Connection between discriminant of ideal and quadratic number field. Problem:
Let $K$ be a quadratic field, $\mathcal{O}_K$ its ring of integers. Now I want to prove the equality
$$D(\mathfrak{a}) = \mathcal{N}(\mathfrak{a})^2 \cdot \Delta_K $$
where $\mathfrak{a} \subseteq K$ is a fractional ideal, $D(\mathfrak{a})$ is the Idealdiscriminant, $\mathcal{N}(\mathfrak{a})$ is the Idealnorm and $\Delta_K$ is the discriminant of $K$. 
Attempt:
Let $(1, \omega)$ be an integral basis of $\mathcal{O}_K$, $n \in \mathbb{N}$ such that $n \mathfrak{a} \subseteq \mathcal{O}_K$, $(x, y)$ an integral basis of $n\mathfrak{a}$. (Disclaimer: I'm not required to show that such a basis, $n$ exist). 
$$\implies \hspace{15pt}(\frac{x}{n}, \frac{x}{n}) \text{ is an integral basis of } \mathfrak{a}$$
There are also $a_1, a_2, b_1, b_2 \in \mathbb{Z}$ satisfying: 
$$x = a_1 + a_2 \omega \hspace{30pt} y = b_1 + b_2\omega$$
Using the definition of $D(\mathfrak{a})$:
$$
D(\mathfrak{a}) = 
\text{det}\left[ {\begin{array}{cc}
\frac{x}{n} & \frac{y}{n} \\
\frac{\bar x}{n} & \frac{\bar y}{n} \\
\end{array} } \right]^2 = \frac{1}{n^4} \cdot \text{det}\left[ {\begin{array}{cc}
x & y \\
\bar x & \bar y \\
\end{array} } \right]^2 = \frac{(x\bar y - \bar x y)^2}{n^4} \\ 
= \frac{((a_1 + a_2\omega)(b_1 + b_2\bar\omega) - (a_1 + a_2\bar\omega)(b_1 + b_2\omega))^2}{n^4} \\
= \frac{(a_1b_2(\bar\omega - \omega) + a_2b_1(\omega - \bar\omega))^2}{n^4} \\
= \frac{(\bar\omega - \omega)^2(a_1b_2 - a_2b_1)^2}{n^4} \\
= \frac{\Delta_K(a_1b_2 - a_2b_1)^2}{n^4}
$$
At this point it would be really nice if $\mathcal{N}(\mathfrak{a}) =  \frac{a_1b_2 - a_2b_1}{n^2}$. This is the case iff $\mathcal{N}(n\mathfrak{a}) = a_1b_2 - a_2b_1$. Sadly I have no clue how to prove this last equality. I would be happy as well if someone provided a simpler proof to the initial equality. 
 A: It is true that $\mathcal N(n\mathfrak a) = a_1 b_2 - a_2 b_1$.
This follows from the following fact about abelian groups: if $G = \mathbb Z \oplus \mathbb Z$, and if $H$ is a subgroup of $G$ generated by the linearly-independent elements $(a_1, a_2), (b_1, b_2) \in G$, then the order of the quotient group $G/H$ is
$$ |G/H| = \left| \det \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix}\right|.$$
(See this post, which derives this result by considering the Smith normal form of the matrix describing the embedding of $H$ into $G$ with respect to your chosen bases.)
Now apply this with $G = \mathcal O_K$ and $H = n\mathfrak a$ (both viewed as abelian groups)…

By the way, you probably don't need to be so explicit in your matrix manipulations. The key idea that makes your proof work is that
$$ \begin{bmatrix} x & y \\ \bar x & \bar y \end{bmatrix} =  \begin{bmatrix} 1 & \omega \\ 1 & \bar\omega \end{bmatrix}\begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix} . $$
Now observe that
$$ D(n\mathfrak a) = \det\begin{bmatrix} x & y \\ \bar x & \bar y \end{bmatrix}^2, \ \ \ \Delta_K = \det \begin{bmatrix} 1 & \omega \\ 1 & \bar\omega \end{bmatrix}^2, \ \ \ \mathcal N(n\mathfrak a)^2 = \det\begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix}^2 .$$
