## Trace and Norm Generalized

Let $$K, L$$ be number fields satisfying $$K \subset L$$. Let $$\sigma_1,...,\sigma_n$$ be the $$n = [L:K]$$ embeddings of $$L$$ in $$\mathbb{C}$$ that fix $$K$$. Let $$\alpha \in L$$. Then define the relative trace and relative norm as follows

$\array { T^L_K(\alpha)=\sigma_1(\alpha)+...+\sigma_n(\alpha) \\ N^L_K(\alpha)=\sigma_1(\alpha)...\sigma_n(\alpha) }$

Thus $$T^K = T^K_\mathbb{Q}$$ and $$N^K = N^K_\mathbb{Q}$$. As before we can show:

Theorem: Let $$\alpha \in L$$ and let $$d$$ be the degree of $$\alpha$$ over $$K$$. Let $$t(\alpha)$$ and $$n(\alpha)$$ be the sum and product of the $$d$$ conjugates of $$\alpha$$ over $$K$$. Then

$\array { T^L_K(\alpha) = \frac{n}{d}t(\alpha) \\ N^L_K(\alpha) = n(\alpha)^\frac{n}{d} }$

Corollary: $$T^L_K(\alpha), N^L_K(\alpha) \in K$$. If $$\alpha$$ lies in the number ring of $$L$$, then they lie in the number ring of $$K$$.

Theorem: Let $$K,L,M$$ be number fields with $$K\subset L \subset M$$. Then for all $$\alpha \in M$$ we have transitivity in the following sense

$\array { T^L_K(T^M_L(\alpha)) &=& T^M_K(\alpha) \\ N^L_K(N^M_L(\alpha)) &=& N^M_K(\alpha) }$

Proof: Let $$\sigma_1,...,\sigma_n$$ be the embeddings of $$L$$ in $$\mathbb{C}$$ that fix $$K$$, and let $$\tau_1,...,\tau_n$$ be the embeddings of $$M$$ in $$\mathbb{C}$$ that fix $$L$$. We first need to extend the embeddings to automorphisms of some field so that we may compose them. Hence fix a normal extension $$N$$ of $$\mathbb{Q}$$ such that $$M \subset N$$. Then all the $$\sigma_i, \tau_i$$ may be extended to automorphisms of $$N$$. Fix one extension of each and keep the labels $$\sigma_i, tau_i$$. Now the mappings can be composed:

$\array { T^L_K(T^M_L(\alpha)) &=& \sum_{i=1}^{n}\sigma_i{(\sum_{j=1}^m\tau_j(\alpha))} &=& \sum_{i,j}\sigma_i\tau_j(\alpha) \\ N^L_K(N^M_L(\alpha)) &=& \prod_{i=1}^{n}\sigma_i{(\prod_{j=1}^m\tau_j(\alpha))} &=& \prod_{i,j}\sigma_i\tau_j(\alpha) }$

We now need to show that the $$mn$$ mappings $$\sigma_i \tau_j$$ restricted to $$M$$ are the embeddings of $$M$$ in $$\mathbb{C}$$ which fix $$K$$. Since all $$\sigma_i\tau_j$$ fix $$K$$ and there are $$mn = [M:L][L:K] = [M:K]$$ of them, it remains to show they are all distinct when restricted to $$M$$.

Suppose two of the mappings agreed on $$M$$. Then they also agree on $$L$$. The $$\tau$$ maps fix $$L$$, so this means we have $$\sigma_i \ne \sigma_j$$ agreeing on all of $$L$$, which is a contradiction.∎

We may interpret the trace an norm as follows. Let $$K \subset L$$ be fields and let $$\alpha \in L$$. Then considering $$L$$ as a vector space over $$K$$, multiplication by $$\alpha$$ is a linear mapping. Let $$A$$ be a matrix representing this map with respect to some basis $$\alpha_1,\alpha_2,...$$ for $$L$$ over $$K$$. Then $$T^L_K(\alpha)$$ and $$N^L_K(\alpha)$$ are the trace and determinant of $$A$$.

Ben Lynn blynn@cs.stanford.edu 💡