Basis for an ideal in $\Bbb{Q}(\sqrt[3]{m})$ Let $m$ be a cube-free integer, so that $K=\Bbb{Q}(\sqrt[3]{m})$ is a cubic extension of the rationals. Then the ring of integers in $K$ is a free $\Bbb{Z}$-module of rank $3$, with a basis of the form $\{1,\alpha_1,\alpha_2\}$, and I know how to find $\alpha_1$ and $\alpha_2$ given a specific value of $m$. An ideal in this ring is a $\Bbb{Z}$-submodule of full rank, and I'm trying to say something about its generators.
I'm fairly certain we can say that the ideal has a basis where one element is a rational integer, namely the smallest rational integer in the ideal. I want to extend this, and say that there will always be a basis of the form $\{a, b+c\alpha_1, d+e\alpha_1+f\alpha_2\}$. I suspect this is true, but I don't really know how to show it.
Is this true for $\Bbb{Z}$-modules in general, that given a rank $3$ submodule of a rank $3$ module, we can find generators for it so that the first comes from a specified rank $1$ submodule, and the second comes from a specified rank $2$ submodule?
I hope this question makes sense.
 A: Indeed, it is true over any principal ideal domain (such as $\mathbb Z$). See http://en.wikipedia.org/wiki/Elementary_divisors
Let me explain how this works in your example.
You have two free rank-$3$ ${\mathbb Z}$-modules $M \subseteq M'$,
and $(m_1,m_2,m_3)$ is a ${\mathbb Z}$-basis of $M'$. You want
a basis of $M$ of the form 
$(c_1m_1,c_2m_1+c_3m_2,c_4m_1+c_5m_2+c_6m_3)$, where the $c_i$ 
($1 \leq i \leq 6$) are integers.
Here is how the $c_i$ can be constructed : let 
$$
\begin{array}{lcl}
I_1 &=& \lbrace p\in {\mathbb Z} \ | \ pm_1\in M\rbrace \\
I_2 &=& \lbrace p\in {\mathbb Z} \ | \ pm_2\in M+{\mathbb Z}m_1\rbrace \\
I_3 &=& \lbrace p\in {\mathbb Z} \ | \ pm_3\in M+
{\mathbb Z}m_1+{\mathbb Z}m_2\rbrace \\
\end{array}
$$
It is easy to check that these are all ideals of $\mathbb Z$, so there
are integer constants $c_1,c_3,c_6$ such that 
$I_1=c_1{\mathbb Z},I_2=c_3{\mathbb Z},I_1=c_6{\mathbb Z}$.
By definition of $I_1$, the vector $v_1=c_1m_1$ is in $M$.
By definition of $I_2$, there is a constant $c_2\in{\mathbb Z}$ such 
that $v_2=c_2m_1+c_3m_2\in M$. 
By definition of $I_3$, there are constants $c_4,c_5\in{\mathbb Z}$ such 
that $v_3=c_4m_1+c_5m_2+c_6m_3\in M$. 
Now, consider the $\mathbb Z$-module
$$
N={\mathbb Z}v_1+{\mathbb Z}v_2+
{\mathbb Z}v_3
$$
By construction we have $N\subseteq M$. Conversely, let $m\in M$.
Since $M\subseteq M'$ we have $x_1,x_2,x_3\in\mathbb Z$ such that
$m=x_1m_1+x_2m_2+x_3m_3$. 
By definition of $I_3$, there is a $q_3\in\mathbb Z$ such that
$x_3=q_3c_6$. Then the element 
$$m'=m-q_3v_3=(x_1-q_3c_4)m_1+(x_2-q_3c_5)m_2$$
is in $M$. By definition of $I_2$, there is a $q_2\in\mathbb Z$ such that
$x_2-q_3c_5=q_2c_3$.  Then the element 
$$m''=m'-q_2v_2=(x_1-q_3c_4-q_2c_2)m_1$$
is in $M$. Finally, by definition of $I_1$ there is a $q_1\in\mathbb Z$ such that
$x_1-q_3c_4-q_2c_2=q_1c_1$, so that $m=q_1v_1+q_2v_2+q_3v_3\in N$.
We have shown that $M=N$ and therefore finished the proof.
