Let \(K\) be a number field. Define the 'trace' \(T^K\) and the 'norm' \(N^K\) as follows. Let \(\sigma_1,...,\sigma_n\) be the embeddings of \(K\) in \(\mathbb{C}\) where \(n = [K:\mathbb{Q}]\). For \(a \in K\), define

(we omit the field when it is clear which one we are referring to).

For the next part, let \(a\) have degree \(d\) over \(\mathbb{Q}\). Let \(t, n\) denote the sum and product of the \(d\) conjugates of \(a\) over \(\mathbb{Q}\).

Theorem:

where \(n = [K:\mathbb{Q}]\). (Note \(n / d = [K:\mathbb{Q}[a]]\)).

Proof: \(t, n\) are simply \(T^{\mathbb{Q}[a]}, N^{\mathbb{Q}[a]}\), and each embedding of \(\mathbb{Q}[a]\) in \(\mathbb{C}\) extends to \(n /d\) embeddings of \(K\) in \(\mathbb{C}\).

Corollary: \(T(a), N(a) \in \mathbb{Q}\).

Proof: This follows from the fact that \(-t(a), \pm n(a)\) are coefficients of the minimal polynomial of \(a\).

Corollary: \(T(a), N(a) \in \mathbb{Z}\) if \(a\) is an algebraic integer.

For example, in the quadratic field \(K = \mathbb{Q}[\sqrt{m}]\) we have \(T(a + b\sqrt{m}) = 2a, N(a + b\sqrt{m}) = a^2 - m b^2\) (where \(a,b\in\mathbb{Q}\)). For this case, the converse of the corollary is also true.