Problem with the ring $R=\begin{bmatrix}\Bbb Z & 0\\ \Bbb Q &\Bbb Q\end{bmatrix}$ and its ideal $D=\begin{bmatrix}0&0\\ \Bbb Q & \Bbb Q\end{bmatrix}$

Question

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.

Answer 1

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?

Answer 2

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 $$ x_1 \begin{pmatrix} 0 & 0 \\ 1 & 1 \end{pmatrix} + x_2 \begin{pmatrix} 0 & 0 \\ 1 & 2 \end{pmatrix}. $$