2
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – 123 Oct 23 '16 at 23:37
  • $\begingroup$ The embeddings only send $\alpha$ to $\omega \alpha$ and $\omega^2\alpha$ if $a=0$ in your polynomial. Otherwise the conjugates are more complicated. $\endgroup$ – carmichael561 Oct 23 '16 at 23:39
  • $\begingroup$ So would I need Cardano's formula (or something similar) to recover the discriminant? $\endgroup$ – 123 Oct 23 '16 at 23:42
  • 1
    $\begingroup$ 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 $[(\alpha_1-\alpha_2)(\alpha_1-\alpha_3)(\alpha_2-\alpha_3)]^2$. $\endgroup$ – carmichael561 Oct 23 '16 at 23:45
  • $\begingroup$ (continued) which is the discriminant of the minimal polynomial of $\alpha_1$. $\endgroup$ – carmichael561 Oct 23 '16 at 23:46
2
$\begingroup$

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 $\alpha$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.