Trace Form on Product of Fields

Question

I am studying algebraic number theory and I am having trouble understanding something. Let $K$ be a number field with ring of integers $\mathcal{O}_K$. Suppose a prime $p$ does not ramify in $K$. Then we can write $p\mathcal{O}_K = \mathfrak{p}_1...\mathfrak{p}_g$ where this is a product of distinct prime ideals in $\mathcal{O}_K$. Thus by CRT we can see $$\mathcal{O}_K/p\mathcal{O}_K \cong B/\mathfrak{p}_1 \times ... \times B/\mathfrak{p}_g $$

Now we can look at the trace pairing $\mathcal{O}_K/p\mathcal{O}_K \times \mathcal{O}_K/p\mathcal{O}_K \to \mathbb{Z}/p\mathbb{Z}$ via $(x,y) \mapsto Tr(xy)$. I am wondering how to interpret this map in light of chinese remainder theorem. How do I interpret

\begin{align*} \Big{(}(x+\mathfrak{p}_1, ...,x+\mathfrak{p}_g), (y+\mathfrak{p}_1, ...,y+\mathfrak{p}_g)\Big{)} \mapsto Tr(xy+\mathfrak{p}_1, ...,xy+\mathfrak{p}_g) \end{align*} What does the right handside look like? I know it must be the same as $Tr(xy)$ but I am not sure how you get there simply by manipulating the right.

Answer 1

$(p)=\prod_{j=1}^g P_j$ is unramified in $O_K$. There is $\pi_j\in O_K/(p), \pi_j\equiv 1\bmod P_j,\pi_j\equiv 0\bmod P_i$ and $$O_K/(p)=\sum_{j=1}^g \pi_j (O_K/P_j)$$

Take some basis

• $O_K = \sum_{l=1}^n b_l Z,O_K/(p)=\sum_{l=1}^n b_l F_p$

• $O_K/P_j=\sum_{l=1}^{f_j} u_{j,l} F_p$ $$O_K/(p)=\sum_{j=1}^g \pi_j \sum_{l=1}^{f_j} u_{j,l} F_p$$

For $a\in O_K$ let $A\in M_n(Z)$ be the matrix of the multiplication by $a\in O_K$ such that $a b_i=\sum_{l=1}^n A_{i,l} b_l$. Then $$Tr_{O_K/Z}(a) = tr(A)\in Z,\qquad Tr_{(O_K/(p))/F_p}(a)=tr(A)\in F_p$$ Let $A_j'$ be the matrix of the multiplication by $a\in O_K/P_j$ and $A'$ the matrix of the multiplication by $a\in O_K/(p)$ in the basis of the $\pi_j u_{j,l}$. Then $$A' = \pmatrix{ A_1' & & \\ & A_2' & \\ & & \ddots}, \qquad A=QA'Q^{-1}, \qquad Q\in GL_n(F_p)$$ $$Tr_{(O_K/(p))/F_p}(a)=tr(A)=tr(A')=\sum_{j=1}^g tr(A_j')=\sum_{j=1}^g Tr_{(O_K/P_j)/F_p}(a) $$