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


Ben Lynn blynn@cs.stanford.edu 💡