How does Neukirch's definition of the discriminant relate to the discriminant of a polynomial?

Question

I am trying to understand how Neukirch's definition of the discriminant relates to the discriminant of a polynomial.

He defines the discriminant of a basis of a (separable) field extension as follows;

The $\mathbf{discriminant}$ of a basis $\alpha_1,\dots,\alpha_n$ be a basis of the separable extension $L\mid K$ is defined by $$d(\alpha_1,\dots,\alpha_n)=\det((\sigma_i\alpha_j))^2 $$ where $\sigma_i$, $i=1,\dots,n$ varies over the $K$-embeddings $L\to \bar{K}$

It appears that the discriminant is a property of a field extension. If this is the case, then what is the field extension that the discriminant of the polynomial decribes?

Answer 1

If we choose a power basis for $L/K$, i.e. a basis of the form $1,\alpha,\dots,\alpha^{n-1}$, then the discriminant of this basis is equal to the discriminant of the minimal polynomial of $\alpha$. I believe Neukirch proves this shortly after he introduces the discriminant.

Answer 2

Apply his definition to the case of a basis of the form $\alpha, \alpha^2,\ldots ,\alpha^{n-1}$. Then the matrix is a Vandermond matrix and its determinant is the discriminant of the irreducible polynomial of $\alpha$.