Ideal in the ring of upper triangular matrices I'm trying to describe an ideal of the ring $R=\left\{ \begin{pmatrix}a & b\\
0 & c
\end{pmatrix}:a,b,c \in \mathbb{R}\right\} $
It's easy to prove that $I=\left\{ \begin{pmatrix}0 & a\\
0 & 0
\end{pmatrix}:a\in\mathbb{R}\right\} $ and $J=\left\{ \begin{pmatrix}a & b\\
0 & 0
\end{pmatrix}:a,b\in\mathbb{R}\right\} $ are ideals of $R$
My question is: how can I find other ideals?
Any help would be appreciated.
Thanks.
 A: I assume that by "ideal" you mean "two sided ideal".
Note that you can independently scale the two columns or the two rows of a matrix by multiplying on one side or the other by a diagonal matrix.  So in a one dimensional ideal the matrices can only have one nonzero entry.  There are three locations for this entry and you can check that of the three, only
$$\begin{pmatrix} 0 & \ast \\ 0 & 0 \end{pmatrix}$$
gives an ideal, so this is the only possible $1$ dimensional ideal.  The ring is three dimensional so all that's left are the two dimensional ideals.
A matrix in $R$ is a unit if and only if it has non-zero entries on the diagonal and we're looking for a proper ideal so all our matrices must have a zero on the diagonal.  The ideal must also contain a matrix which has a nonzero on the diagonal (otherwise the ideal is the one dimensional ideal above).  We can scale this nonzero entry to $1$ so our ideal must contain a matrix of the form
$$\begin{pmatrix} 1 & \ast \\ 0 & 0 \end{pmatrix} \qquad \text{or} \qquad \begin{pmatrix} 0 & \ast \\ 0 & 1 \end{pmatrix}$$
Note it cannot contain one of each because adding them would give a unit.  So all the matrices in the ideal must be of the form
$$\begin{pmatrix} \ast & \ast \\ 0 & 0 \end{pmatrix} \qquad \text{or} \qquad \begin{pmatrix} 0 & \ast \\ 0 & \ast \end{pmatrix}$$
Now you just have to check that those actually give ideals and then you'll have shown that there are exactly three proper non-trivial ideals.
A: $\begin{pmatrix} a & b\\ 0 & c \end{pmatrix}
\begin{pmatrix} x & y\\ 0 & z \end{pmatrix} =
\begin{pmatrix} ax & ay+bz\\ 0 & cz \end{pmatrix}$.
If $a\neq 0$, you can vary $ax$ and $ay$ over $\mathbb{R}$, by varying $x$ and $y$. Similarly, if $b\neq 0$, you can vary $bz$ over $\mathbb{R}$, and if $c\neq 0$, you can vary $cz$ over $R$ by varying $z$. Thus, the different right ideals of $R$ are:
$\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & \ast \end{pmatrix}, \begin{pmatrix} 0 & \ast \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & \ast \\ 0 & \ast \end{pmatrix}, \begin{pmatrix} \ast & \ast \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} \ast & \ast \\ 0 & \ast \end{pmatrix}$
