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Let $\alpha$ be a root of the irreducible cubic polynomial $x^{3}+px+q$, $p,q\in \mathbb{Q}$. How can I compute the discriminant $\Delta(1,\alpha,\alpha^{2})$ relative to $\mathbb{Q}(\alpha)$?

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2 Answers 2

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If $\alpha=\alpha_1,\alpha_2, \alpha_3$ are the different roots of $f=x^3+px+q$, then \begin{align*} \Delta(1,\alpha,\alpha^2) &= \det \left( \matrix{1&\alpha_1&\alpha_1^2\\1&\alpha_2 &\alpha_2^2\\ 1&\alpha_3&\alpha_3^2} \right)^2 \quad \text{by definition of the discriminant}\\ &= \prod_{i<j} (\alpha_i-\alpha_j)^2 \quad \text{(Vandermonde determinant)}\\ &= \operatorname{Disc}(f) \quad \text{by definition of the discriminant of a polynomial}\\ &= -\operatorname{Res}(f,f') \quad \text{by a formula for the discriminant in terms of the resultant}\\ &= -\det\left(\matrix{1&0&p&q&0\\0&1&0&p&q\\3&0&p&0&0\\0&3&0&p&0\\0&0&3&0&p}\right) \quad \text{by definition of the resultant}\\ &= -27q^2-4p^3 \quad \text{by computation of the determinant.} \end{align*}

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The easiest would be to not use the discriminant, but to rather draw up a variation table with the help of differentiation.

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