Tensoring does not change the determinant (norm map) Let $ K $ be a number field, and let $ L $ be a finite extension of $ K $. Let $ p $ be a prime ideal in $ \mathcal{O}_K $, then we know that there is a canonical isomorphism 
$$ L \otimes_K K_p \simeq \prod_{q \mid p} L_q. $$
($ q $ runs over all prime ideals in $ L $ lying over $ p $.)
Why does it follow from here that for any $ b \in L $, 
$$ N_{L/K} (b) = \prod_{q \mid p} N_{L_q / K_p} (b)? $$
More specifically, why is that tensoring over $ K $ with $ K_p $ does not change the determinant of the $K$-linear map $ x \mapsto bx $?
 A: For $a\in L$, you have two instances of the regular representation. You have $L\to L$ by $x\mapsto ax$, and you have $L\otimes_KR\to L\otimes_KR$ by $\xi\mapsto(a\otimes1)\xi$, where $\xi$ stands for an arbitrary element of the tensor product. Here, $R$ is any old $K$-algebra.
In the former case, the map is described completely by the result of multiplying $a$ to the elements $\{x_1,\cdots,x_n\}$ of a $K$-basis of $L$. You get a matrix, and you take its determinant. And in the latter case, the map is decribed completely by the result of multiplying $a\otimes1$ to the elements $\{x_1\otimes1,\cdots,x_n\otimes1\}$ of the corresponding $R$-basis of $L\otimes_KR$. You get the “same” matrix, and the determinant is $(\det a)\otimes1$. In the two cases, you have calculated the norm of $a$, respectively the norm of $a\otimes1$.
In the case you (and I) are interested in, where $L\otimes K_p$ splits into a sum of $K_p$-algebras, one is tempted to take a basis of the tensor product that more nearly reflects the splitting. But taking a different basis doesn’t change the determinant.
I hope this addressed your worries; it certainly was not as efficient an explanation as I would have liked.
