Editing Adele ring (section)

===Trace and norm on the adele ring===

Let <math>L/K</math> be a finite extension. Since <math>\mathbb{A}_K=\mathbb{A}_K \otimes_K K</math> and <math>\mathbb{A}_L=\mathbb{A}_K \otimes_K L</math> from the Lemma above, <math>\mathbb{A}_K</math> can be interpreted as a closed subring of <math>\mathbb{A}_L.</math> For this embedding, write <math>\operatorname{con}_{L/K}</math>. Explicitly for all places <math>w</math> of <math>L</math> above <math>v</math> and for any <math>\alpha \in \mathbb{A}_K, (\operatorname{con}_{L/K}(\alpha))_w=\alpha_v \in K_v.</math>

Let <math>M/L/K</math> be a tower of global fields. Then:

:<math>\operatorname{con}_{M/K}(\alpha)=\operatorname{con}_{M/L}(\operatorname{con}_{L/K}(\alpha)) \qquad \forall \alpha \in \mathbb{A}_K.</math>

Furthermore, restricted to the principal adeles <math>\operatorname{con}</math> is the natural injection <math>K \to L.</math>

Let <math>\{\omega_1,\ldots,\omega_n\}</math> be a basis of the field extension <math>L/K.</math> Then each <math>\alpha \in \mathbb{A}_L</math> can be written as <math>\textstyle \sum_{j=1}^n \alpha_j \omega_j,</math> where <math>\alpha_j \in \mathbb{A}_K</math> are unique. The map <math>\alpha \mapsto \alpha_j</math> is continuous. Define <math>\alpha_{ij}</math> depending on <math>\alpha</math> via the equations:

:<math>\begin{align}
\alpha \omega_1 &=\sum_{j=1}^n \alpha_{1j} \omega_j \\
&\vdots \\
\alpha \omega_n &=\sum_{j=1}^n \alpha_{nj} \omega_j
\end{align}</math>

Now, define the trace and norm of <math>\alpha</math> as:

:<math>\begin{align}
\operatorname{Tr}_{L/K}(\alpha) &:= \operatorname{Tr} ((\alpha_{ij})_{i,j})=\sum_{i=1}^n \alpha_{ii}\\
N_{L/K}(\alpha)                 &:= N ((\alpha_{ij})_{i,j})=\det((\alpha_{ij})_{i,j})
\end{align}</math>

These are the trace and the determinant of the linear map

:<math>\begin{cases} \mathbb{A}_L \to \mathbb{A}_L \\ x \mapsto \alpha x\end{cases}</math>

They are continuous maps on the adele ring, and they fulfil the usual equations:

: <math>\begin{align}
\operatorname{Tr}_{L/K}(\alpha+\beta)&=\operatorname{Tr}_{L/K}(\alpha) + \operatorname{Tr}_{L/K}(\beta) && \forall \alpha, \beta \in \mathbb{A}_L\\
\operatorname{Tr}_{L/K}(\operatorname{con}(\alpha))&=n\alpha && \forall \alpha \in \mathbb{A}_K\\
N_{L/K}(\alpha \beta)&=N_{L/K}(\alpha) N_{L/K}(\beta) && \forall \alpha, \beta \in \mathbb{A}_L\\
N_{L/K}(\operatorname{con}(\alpha))&=\alpha^n && \forall \alpha \in \mathbb{A}_K
\end{align}</math>

Furthermore, for <math>\alpha \in L, </math><math>\operatorname{Tr}_{L/K}(\alpha)</math> and <math>N_{L/K}(\alpha)</math> are identical to the trace and norm of the field extension <math>L/K.</math> For a tower of fields <math>M/L/K,</math> the result is:

: <math>\begin{align}
\operatorname{Tr}_{L/K}(\operatorname{Tr}_{M/L}(\alpha)) &= \operatorname{Tr}_{M/K}(\alpha) && \forall \alpha \in \mathbb{A}_M\\
N_{L/K} (N_{M/L}(\alpha))&=N_{M/K}(\alpha) && \forall \alpha \in \mathbb{A}_M
\end{align}</math>

Moreover, it can be proven that:<ref>See {{harvnb|Weil|1967|p=64}} or {{harvnb|Cassels|Fröhlich|1967|p=74}}.</ref>

:<math>\begin{align}
\operatorname{Tr}_{L/K}(\alpha) &= \left (\sum_{w | v}\operatorname{Tr}_{L_w/K_v}(\alpha_w) \right )_v && \forall \alpha \in \mathbb{A}_L\\
N_{L/K}(\alpha) &= \left (\prod_{w | v}N_{L_w/K_v}(\alpha_w) \right )_v && \forall \alpha \in \mathbb{A}_L
\end{align}</math>