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Let's consider $p(x)=x^p-1$, assuming $p$ as a prime number. Its degree over $\mathbb{Q}$ is $p-1$ and its Galois Group is $\mathcal{C}_{p-1}$, so there is a unique subextension of degree 2. Let's call $L$ this subextension. In which cases $\sqrt{p} \in L$?

For $p=5$ and $p=7$ it happens.

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Made a few edits to use more standard terminology: "degree" (instead of grade) and "subextension" (instead of underextension). – Ted Dec 30 '12 at 16:38
up vote 2 down vote accepted

This only happens when $p \equiv 1$ (mod 4). When $p \equiv 3$ (mod 4), the unique quadratic extension contained in $\mathbb{Q}(\zeta)$ is $\mathbb{Q}(\sqrt{-p})$ (where $\zeta$ is a primitive $p$th root of unity), and this won't be contained in $L$, which is the maximal real subextension of $\mathbb{Q}(\zeta)$.

For a proof, see quadratic Gauss sums. The formula for $g(1,p)$ gives an explicit expression of $\sqrt{p}$ or $\sqrt{-p}$ in terms of primitive $p$th roots of unity.

share|cite|improve this answer, talking about g(1,p), how do I show that sum is real? – Ivan Dec 30 '12 at 18:49
It's only real when $p \equiv 1$ (mod 4). You can see directly that $g(1,p)$ is real in this case because -1 is a square mod $p$, so for every term $\zeta^{n^2}$ that appears, so does its conjugate $\zeta^{-n^2}$. – Ted Dec 30 '12 at 18:57

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