Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
add comment

1 Answer

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|improve this answer
    
en.wikipedia.org/wiki/Quadratic_Gauss_sum#Properties, 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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.