Can every ideal in ring of integers be made principal? Let $K$ be a number field, $\mathfrak{a} \subset O_K$ an integral ideal. Can we find an integral ideal $\mathfrak{b} \subset O_K$ so that $\mathfrak{ab}=(\alpha)$ for some $0\neq \alpha \in O_K$? 
I know that if $K$ is a quadratic field, given $\mathfrak{a} =(c,d)$ then $\mathfrak{b}=(\bar{c},\bar{d})$ is the desired ideal but I wonder whether it is correct or not in a general number field.
 A: The fact you stated about quadratic fields is true, but not obvious. 
What you are saying is that the product of the ideals $(c,d)$ and $(\bar c, \bar d)$ is a principal ideal generated by an element from $\mathbb{Z}$. Is that obvious? Apriori, the product $(c,d) \cdot (\bar c, \bar d)$ equals $(c \bar c, c \bar d, d \bar c, d\bar d)$. It turns out that is also equals $( c \bar c, c \bar d + d \bar c, d \bar d)$, that is apriori smaller. The latter one is indeed generated by elements from $\mathbb{Z}$, so principal.
This equality can be generalized as follows: consider $I=(a_0, \ldots, a_m)$, $J=(b_0, \ldots, b_n)$ ideals in $\mathcal{O}_K$. Consider the polynomial
$$c_0 + a_1 x + \cdots c_{m+n} x^{m+n} = ( a_0 + a_1 x + \cdots a_m x^m)(b_0 + b_1 x + \cdots b_n x^n)$$
Then the ideal $K=(c_0, \ldots, c_{m+n})$ is the product of the ideals $I=(a_0, \ldots, a_m)$ and $J=(b_0, \ldots, b_n)$. For $\mathcal{O}_K= \mathbb{Z}$ this is Gauss' lemma. Apriori, we have the inclusion $K \subset I \cdot J$.  This can be shown as a byproduct of the theory of Dedekind rings, or, like in Hecke's book on algebraic numbers, as a foundational result in some particular case that is relevant to us. We now follow Hecke's exposition.
Given $I = (a_0, \ldots, a_n)$, one can find an "inverse" of $I$ as follows. Consider the polynomial $P(x) = \sum a_k x^k$. Embed $K$ into a Galois extension. Let $P^{(i)}$ all the possible conjugates of $P$, with $P = P^{(0)}$. Then $R=\prod_i P^{(i)}$ is a polynomial with coefficients in $\mathbb{Z}$. Let $Q = \prod_{i \ne 0} P^{(i)}$. It is a polynomial with integral coefficients, and $P \cdot Q \in \mathbb{Z}[x]$, so $Q\in \mathcal{O}_K[x]$. Recall that  $I$ the ideal generated by the coefficients of $P$. Let  $J$ the ideal generated by the coefficients of $Q$. Then $I \cdot J$ is the ideal generated by the coefficients of $R$, hence of form $(d)$, with $d \in \mathbb{Z}$. 
In fact, to find such a $J$, it is enough to find a (non-zero) polynomial $Q$ in $\mathcal{O}_K[x]$ so that $P\cdot Q \in \mathbb{Z}[x]$. 
