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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.

learning_math
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Daniel
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2 Answers2

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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 $$\Phi_p(X) = \prod_i (X-\zeta_i) = 1 + X + \ldots + X^{p-1}.$$ Altogether, this shows $$\Delta = (-1)^{(p-1)/2} p^{p-2}.$$

marlu
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  • Yes, thanks, I fixed it – marlu Nov 14 '17 at 00:04
  • The step where you plug in $\zeta$ and take the norm of both sides, shouldn't that give you $N(\zeta - 1)N(\Phi_p(\zeta)) = N(\zeta^p -1)$, I don't see why on the right hand side you take the derivative of $\Phi_p(\zeta)$ and I have no idea how you get $N(p\zeta^{-1}) = p^{p-1}$ on the left hand side. – Samantha Wyler Sep 18 '20 at 17:15
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    Starting from $(X-1)\Phi_p(X) = X^p-1$, you first take derivatives on both sides, then plug in $\zeta$ and then take norms. This gives you $N(\zeta-1) N(\Phi_p'(\zeta)) = N(p \zeta^{-1})$. For the equality $N(p\zeta^{-1}) = p^{p-1}$ one has to use the fact that the norm of $\zeta$ equals 1 (the constant coefficient of the minimal polynomial, with positive sign since the degree $p-1$ is even), and the fact that $N(p) = p^{p-1}$ which follows from that fact that $p$ is contained in $\mathbb Q$ and the extension $\mathbb Q(\zeta)/\mathbb Q$ has degree $p-1$. – marlu Dec 04 '20 at 02:40
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    Initially I was confused on why taking the derivative of the left had side and then plugging in $\zeta$ and then taking the norm, resulted in $N(\zeta - 1)N(\Phi_p'(\zeta)$ instead of $N(\zeta)N(\Phi_p(\zeta)) + N(\zeta - 1)N(\Phi_p'(\zeta)$. But now that I am re looking at this, I am some what embarrassed to realize its because $N(\Phi_p(\zeta)) = 0$, since $\Phi_p(\zeta) = 0$, being the minimal polynomial of $\zeta$. – Samantha Wyler Dec 04 '20 at 16:01
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Thought I would add my solution, which is less elegant, but doesn't require too much creativity.

It suffices to find the discriminant of the number field $L = \mathbb Q(\zeta_p)/\mathbb Q$. We will use that $\mathcal O_L = \mathbb Z[\zeta_p]$, so $1, \zeta_p, \zeta_p^2, \ldots, \zeta_p^{p-2}$ is an integral basis.

We know $\zeta_p$ has min. poly $f(X) = X^{p-1}+X^{p-2}+\ldots+1$ over $\mathbb Q$. Moreover, this is the min. poly for all its roots, thus each of this roots have trace $-1$ (reading off from the coefficient of $X^{p-2}$). So we've found:

$$ \mathrm{Tr}(1) = [\mathbb Q(\zeta_p) : \mathbb Q] = p-1 $$ $$ \mathrm{Tr}(\zeta_p^i) = -1 \quad (1 \le i \le p-1) $$

Now applying the definition of the discriminant, we need to compute the determinant of a $(p-1)\times(p-1)$ matrix, $$\mathcal D_L = \det([\mathrm{Tr}(\zeta_p^i\zeta_p^j)]_{0 \le i,j \le (p-2)}) = \det \small \begin{pmatrix} p-1 & -1 &-1 & \ldots & -1 \\ -1 & -1 & -1 & \ldots & -1 \\ \vdots & \vdots & -1 && p-1 \\ \vdots & \vdots & & \unicode{x22F0} \\ -1 & -1 & p-1 & & -1 \end{pmatrix}$$

Now for some row and column operations. First subtract row 2 from every other row; then subtract column 2 from every other column. We obtain

$$ \small\begin{pmatrix}p \\ & -1 \\ && & & p \\ & && \unicode{x22F0} \\ & & p\end{pmatrix} $$

With $(p-3)/2$ column swaps we make this matrix diagonal, with determinant $p \times (-1) \times p^{p-3}$. Since each column swap introduced a factor of $-1$, we've arrived at:

$$ \mathcal D_L = (-1)^{(p-3)/2} \times p \times (-1) \times p^{p-3} = (-1)^{(p-1)/2}p^{p-2} $$

FlipTack
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