## The Discriminant

Let $$K$$ be a number field of degree $$n$$ over $$\mathbb{Q}$$. Let $$\sigma_1,...,\sigma_n$$ be the embeddings of $$K$$ in $$\mathbb{C}$$. We write $$[a_{i j}]$$ for the matrix with $$a_{i j}$$ in the $$i$$th row and $$j$$ column, and $$|a_{i j}|$$ for the determinant of this matrix. Define the discriminant of $$\alpha_1,...,\alpha_n \in K$$ by

$disc(\alpha_1,...,\alpha_n) = |\sigma_i(\alpha_j)|^2$

As with the trace and norm, the discriminant may be generalized by replacing $$\mathbb{Q}$$ with any number field.

Theorem: $$disc(\alpha_1,...,\alpha_n) = |T(\alpha_i \alpha_j)|$$

Proof: Recall some basic facts about determinants:

$|a_{i j}| = |a_{j i}|, |A B| = |A||B|$

Then the theorem follows from

$[\sigma_j(\alpha_i)][\sigma_i(\alpha_j)] = [\sigma_1(\alpha_i\alpha_j) +...+ \sigma_n(\alpha_i\alpha_j)] = [T(\alpha_i\alpha_j)]$

Corollary: $$disc(\alpha_1,...,\alpha_n) \in \mathbb{Q}$$, and if the $$\alpha_i$$ are integral, then $$disc(\alpha_1,...,\alpha_n) \in \mathbb{Z}$$.

Theorem: $$disc(\alpha_1,...,\alpha_n) = 0$$ if and only if $$\alpha_1,...,\alpha_n$$ are linearly dependent over $$\mathbb{Q}$$.

Proof: If the $$\alpha_j$$ are linearly dependent over $$\mathbb{Q}$$ then so are the columns of the matrix $$[\sigma_i(\alpha_j)]$$ and its determinant will be zero. Conversely if $$disc(\alpha_1,...,\alpha_n) = 0$$, then the rows $$R_i$$ of the matrix $$[T(\alpha_i \alpha_j)]$$ are linearly dependent. Suppose the $$\alpha_i$$ are linearly independent. Then let $$a_1,..,a_n$$ be rationals, not all zero, such that $$a_1 R_1 + ... + a_n R_n = 0$$. Then consider $$\alpha = a_1 \alpha_1 + ... + a_n \alpha_n \ne 0$$. Note we have $$T(\alpha \alpha_j) = 0$$ for all $$j$$. The $$\alpha_j$$ form a basis for $$K$$ over $$\mathbb{Q}$$, and since $$\alpha \ne 0$$, so do the $$\alpha \alpha_j$$. But this implies $$T(\beta) = 0$$ for all $$\beta \in K$$, a contradiction since there exist elements of nonzero trace (for example $$T(1) = n$$).

Theorem: Let $$K = \mathbb{Q}[\alpha]$$ and let $$\alpha_1,...,\alpha_n$$ be the conjugates of $$\alpha$$ over $$\mathbb{Q}$$. Let $$f$$ be the minimal polynomial of $$\alpha$$. Then

$disc(1,\alpha,...,\alpha^{n-1}) = {\prod_{1\le r < s \le n} (\alpha_r -\alpha_s)^2} = {\pm N^K(f'(\alpha))}$

where the sign is positive if and only if $$n = 0,1 (mod 4)$$.

Proof: First observe $$|\sigma_i(\alpha^{j-1})| = |(\sigma_i(\alpha))^{j-1}| =|\alpha^{j-1}_i|$$ is a Vandermonde determinant, yielding the first equality.

Next we have

${\prod_{r< s}(\alpha_r - \alpha_s)^2} = { \pm {\prod_{r\ne s}(\alpha_r-\alpha_s)}}$

where the plus sign holds if and only if $$n = 0,1 (mod 4)$$. Since $$f'$$ has rational coefficients, we have

$N^K(f'(\alpha)) = {\prod_{r=1}^n \sigma_r(f'(\alpha))} = {\prod_{r=1}^n f'(\sigma_r(\alpha))} = {\prod_{r=1}^n f'(\alpha_r)}$

Then the second equality follows from the fact that for all $$r$$

$f'(\alpha_r) = {\prod_{s \ne r} (\alpha_r - \alpha_s)}$

(To see this, let $$f(x) = (x-\alpha_r)g(x)$$. Then by the product rule, $$f'(\alpha_r) = g(\alpha_r)$$, and the roots of $$g$$ are all the roots of $$f$$ except for $$\alpha_r$$.)

Let us now apply this theorem to compute $$disc(1,\omega,...,\omega^{p-2})$$ for $$\omega = e^{2 \pi i / p}$$ where $$p$$ is an odd prime. We have $$f(x) = 1 + x +... + x^{p-1}$$. Since $$x^p - 1 = (x-1)f(x)$$, we have $$p x^{p-1} = f(x) + (x-1)f'(x)$$ showing that

$f'(\omega) = \frac{p}{\omega(\omega-1)}$

Taking norms gives

$N(f'(\omega))=\frac{N(p)}{N(\omega)N(\omega-1)} = \frac{p^{p-1}}{N(\omega - 1)}$

Since $$(1-\omega)...(1-\omega^{p-1}) = p$$, we have $$N(\omega - 1) = N(1 - \omega) = p$$, hence $$N(f'(\omega)) = p^{p-2}$$.

Write $$disc(\alpha)$$ for $$disc(1,\alpha,...,\alpha_n)$$ where $$\alpha$$ is a degree $$n$$ algebraic integer over $$\mathbb{Q}$$ and $$K$$ is taken to be $$\mathbb{Q}[\alpha]$$.

We now know that $$disc(\omega) = {\pm p^{p-2}}$$ for prime $$p$$. For $$\omega = e^{2 \pi i / m}$$ for general $$m$$, the expression is more complicated, but it can be readily shown that $$disc(\omega) | m^{\phi(m)}$$ as follows. Let $$f$$ be the minimal polynomial of $$\omega$$. Then Since $$x^m - 1 = f(x)g(x)$$ for some $$g \in \mathbb{Z}[x]$$ we have $$m = \omega f'(\omega)g(\omega)$$. Taking norms gives

$m^{\phi(m)} = {\pm disc(\omega)N(\omega g(\omega))}$

and we are done since $$N(\omega g(\omega)) \in \mathbb{Z}$$.

Ben Lynn blynn@cs.stanford.edu 💡