Special basis of an ideal as a $\mathbb{Z}$-module in number fields I was speculating that the following may be true (but do not see any easy way to settle it; it must be known, I suppose):
Given a (say, prime) ideal $\mathfrak{p}$ of the ring of integers $\mathcal{O}_K$ of a number field $K$ of degree $d$, do there always exist a basis $a_1, ..., a_d$ of  $\mathcal{O}_K$ as a $\mathbb{Z}$-module and elements $x_1, ..., x_d \in \mathcal{O}_K$ such that $x_1a_1, ..., x_da_d$ were a basis for $\mathfrak{p}$ as a $\mathbb{Z}$-module?
I would appreciate if you could provide a solution or a reference.
 A: Yes. You are right.
It follows from the structure theorem on subgroup of a finitely generated free abelian group, in fact, we can find a base $e_1,\ldots,e_d$ of $\mathfrak{O}_k$ and integers $n_1,\ldots,n_d$ with $n_1|n_2|\ldots$ such that $\mathfrak{p}=n_1\mathbb{Z}e_1+\cdots+n_d\mathbb{Z}e_d$. 
You can find a structrue theorem about modules over principal ideal domain (Theorem 1) in page 21 in the P.Samuel's book "algebraic theory on numbers". While in general case, a subgroup of a finitely generated free abelian is not necessary of full rank. But your case, a non-zero ideal must be of rank $d$, so all here $n_i$ can be choosen as positive integers.
A: Yes. Notice that $\mathfrak p$ is a $\mathbb Z$-submodule of $\mathcal O_K$. We know that $\mathcal O_K$ is a free $\mathbb Z$-module of rank $d$. Since $\mathbb Z$ is principal, we can use the following result (from Samuel's  Algebraic theory of numbers, page 21):
Theorem Let $A$ be a principal ideal ring, $M$ a free $A$-module of rank $n$, and $M'$ a submodule of $M$. Then:


*

*$M'$ is free of rank $q$, $0\leq q \leq n$.

*If $M'\neq 0$, there exists a base $(e_1,\dots,e_n)$ of $M$ and non-zero elements $a_1,\dots,a_q\in A$ such that $(a_1e_1,\dots,a_qe_q)$ is a basis of $M'$ and such that $a_i$ divides $a_{i+1}$, $1\leq i \leq q-1$.


In particular, $\mathfrak p$ has rank $d$, so there exists a basis $(e_1,\dots,e_d)$ of $\mathcal O_K$ and $a_1,\dots,a_d \in \mathbb Z$ such that $(a_1e_1,\dots,a_de_d)$ is a basis $\mathfrak p$.
