Let $K$ be an algebraic number field and $\mathfrak{a} \neq 0$ an ideal of $\mathcal{O}_K$, with prime factorization $\mathfrak{a} = \mathfrak{p}_1^{v_1}*...*\mathfrak{p}_r^{v_r}$. Assume $r=2$.
By the chinese reminder theorem we see that $\mathcal{O}_K/\mathfrak{a} \cong \mathcal{O}_K/\mathfrak{p}_1^{v_1}\bigoplus\mathcal{O}_K/\mathfrak{p}_2^{v_2}$.
If I know the absolute norm $\mathcal{R}$($\mathfrak{p}_i)^{v_i}$ for $i \in \{1, 2\}$, why then is $\mathcal{R}$($\mathfrak{a})=$$\mathcal{R}$($\mathfrak{p}_1)^{v_1}*\mathcal{R}$($\mathfrak{p}_2)^{v_2}$
Thanks for your help, or any hints!
By defintion, the absolute norm $\mathcal R(\mathfrak a)$= card $(\mathfrak o_K/\mathfrak a)$, so that $\mathcal R(\mathfrak a)$ is obviously the product of the $\mathcal R({\mathfrak p_i}^{n_i})$. To show that $\mathcal R({\mathfrak p}^{n})= \mathcal R({\mathfrak p})^n$, consider, in the chain of descending ideals $o_K > \mathfrak p > ... > \mathfrak p^i > ... > {\mathfrak p}^n$, the successive quotients $\mathfrak p^i / \mathfrak p^{i+1}$. Multiplication by $\mathfrak p$ induces readily an isomorphism $o_K / \mathfrak p \cong \mathfrak p^i / \mathfrak p^{i+1}$, and we are done.