Simple Answer to Showing $N_{k/\mathbb{Q}}(\alpha)=N((\alpha))$ So I'm trying to show that if we have some number field $k/\mathbb{Q}$ and ring of integers $R_k\subset k$, and an element of $R_k$, say $\alpha$, that the field norm of $\alpha$ is equal to the "norm" of the ideal $(\alpha)$, which from what I can understand is defined to be the number of elements of $R_k/(\alpha)$.  I have found a proof of this fact in W. Narkewiecz's "Elementary and Analytic Theory of Algebraic Numbers" (pp. 57-58) which uses a huge number of symbols, defining scads of sums and little niggling elements all over the place apparently arbitrarily (aside from the fact that it all works out in the end). I guess I could spend an hour trying to figure all this out, but this is a sort of homework assignment and I don't really want to just copy this down.  Is there any sensible, intuitive way of understanding what is going on here, or is truly just a nice effect of all the symbols coming out the right way? (A better, more elementary reference might be helpful as well...)
Thanks
 A: Prove that the ideal in $\mathbb{Q}$ generated by $N_{k/\mathbb{Q}} (\alpha)$ is the absolute norm of the ideal $(\alpha)$ in $R_k$.  If $k$ is Galois over $\mathbb{Q}$, this can be done cleanly and simply.  Prove that the product of the conjugates of a prime power $\mathfrak{p}^k$ in $R_k$ is principal, generated by the rational integer $\left| R_k/\mathfrak{p}^k \right|$.  Then show that this result extends (by the Chinese Remainder theorem and the Unique Factorization Theorem for ideals in $R_k$) to arbitrary nonzero ideals of $R_k$.
I would imagine the mess you saw in your reading is due to the fact that multiplying the conjugates of a prime ideal doesn't make any sense if the field is not Galois, so a lot of complication is added dealing with the more general case.
A: Sketch of the Proof: 
The transition matrix $A$ from an integral basis $w_1,...,w_n$ of $\mathcal{O}_K$ to the $\mathbb{Z}$-basis $\alpha w_1,...,\alpha w_n$ of $(\alpha)\subset \mathcal{O}_K$ is the same as the representation matrix of the map $T_\alpha:K\rightarrow K$. 
Hence, the absolute value of the determinant of $A$ is equal to both $(\mathcal{O}_K:(\alpha))$ (i.e, the norm of $(\alpha)$) and $N_{K/\mathbb{Q}}(\alpha)$  
