Superfluous right ideals of a formal triangular matrix ring. Definition A right ideal $I$ of a ring $R$ is said to be superfluous (or small) if there is no proper right ideal $J$ of $R$ such that $I+J=R$.
I am stuck in finding superfluous right ideals of a formal triangular matrix ring
$\begin{bmatrix}A&M\\0&B\end{bmatrix}$.
Please suggest me any source. I will be highly thankful to you.
 A: For any right ideal $I\subseteq R$, let $I_A\subseteq A$ denote the right ideal formed from the top left entries of elements of $I$.  Similarly let $I_B\subseteq B$ denote the right ideal formed from the bottom right entries of elements of $I$.
Then a right ideal $I\subseteq R$ is superfluous if and only if the ideals $I_A\subseteq A, I_B\subseteq B$ are both superfluous.
Proof:
If $I\subseteq R$ is a superfluous ideal, then we know that $I_A$ is superfluous, as otherwise we would have $J_A+I_A=A$ for some proper ideal $J_A\subset A$.  Then the ideal:
$$J=\left\{\left(\begin{array}{cc}
a&m\\0&b
\end{array}\right)
|a\in J_A, b\in B, m\in M \in M
\right\},
$$
is a proper ideal of $R$, but $I+J=R$, contradicting that $I$ was superfluous.
Thus we know that $I_A$ is superfluous, and the same argument gives that $I_B$ is superfluous.
Conversely suppose that $I_A,I_B$ are superfluous.  We will show that $I$ is superfluous.
If $I+J=R$ then the top left entries of elements of $J$ form an ideal $J_A\subseteq A$, and $I+J_A=A$ so $J'=A$.  Similarly the bottom right entries of elements of $J$ are all of $B$. We conclude that $J=R$.

It may be helpful to note that in the above proof we use that if an ideal $K\subseteq R$ contains elements $$\left(\begin{array}{cc}
1&m\\0&b
\end{array}\right),\qquad\left(\begin{array}{cc}
a&m'\\0&1
\end{array}\right)
$$
then $K=R$.  This follows from the fact that right multiplication by an upper triangular matrix can zero out either column, leaving the other column the same.
A: This can be easily done by noting that the Jacobson radical is the largest superfluous right ideal (and the same can be said as a left ideal.). And I do mean "largest" in the sense that it contains all other superfluous right ideals.
So all you have to do is compute the Jacobson radical and then look at its right submodules. Here is a previous post where I was explaining just that, that the radical is $\begin{bmatrix}J(A)&M\\0&J(B)\end{bmatrix}$.
In a different answer I covered the characterization of right ideals in rings like this. So between the two, you'll be able to work out exactly what the submodules of the Jacobson radical look like, and hence you have all superfluous right ideals.
