Discriminant in a cubic extension

Question

There is a question: Compute the discriminant $\triangle(1,\alpha,\alpha^2)$, relative to $\mathbb{Q}(\alpha)$, where $\alpha$ is a root of the reducible cubic $x^3+px+q$, $p,q\in\mathbb{Q}$. Is this answer the same as the irreducible cubic $x^3+px+q$, namely, $-4p^3-27q^2$? I don't realize the difference between them.

Answer 1

Lemma: For an algebraic number field $\mathbb{Q}(\alpha)$ with $d=[\mathbb{Q}(\alpha):\mathbb{Q}]$, we have: $$\Delta(1,\alpha,\ldots,\alpha^{d-1}) =(-1)^{\frac{d(d-1)}{2}}N(\operatorname{Min}_\alpha'(\alpha)).$$ Proof: Let $\alpha_1,\ldots,\alpha_d$ be the $d$ roots of $\operatorname{Min}_\alpha$, then: \begin{align*} \operatorname{Min}_\alpha(X) =\prod_{j=1}^d(X-\alpha_i) &\Rightarrow \operatorname{Min}_\alpha'(X) =\sum_{i=1}^d\prod_{\stackrel{j=1}{j\neq i}}^d(X-\alpha_i) \Rightarrow \operatorname{Min}_\alpha'(\alpha_k) =\prod_{\stackrel{j=1}{j\neq k}}^d(\alpha_k-\alpha_j) \\ &\Rightarrow N(\operatorname{Min}_\alpha'(\alpha)) =\prod_{k=1}^d\operatorname{Min}_\alpha'(\alpha_k) =\prod_{\stackrel{j,k=1}{j\neq k}}^d(\alpha_k-\alpha_j) \end{align*} and therefore: $$\Delta(1,\alpha,\ldots,\alpha^{d-1}) =\prod_{\stackrel{j,k=1}{j<k}}^d(\alpha_k-\alpha_j)^2 =(-1)^{\frac{d(d-1)}{2}}\prod_{\stackrel{j,k=1}{j\neq k}}^d(\alpha_k-\alpha_j) =(-1)^{\frac{d(d-1)}{2}}N(\operatorname{Min}_\alpha'(\alpha)),$$ which concludes the proof. $\square$

Every cubic number field $\mathbb{K}$ (with $[\mathbb{K}:\mathbb{Q}]=3$) can be written as $\mathbb{K}=\mathbb{Q}(\alpha)$ with a root $\alpha$ of an irreducible cubic polynomial $x^3+px+q$. Let $\alpha_1$, $\alpha_2$ and $\alpha_3$ be the three roots of $x^3+px+q$, then we get: \begin{align*} \operatorname{Min}_\alpha(X) &=(X-\alpha_1)(X-\alpha_2)(X-\alpha_3) \\ &=X^3-\underbrace{(\alpha_1+\alpha_2+\alpha_3)}_{=s_1}X^2 +\underbrace{(\alpha_1\alpha_2+\alpha_2\alpha_3+\alpha_1\alpha_3)}_{=s_2}X -\underbrace{\alpha_1\alpha_2\alpha_3}_{=s_3}, \end{align*} therefore $s_1=0$, $s_2=p$ and $s_3=-q$ and using according to the upper lemma: \begin{align*} \Delta(1,\alpha,\alpha^2) &=-N(\operatorname{Min}_\alpha'(\alpha)) =-N(3\alpha^2+p\alpha) =-(3\alpha_1^2+p\alpha_1)(3\alpha_2^2+p\alpha_2)(3\alpha_3^2+p\alpha_3) \\ &=-27(\alpha_1\alpha_2\alpha_3)^2 -9p(\alpha_1^2\alpha_2^2+\alpha_2^2\alpha_3^2+\alpha_1^2\alpha_3^2) -3p^2(\alpha_1^2+\alpha_2^2+\alpha_3^2)-p^3 \\ &=-27s_3^2 -9p(s_2^2-2s_1s_3) -3p^2(s_1^2-2s_2) -p^3 \\ &=-27q^2-9p^3+6p^3-p^3 =-27q^2-4p^3. \end{align*}