Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $R$?

share|improve this question

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:

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

share|improve this answer

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.

share|improve this answer
    
Edit: for a moment there, I confused left and right.. ;D –  student Dec 4 '11 at 16:45
    
thanks a lot for the help Leandro –  simran Dec 8 '11 at 11:21
    
Aren't the matrices with only the bottom-right entry nonzero also a right ideal? –  Steven Stadnicki Jan 3 '12 at 18:46

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.