If $$R=\left\{ \begin{pmatrix} a &b\\ 0 & c \end{pmatrix} \ : \ a \in \mathbb{Z}, \ b,c \in \mathbb{Q}\right\} $$ under usual addition and multiplication, then what are the left and right ideals of $R$?


2 Answers 2


This is covered in full on page 17 of Lam's First course in noncommutative rings. In general the "triangular ring" where $R$ and $S$ are rings and $M$ is an $R-S$ bimodule looks like:

$$T= \begin{pmatrix} R &M\\ 0 & S \end{pmatrix} $$

You can also visualize the ring as $R\oplus M\oplus S$ with funny multiplication, but don't ever confuse this with ordinary direct sums. Lam explains:

1) The right ideals are all of the form $J_1\oplus J_2$, where $J_1$ is a right ideal of $R$ and $J_2$ is a right $S$ submodule of $M\oplus S$ which contains $J_1M$.

2) Analogously the left ideals are all of the form $I_1\oplus I_2$ where $I_2$ is a left ideal of $S$, and $I_1$ is a left $R$ submodule of $R\oplus M$ which contains $MI_2$.

3) The ideals of $T$ look like $K_1\oplus K_0\oplus K_2$ where $K_1$ is an ideal of $R$, $K_2$ is an ideal of $S$, and $K_0$ is a subbimodule of $M$ containing $K_1M+MK_2$.

As a bonus, believe I remember later somewhere he also shows that the radical of this ring is:

$$ rad(T)= \begin{pmatrix} rad(R) &M\\ 0 & rad(S) \end{pmatrix} $$


This is a partial answer that is too long for a comment. I'm not sure about all ideals, but you have at least two big families of left ideals.

Given $q \in \mathbb{N}$, define: $$I_q = \bigg\{ \left(\begin{matrix} 0 & a/q \\ 0 & 0 \end{matrix}\right)~:~ a \in \mathbb{Z}\bigg\}.$$ This is a left ideal, as you can easily check. Moreover, you have $I_r \subseteq I_q$ if and only if $r$ divides $q$. Then there is the union of all these, which is the left ideal $$I_{\mathbb{Q}} = \bigg\{ \left(\begin{matrix} 0 & x \\ 0 & 0 \end{matrix}\right)~:~ x \in \mathbb{Q}\bigg\}.$$ Moreover, given any $n \in \mathbb{N}$, consider (I'm not good with notation, as you can see): $$I^{(n)} = \bigg\{ \left(\begin{matrix} na & 0 \\ 0 & 0 \end{matrix}\right)~:~ a \in \mathbb{Z}\bigg\}.$$ This is another family of left ideals, satifying $I^{(n)} \subseteq I^{(m)}$ if and only if $m$ divides $n$. Again, the union $$I^{(1)} \equiv I^{(\mathbb{N})} = \bigg\{ \left(\begin{matrix} a & 0 \\ 0 & 0 \end{matrix}\right)~:~ a \in \mathbb{Z}\bigg\}$$ is a left ideal. You can also consider combinations of these left ideals to generate other ones. For any pair $(n, q) \in \mathbb{N} \times \mathbb{N}$, you get the left ideal $$I_q^{(n)} = \bigg\{ \left(\begin{matrix} na & b/q \\ 0 & 0 \end{matrix}\right)~:~ a,b \in \mathbb{Z}\bigg\}.$$ Perhaps those are the only ones, but I'm not sure. Hope this helps.

  • $\begingroup$ Edit: for a moment there, I confused left and right.. ;D $\endgroup$
    – student
    Commented Dec 4, 2011 at 16:45
  • $\begingroup$ thanks a lot for the help Leandro $\endgroup$
    – simran
    Commented Dec 8, 2011 at 11:21
  • $\begingroup$ Aren't the matrices with only the bottom-right entry nonzero also a right ideal? $\endgroup$ Commented Jan 3, 2012 at 18:46

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