Given $R$ be commutative ring and $$ T =\begin{pmatrix} R & R \\ 0 & R \end{pmatrix} = \left\{ \left. \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} \right| a,b,c \in R \right\} $$ be a ring of matrix over $R$.

In this paper https://www.researchgate.net/publication/312907984_On_Right_S-Noetherian_Rings_and_S-Noetherian_Modules on Example 1 said that if $J$ is an ideal of $T$, $J$ can be written as $$ \begin{pmatrix} I_1 & I_2 \\ 0 & I_3 \end{pmatrix}$$ where $I_1, I_2$, and $I_3$ are ideals of $R$ satisfying $ I_1 \subseteq I_2$.

I've read Proposition 1.17 in Lam T.Y, but I can't find any relation between the bold statement and those proposition.

Proposition 1.17. The right ideals of $A$ are the form $J_1 \oplus J_2$, where $J_1$ is right ideal in $R$, and $J_2$ is right $S$-submodule of $M \oplus S$ containing $J_1M$. Where $R,S$ be two rings, $M$ be an $(R,S)$-bimodule, and $$ A= \begin{pmatrix} R & M \\ 0 & S \end{pmatrix} = \left\{ \left. \begin{pmatrix} r & m \\ 0 &s \end{pmatrix} \right| r \in R, m \in M, s \in S \right\}.$$

How to find the relation between the bold statement and those Proposition?

  • $\begingroup$ Simply take $S = M = R$ in the Proposition 1.17. More specifically, first take $S = R$ (as a ring) and then take $M = R$ (as an $(R, R)$-bimodule). $\endgroup$ – WhatsUp Dec 27 '19 at 5:15
  • $\begingroup$ Then why $I_1 \subseteq I_2$ ? $\endgroup$ – RANGGAJAYA CIPTAWAN Dec 27 '19 at 5:18
  • $\begingroup$ You should translate the sentence "$J_2$ is right $S$-submodule of $M \oplus S$ containing $J_1M$". It's really straightforward... $\endgroup$ – WhatsUp Dec 27 '19 at 5:20
  • $\begingroup$ So you take $J_1$ is $I_1$ and $J_2$ is $I_2$? $\endgroup$ – RANGGAJAYA CIPTAWAN Dec 27 '19 at 5:25
  • 1
    $\begingroup$ $J_1$ is $I_1$ and $J_2$ is $I_2 \oplus I_3$. $\endgroup$ – WhatsUp Dec 27 '19 at 5:27

As mentioned in the comments, in your case, $R=S=M$ and $I_1=J_1$ and $I_2\oplus I_3=J_2$.

As for the converse, it is not really a trivial corollary that all ideals have this form from the characterization of left ideals that you are citing, but it is an easy extension to make.

Suppose $I$ is an ideal of $A$, and we write $E_{11}=\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}$ and $E_{22}=\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}$. Then with $I_1=E_{11}AE_{11}$, $I_2=E_{11}AE_{22}$ and $I_3=E_{22}AE_{22}$, you can work to show that $I_1\lhd R$, $I_3\lhd S$ and $I_2$ is a sub-bimodule of $M$ containing both $I_1M$ and $MI_3$.

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