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- norm map
- dualizable
- Adams isomorphism
- Pontrjagin-Thom collapse
Summary
Let \(A\in {}_{k} \mathsf{Alg} ^{\mathrm{fd}}\) and let \([m_a]\) be the matrix of \(x\mapsto ax\) in a \(k{\hbox{-}}\)basis. The trace of \(a \in A\) is the trace of \(m_{a}\). The norm of \(a\) is the determinant of \(m_{a}\).
trace map
Relation to separable algebras: 
norm map
Results
- \(a \in A^{\times} \text {if and only if } \mathrm{N}_{A / k}(a) \neq 0\).
- Let \(M_a\) be the minimal polynnomial of \(a\in A\), then If \(D\) is a finite-dimensional \(k\)-division algebra, then for all \(a \in D\), \begin{align*} \chi_{a}=M_{a}^{\operatorname{deg}\left(\chi_{a}\right) \over \operatorname{deg}\left(M_{a}\right)}=M_{a}^{[D: k] \over [k[a]: k]} . \end{align*} Thus, \(\mathrm{N}_{D / k}(a)=(-1)^{[D: k]} M_{a}(0)^{[D: k] /[k[a]: k]}\).
- \(\operatorname{Tr}_{K / k}(a)=\sum_{i} \sigma_{i}(a) \text { and } \mathrm{N}_{K / k}(a)=\prod_{i} \sigma_{i}(a)\).
- For \(a \in \mathbf{F}_{q^{n}}\), \begin{align*} \operatorname{Tr}_{\mathbf{F}_{q^{n}} / \mathbf{F}_{q}}(a)=a+a^{q}+\cdots+a^{q^{n-1}} \text { and } \mathbf{N}_{\mathbf{F}_{q^{n}} / \mathbf{F}_{q}}(a)=a \cdot a^{q} \cdots a^{q^{n-1}} . \end{align*}
Examples
