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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.

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up vote 8 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 $$\Delta = (-1)^{(p-1)/2} p^{p-2}.$$

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

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