Discriminant in a cubic extension 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.
 A: 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*}
