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

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?

## 2 Answers

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.

• So how can I recover the determinant of a polynomial (say a cubic) from this definition? Is this the right line of thinking? If $x^3+ax+b$ is the minimal polynomial for $\alpha$, then a basis for $K(\alpha)$ is $\{1,\alpha,\alpha^2\}$. The three embeddings send $\alpha$ to $\alpha,\omega\alpha$ and $\omega^2\alpha$ respectively. I don't think the determinant of the resulting matrix does not recover the expected result. – 123 Oct 23 '16 at 23:37
• The embeddings only send $\alpha$ to $\omega \alpha$ and $\omega^2\alpha$ if $a=0$ in your polynomial. Otherwise the conjugates are more complicated. – carmichael561 Oct 23 '16 at 23:39
• So would I need Cardano's formula (or something similar) to recover the discriminant? – 123 Oct 23 '16 at 23:42
• No: use a Vandermonde determinant: if $\alpha_1,\alpha_2,\alpha_3$ are the roots of the cubic, and we use $1,\alpha_1,\alpha_1^2$ for the power basis, then the matrix in the definition of the discriminant becomes $\begin{bmatrix}1&\alpha_1&\alpha_1^2\\1&\alpha_2&\alpha_2^2\\1&\alpha_3&\alpha_3^2\end{bmatrix}$, and the determinant of this matrix is $[(\alpha_1-\alpha_2)(\alpha_1-\alpha_3)(\alpha_2-\alpha_3)]^2$. – carmichael561 Oct 23 '16 at 23:45
• (continued) which is the discriminant of the minimal polynomial of $\alpha_1$. – carmichael561 Oct 23 '16 at 23:46

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 $\alpha$.