Is there a nice way to classify the ideals of the ring of lower triangular matrices?

Suppose $T$ is the subset of $M_2(\mathbb{Z})$ of lower triangular matrices, those of form $\begin{pmatrix} a & 0 \\ b & c\end{pmatrix}$. So $T$ is a subring. Now I know that the ideals of $M_2(\mathbb{Z})$ are all of the form $M_2(I)$ for $I$ and ideal of $\mathbb{Z}$.

However, is there a nice way to describe all the ideals in $T$ specifically, or does it not behave quite as nicely?

• They don't behave quite as nicely, but still we can describe them all (see the link in my answer.) – rschwieb Jun 16 '12 at 12:20

This is a special case of a "triangular ring" construction, and you can find a detailed answer here about its left/right/two-sided ideal lattices.

Adjustments will have to be made if you really want to use lower triangular matrices, but the answer will be similar.

Added: Let's try to interpret this through the help given in that post. Let $T= \begin{pmatrix} R &0\\ M & S \end{pmatrix}$ be your ring, with $R=M=S=\mathbb{Z}$. Under ordinary matrix multiplication, $M$ is an $(S,R)$ bimodule. We may think of this ring as $R\oplus M\oplus S$ with funny multiplication.

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

To see the motivation for the somewhat cryptic conditions given in the other solution, just think: if I have a right ideal and I multiply on the right by $\begin{pmatrix}z&0\\0&0\end{pmatrix}$, what would be included in my ideal? Do the same with a few other sparse matrices and I think you'll see how the conditions work.

So, let us take $12\mathbb{Z}$ to be $J_1$, and pick a $J_2\supseteq 12\mathbb{Z}(\mathbb{Z})=12\mathbb{Z}$. You could pick, for example, $J_2=7\mathbb{Z}\oplus 6\mathbb{Z}\subseteq R\oplus M$. So our candidate ideal is $7\mathbb{Z}\oplus 6\mathbb{Z}\oplus 12\mathbb{Z}\subseteq R\oplus M\oplus S$. Written out properly with matrices it looks like: $$I=\begin{pmatrix} 7\mathbb{Z} &0\\ 6\mathbb{Z} & 12\mathbb{Z} \end{pmatrix}$$

I have to warn you though, that $J_2$ need not be a direct sum of two submodules of $R$ and $M$ like that. You could have $J_2=(0,6\mathbb{Z})+\{(a,a)\mid a\in 7\mathbb{Z}\}=\{(a,a+b)\mid a\in 7\mathbb{Z}, b\in 6\mathbb{Z}\}\subseteq R\oplus M$.

But nevertheless, according to the rules, $$I=\begin{pmatrix} m\mathbb{Z} &0\\ n\mathbb{Z} & t\mathbb{Z} \end{pmatrix}$$ will be a right ideal as long as $n$ divides $t$.

I'll encourage you to try working out the left ideals (but you can summon me again if you get stuck.)

• Thanks rschwieb. I don't have a copy of that book right now, do you mind explaining even briefly how to interpret the third result you linked to for lower triangular matrices? – Camilla Vaernes Jun 16 '12 at 21:05
• @CamillaVaernes Well that post contains pretty much everything you would need from the book. I'll try to expand my answer a little to give some examples. – rschwieb Jun 16 '12 at 21:06
• Are guess my main confusion is, are the ideals of $T$ here just those of form $K_1\oplus K_0\oplus K_2$ where $K_1,K_2$ are ideals of $\mathbb{Z}$, and $K_0$ is a submodule of $\mathbb{Z}$ containing $K_1+K_2$? Or does something change slightly? – Camilla Vaernes Jun 16 '12 at 21:10
• Erroneous thing fixed! Hope it helps. – rschwieb Jun 16 '12 at 22:18
• Thanks. I'm curious about what the entries for $M$ should look like for right ideals. Right multiplying I found $$\begin{pmatrix} r & 0 \\ m & s\end{pmatrix}\begin{pmatrix} z & 0 \\ 0 & 0\end{pmatrix} =\begin{pmatrix} rz & 0 \\ mz & 0\end{pmatrix}$$ $$\begin{pmatrix} r & 0 \\ m & s\end{pmatrix}\begin{pmatrix} 0 & 0 \\ z & 0\end{pmatrix} =\begin{pmatrix} 0 & 0 \\ sz & 0\end{pmatrix}$$ and $$\begin{pmatrix} r & 0 \\ m & s\end{pmatrix}\begin{pmatrix} 0 & 0 \\ 0 & z\end{pmatrix} =\begin{pmatrix} 0 & 0 \\ 0 & sz\end{pmatrix}$$. – Camilla Vaernes Jun 16 '12 at 23:05