5^2$?
Thanks for your help.
Best Answer
By Galois theory, $\Bbb Q(\omega)$ contains a unique quadratic extension. This extension is unramified at all primes other than $p$. The only quadratic field for which this holds is $\Bbb Q(\sqrt p)$ for $p\equiv1\pmod 4$ and $\Bbb Q(\sqrt{- p})$ for $p\equiv3\pmod 4$.
Alternatively use Gauss sums. Prove that $\gamma^2=\pm p$ where $$\gamma=\sum_{a=0}^{p-1}\omega^{a^2}.$$