## 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 $A$.