I'm trying to solve the following problem :
Let $K$ be a quadratic extension of $\mathbb{Q}$ and $\mathfrak{a}$ be a fractional ideal of $K$. Then inverse of $\mathfrak{a}$ is given by $$\mathfrak{a}^{-1}=\frac{1}{N(\mathfrak{a})}\mathfrak{a}'$$ where $N(\mathfrak{a})$ is absolute norm of ideal $\mathfrak{a}$ and $\mathfrak{a}'=\sigma(\mathfrak{a})$ where $\sigma\in Gal(K/\mathbb{Q})$ is nontrivial element. i.e. $\mathfrak{a}'$ is conjugate of $\mathfrak{a}$.
*edited : Norm of integral ideal $\mathfrak{a}\subset O_{K}$ is defined by $N(\mathfrak{a}):=|O_{K}/\mathfrak{a}|$. For a fractional ideal, let $\mathfrak{a}=\lambda \mathfrak{b}$ for some $\lambda \in K^{\times}$ and integral ideal $\mathfrak{b}\subset O_{K}$. Then $N(\mathfrak{a}):=N(\lambda)N(\mathfrak{b})$ where $N(\lambda)=N_{K/\mathbb{Q}}(\lambda)=\lambda\lambda'$.
I showed that it is enough to show for integral ideal $\mathfrak{a}\subset \mathcal{O}_{K}$, but I cannot proceed more because I don't know how to deal with norm of ideal.
I know that norm of principal ideal is given by $N((a))=N(a)$ and I can prove the problem for principal ideal case using this. Is there any result about general case, i.e. $N((a_{1}, \dots, a_{r}))$? My conjecture is for any integral ideal $I$, $N(I)=gcd_{x\in I} N(x)$ but I cannot prove it and even I don't know this will be helpful to solve the problem.
Just use the proof of the equality of the two definitions for the norm of an ideal, one is the absolute value of the product of the Galois conjugates and the other is the index definition. From there it becomes trivial as then you literally have $\mathfrak{a}\sigma(\mathfrak{a}) = N(\mathfrak{a})$ so that your identity is tautological.
But this is just a matter of noting that the usual definition is as the determinant of a certain matrix, namely $\sigma_i(\alpha_j)$ where $\{\alpha_j\}$ is a basis of your ideal as a $\Bbb Z$-module. Using the definition of the product of two ideals as the ideal generated by the products of the generators, the rest follows quickly and is classical.