## 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

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

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

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

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 $N(\omega g(\omega)) \in \mathbb{Z}$.