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.

I'm very bad in computations of this kind :/. I don't know tricks for computing the discriminant of a polynomial, only the definition and using the resultant, but it's very complicated to do only with that tools. I need some help please ._.

I have to prove that the discriminant of $\Phi_p$ is $ (-1)^{\frac{p-1}{2}}p^{p-2}$ I don't know if it's neccesary to assume that $p$ is prime.

share|improve this question

1 Answer 1

up vote 5 down vote accepted

In what follows I will assume that $p$ is an odd prime. Let $\zeta$ be a primitive $p$-th root of unity and denote its conjugates by $\zeta_1,\ldots,\zeta_{p-1}$, so we have $$\Phi_p(X) = \frac{X^p-1}{X-1} = \prod_{i=1}^{p-1} (X-\zeta_i).$$ The discriminant of $\Phi_p$ then can be computed as $$\Delta = \prod_{i<j} (\zeta_i - \zeta_j)^2 = (-1)^{(p-1)/2} \prod_{i\neq j} (\zeta_i - \zeta_j).$$ Note that $$\Phi_p'(X) = \sum_i \prod_{j\neq i} (X-\zeta_j),$$ so $$\prod_{i\neq j} (\zeta_i - \zeta_j) = \prod_i \Phi_p'(\zeta_i) = N_{\mathbb Q(\zeta)/\mathbb Q}(\Phi_p'(\zeta)).$$ To compute this norm, we take the derivative on both sides of $$(X-1)\Phi_p(X) = X^p-1,$$ plug in $\zeta$ and take norms to get $$N(\zeta-1) N(\Phi_p'(\zeta)) = N(p \zeta^{-1}) = p^{p-1}.$$ The norm $N(\zeta-1)$ is given by $$N(\zeta-1) = \prod_i (\zeta_i - 1) = \prod_i (1 -\zeta_i) = p$$ as we see by setting $X = 1$ in $$\Phi_p(X) = \prod_i (X-\zeta_i) = 1 + X + \ldots + X^{p-2}.$$ Altogether, this shows $$\Delta = (-1)^{(p-1)/2} p^{p-2}.$$

share|improve this answer
    
Thanks for the answer :D! –  Daniel Nov 19 '12 at 18:31

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.