An exercise in Neukirchs book about algebraic number theory is to show, that every ideal $\mathfrak{a}$ of a Dedekind domain $\mathcal{O}$ can be generated by $2$ elements [or is principal].
Neukirch suggests to use that $\mathcal{O}/\mathfrak{a}$ is a PIR, which I already showed in the exercise before. [and I want a solution which use these fact]
What I got so far: Take an arbitrary ideal $\mathfrak{a}$ of $\mathcal{O}$, so $\mathcal{O}/\mathfrak{a}$ is a PIR. This means that for $a \in \mathcal{O}$ $a \neq 0$, $(a)/ \mathfrak{a}$ is a principal ideal of $\mathcal{O}/\mathfrak{a}$, so there has to be an $b \in \mathcal{O}$ s.t. $(a) + \mathfrak{a} = (b)$
But actually now I am stuck. I do not even know, if this leads to a solution. Thank you for any hints.
If $\mathfrak{a}=(0)$ or $(1)$ then $\mathfrak{a}$ is principal, so assume that $\mathfrak{a}\neq (0)$ or $(1)$.
Then there is some $a\in\mathfrak{a}$ which is non-zero and not a unit. By what you have already shown, the quotient $\mathcal{O}/a\mathcal{O}$ is a principal ideal ring, so there is some $b\in \mathcal{O}$ such that the image of $b$ in $\mathcal{O}/a\mathcal{O}$ generates the image of $\mathfrak{a}$ in this ring.
Hence if $x\in\mathfrak{a}$ then $x=ya+zb$ for some $y,z\in\mathcal{O}$, which shows that $a$ and $b$ generate $\mathfrak{a}$.