Let $L$ be an extension field of $\mathbb Q$ with $[L:\mathbb Q]=n$. Let $A$ be the ring of integers of $L$ i.e. the Integral closure of $\mathbb Z$ in $L$. Then $A$ is a Dedekind domain. If $I$ is a non-zero ideal of $A$, then I can show that $I$ is a free $\mathbb Z$-module of rank $n$ and $aA \subseteq I$ for some $0 \ne a \mathbb Z$ . Moreover $A/I$ is finite.
My question is : If $I=(x)$ is a non-zero principal ideal, then how to show that $|A/I|=| N_{L|\mathbb Q } (x)|$ ? where $N_{L|\mathbb Q}$ denotes the field Norm function .
Choose an integral basis for $O_L$ (which will denote the integral closure of $\mathbb{Z}$ in the extension $L$), denoted $\{ \alpha_1 , \dots , \alpha_n \}$. Then, certainly $\{ x \alpha_1 , \dots , x \alpha_n \}$ is an integral basis for the principal ideal $(x)$. Then, by definition of discriminant $$\textrm{disc} ( (x)) = \det ( \sigma_i ( x \alpha_j )_{i,j} )^2$$ Where $\sigma_i$ denote embeddings of our $\alpha_j$ into some algebraic closure of $L$. Note that by properties of determinants, $$\det ( \sigma_i ( x \alpha_j )_{i,j} ) = \prod_{i=1}^n \sigma_i (x) \cdot \det ( \sigma_i (\alpha_j)_{i,j} )$$ But of course the product on the right is precisely the norm of $x$. We then see $$N_{L/\mathbb{Q}} (x)^2 = \frac{\textrm{disc} (O_L) }{\textrm{disc}( (x))}$$ But the quotient on the right is already known to be precisely $|O_L /(x) |^2$, whence we conclude that $N_{L/\mathbb{Q}} (x) = |O_L / (x)|$, as desired.