Show $\mathrm{disc}(1,\alpha,\dots,\alpha^{d-1})=(-1)^{d(d-1)/2}N_{\mathbb Q(\alpha)/\mathbb Q}(f'(\alpha))$, with $d=[\mathbb Q(\alpha):\mathbb Q]$

Question

Suppose $K = \mathbb{Q}(\alpha)$ for some $\alpha$ algebraic, and let $f$ be the minimal polynomial of $\alpha$ over $\mathbb{Q}$. Say $d = [K:\mathbb{Q}]$. I want to show that

$$\mathrm{disc}(1, \alpha, \alpha^2, \dots, \alpha^{d-1}) = (-1)^{d(d-1)/2} N_{K/\mathbb{Q}} (f'(\alpha)).$$

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I'm trying to reason as follows.

I know that if $\sigma_1, \dots, \sigma_d: K \to \mathbb{C}$ are the $d$ distinct field embeddings of $K$ into $\mathbb{C}$, then $\sigma_1(\alpha), \dots, \sigma_d(\alpha)$ are the roots of $f$. Say that $\sigma_1(\alpha) = \alpha$.

We have that

$$f(X) = \prod_{i=1}^n (X - \sigma_i(\alpha))$$

so

$$f'(X) = \prod_{i={\color{red} 2}}^n (X - \sigma_i(\alpha)) + (X - \sigma_1(\alpha)) (\text{some other terms})$$

thus

$$f'(\alpha) = \prod_{i=2}^n (\sigma_1(\alpha) - \sigma_i(\alpha)).\tag{1}\label{eq1}$$

And I am also aware of the identity

$$\mathrm{disc}(1, \alpha, \alpha^2, \dots, \alpha^{d-1})=\prod_{1 \leq i < j \leq d}(\sigma_i(\alpha) - \sigma_j(\alpha))^2\tag{2}\label{eq2}$$

which is very similar - so I expect I must have to show $(-1)^{d(d-1)/2} N_{K / \mathbb{Q}}( f'(\alpha))$ is equal to this, i.e. relate $(1)$ and $(2)$.

My issue lies in calculating $N_{K / \mathbb{Q}}(f'(\alpha))$. I know the norm to be defined as the determinant of the linear transform $x \mapsto f'(\alpha)$, viewing $K$ as a $\mathbb{Q}$-vector space. But actually performing this manipulation to show that equality is where I'm confused.

Answer 1

Hopefully you know that we have: $$\mathrm{N}^K_\Bbb Q(x)=\prod_{\sigma\in\mathrm{Gal}(K;\Bbb Q)}\sigma(x)$$

So: $$\begin{align}\mathrm{N}^K_\Bbb Q(f'(\alpha))&=\prod_{\sigma\in\mathrm{Gal}(K;\Bbb Q)}\sigma\left(\prod_{\tau\in\mathrm{Gal}(K;\Bbb Q)\\\quad\,\tau\neq\mathrm{id}}(\alpha-\tau(\alpha))\right)\\&=\prod_{\sigma\in\mathrm{Gal}(K;\Bbb Q)}\prod_{\tau\in\mathrm{Gal}(K;\Bbb Q)\\\quad\,\tau\neq\mathrm{id}}(\sigma(\alpha)-(\sigma\tau)(\alpha))\\&=\prod_{\sigma,\sigma'\in\mathrm{Gal}(K;\Bbb Q)\\\quad\,\sigma\neq\sigma'}(\sigma(\alpha)-\sigma'(\alpha))\\&=\prod_{1\le i<j\le n}(\sigma_i(\alpha)-\sigma_j(\alpha))(\sigma_j(\alpha)-\sigma_i(\alpha))\\&=\prod_{1\le i<j\le n}(-1)(\sigma_i(\alpha)-\sigma_j(\alpha))^2\\&=(-1)^{\frac{n(n-1)}{2}}\prod_{1\le i<j\le n}(\sigma_i(\alpha)-\sigma_j(\alpha))^2\\&=(-1)^{\frac{n(n-1)}{2}}\mathrm{disc}(\alpha)\end{align}$$As desired. The most important point in this proof is the third line, where we use the fact we are dealing with a group! As $\sigma$ is fixed and $\tau$ ranges freely, avoiding only the identity, the product $\sigma\tau$ also ranges freely through the group, avoiding only the element $\sigma$. Recall $g\cdot G=G$ for any $g\in G$ and any group $G$; group multiplication is a bijection. This kind of symmetry is crucial in many proofs. The latter lines just fix an enumeration of the elements of the Galois group, the precise ordering isn't important. What I've done is take every pair $(\sigma,\sigma')$ and reorder the product by combining the factors associated to $(\sigma,\sigma')$ and its friend, $(\sigma',\sigma)$.