Number Fields

Let $K,L$ be number fields satisfying $K\subset L$. Let ${\sigma }_1,...,{\sigma }_n$ be the $n=[L:K]$ embeddings of $L$ in $ℂ$ that fix $K$. Let $\alpha \in L$. Then define the relative trace and relative norm as follows

$$\begin{array}{c}{T}_K^L(\alpha )={\sigma }_1(\alpha )+...+{\sigma }_n(\alpha )\\ N_K^L(\alpha )={\sigma }_1(\alpha )...{\sigma }_n(\alpha )\end{array}$$

Thus $T^K=T_{\mathbb{Q}}^K$ and $N^K=N_{\mathbb{Q}}^K$. 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

$$\begin{array}{c}{T}_K^L(\alpha )=\frac{n}{d}t(\alpha )\\ N_K^L(\alpha )=n(\alpha {)}^{\frac{n}{d}}\end{array}$$

Corollary: $T_K^L(\alpha ),N_K^L(\alpha )\in K$. 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

$$\begin{array}{ccc}{T}_K^L(T_L^M(\alpha ))& =& T_K^M(\alpha )\\ N_K^L(N_L^M(\alpha ))& =& N_K^M(\alpha )\end{array}$$

Proof: Let ${\sigma }_1,...,{\sigma }_n$ be the embeddings of $L$ in $ℂ$ that fix $K$, and let ${\tau }_1,...,{\tau }_n$ be the embeddings of $M$ in $ℂ$ 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 ${\sigma }_i,{\mathrm{tau}}_i$. Now the mappings can be composed:

$$\begin{array}{ccccc}{T}_K^L(T_L^M(\alpha ))& =& \sum _{i=1}^n{\sigma }_i(\sum _{j=1}^m{\tau }_j(\alpha ))& =& \sum _{i,j}{\sigma }_i{\tau }_j(\alpha )\\ N_K^L(N_L^M(\alpha ))& =& \prod _{i=1}^n{\sigma }_i(\prod _{j=1}^m{\tau }_j(\alpha ))& =& \prod _{i,j}{\sigma }_i{\tau }_j(\alpha )\end{array}$$

We now need to show that the $\mathrm{mn}$ mappings ${\sigma }_i{\tau }_j$ restricted to $M$ are the embeddings of $M$ in $ℂ$ which fix $K$. Since all ${\sigma }_i{\tau }_j$ fix $K$ and there are $\mathrm{mn}=[M:L][L:K]=[M:K]$ 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. $\blacksquare$

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 $T_K^L(\alpha )$ and $N_K^L(\alpha )$ are the trace and determinant of $A$.