Problem with the ring $R=\begin{bmatrix}\Bbb Z & 0\\ \Bbb Q &\Bbb Q\end{bmatrix}$ and its ideal $D=\begin{bmatrix}0&0\\ \Bbb Q & \Bbb Q\end{bmatrix}$ Let us consider the ring
$
R:=\begin{bmatrix}\Bbb Z & 0\\ \Bbb Q & \Bbb Q\end{bmatrix}
$
and its two-sided ideal
$
D:=\begin{bmatrix}0 & 0\\ \Bbb Q & \Bbb Q\end{bmatrix}
$.
Let then consider the free right $R$-module $F_R:=\bigoplus_{\lambda\in\Lambda}x_{\lambda}R$.
I must show that
$$
\bigcap_{n\ge1}nF_R=\bigoplus_{\lambda\in\Lambda}x_{\lambda}D=F_RD\;\;.
$$
I proved the first equality using the fact that $\bigcap_{n\ge1}nR=D$.
The second inequality: observing that $D\unlhd R$ (i.e. $D$ is a two-sided ideal of $R$) we have that $D=RD$, from which we would have
$$
\bigoplus_{\lambda\in\Lambda}x_{\lambda}D
=\bigoplus_{\lambda\in\Lambda}x_{\lambda}RD
=\underbrace{\left(\bigoplus_{\lambda\in\Lambda}x_{\lambda}R\right)}_{=F_R}D
$$
my problem is with the last equality in this last line: $"\supseteq"$ is obvious. What I cannot prove is the other inclusion $"\subseteq"$.
I try writing the generic element of LHS, say $\sum_{\lambda\in F}x_{\lambda}r_{\lambda}d_{\lambda}=x_1r_1d_1+\cdots+x_nr_nd_n$, for some finite $F\subseteq\Lambda,\;|F|=n$. Then I should find some $d\in D$ and $r_1',\dots,r_n'\in R$ such that $x_1r_1d_1+\cdots+x_nr_nd_n=(x_1r_1'+\cdots+x_nr_n')d$: in such a way this last element would be in
$
{\left(\bigoplus_{\lambda\in\Lambda}x_{\lambda}R\right)}D
$
which is our RHS and I would have finished.
I tried to write out the matrices to find $d$ and the $r_j$'s, even doing some not elegant computations, but I didn't found any way to go out! Can someone help me? Many thanks!
EDIT: see my answer below.
 A: I don't think it is possible to prove that $\bigoplus x_{\lambda} D \subseteq F_R D$, and here's why:
given an element $x_1d_1 + \dotsc + x_n d_n \in \bigoplus x_{\lambda} D$ we need to find $r_i \in R$ and $d \in D$ such that
$$
d_i = r_i d \quad \text{for } 1 \leq i \leq n. \tag{1} \label{eq:1}
$$
Now let's write
$$
d_i =
\begin{pmatrix}
0 & 0 \\
p_i & q_i
\end{pmatrix}
\quad
d =
\begin{pmatrix}
0 & 0 \\
p & q
\end{pmatrix}
\quad
r_i =
\begin{pmatrix}
z_i & 0 \\
x_i & y_i
\end{pmatrix}
$$
and observe that
$$
r_i d =
\begin{pmatrix}
z_i & 0 \\
x_i & y_i
\end{pmatrix}
\begin{pmatrix}
0 & 0 \\
p & q
\end{pmatrix} =
\begin{pmatrix}
0 & 0 \\
y_i p & y_i q
\end{pmatrix}
$$
so $\eqref{eq:1}$ is equivalent to the system of $2n$ equations
$$
\begin{cases}
p_i = y_i p\\
q_i = y_i q
\end{cases}
\quad
\text{for } 1 \leq i \leq n. \tag{2} \label{eq:2}
$$
This implies $\frac{p}{q} = \frac{p_i}{q_i}$ for every $1 \leq i \leq n$, so $\eqref{eq:2}$ cannot have a solution in general, e.g. for
$$
x_1
\begin{pmatrix}
0 & 0 \\
1 & 1
\end{pmatrix}
+ x_2
\begin{pmatrix}
0 & 0 \\
1 & 2
\end{pmatrix}.
$$
A: Ok I found an answer. Please tell me if I am right!
Call
$A:=\begin{bmatrix}0 & 0\\ \Bbb Q & 0\end{bmatrix}$
and
$B:=\begin{bmatrix}0 & 0\\ 0 & \Bbb Q\end{bmatrix}$.
It's clear that $D=A\oplus B$; moreover $A$ and $B$ are both left ideals of $R$, so $RA=A$ and $RB=B$. Then
\begin{align*}
\bigoplus_{\lambda\in\Lambda}x_{\lambda}D
=&\bigoplus_{\lambda\in\Lambda}x_{\lambda}A\;\oplus\;\bigoplus_{\lambda\in\Lambda}x_{\lambda}B\\
=&\bigoplus_{\lambda\in\Lambda}x_{\lambda}RA\;\oplus\;\bigoplus_{\lambda\in\Lambda}x_{\lambda}RB\\
\stackrel{(*)}{=}&\underbrace{\left(\bigoplus_{\lambda\in\Lambda}x_{\lambda}R\right)}_{=F_R}A\;\oplus\;\underbrace{\left(\bigoplus_{\lambda\in\Lambda}x_{\lambda}R\right)}_{=F_R}B\\
=&F_RA\;\oplus\;F_RB\\
=&F_R(A\oplus B)\\
=&F_RD
\end{align*}
as wanted; the only thing left to prove is $(*)$: proving $\bigoplus_{\lambda\in\Lambda}x_{\lambda}RA=\left(\bigoplus_{\lambda\in\Lambda}x_{\lambda}R\right)A$ will be enough.
$"\supseteq"$ is as above in my post. Let's prove then $"\subseteq"$.
Let $x_1r_1a_1+\cdots+x_nr_na_n\in \bigoplus_{\lambda\in\Lambda}x_{\lambda}RA$.
If now we write $r_i=\begin{bmatrix}z_i & 0\\ x_i & y_i\end{bmatrix}\in R$ and
$a_i=\begin{bmatrix}0 & 0\\ q_i & 0\end{bmatrix}\in A$, we can consider
$r_i'=\begin{bmatrix}0 & 0\\ 0 & q_iy_i\end{bmatrix}\in R$ for every $i=1,\dots,n$ and $a=\begin{bmatrix}0 & 0\\ 1 & 0\end{bmatrix}\in A$.
In this way we would have
$$
x_1r_1a_1+\cdots+x_nr_na_n=(x_1r_1'+\cdots+x_nr_n')a\in\left(\bigoplus_{\lambda\in\Lambda}x_{\lambda}R\right)A
$$
Finally the identity $\bigoplus_{\lambda\in\Lambda}x_{\lambda}RB=\left(\bigoplus_{\lambda\in\Lambda}x_{\lambda}R\right)B$ can be proved in a similar way.
This proves $(*)$ and thus the original equality.
I think I'm right. Do you agree?
