Infinite minimal left ideals Let $R$ be an infinite field and let $S$ be the ring
$$S = \begin{pmatrix}
R & R & R\\
0 & R & 0\\
0 & 0 & R
\end{pmatrix}$$
Show that there are infinite minimal left ideals of $S$.
I tried to consider the possible minimal left ideals of $S$. Given a nonzero matrix $A$ in one of these ideals, say $I$, I think I could multiply only once, or maybe twice in the left by some matrices of $S$ to get a matrix $B$ whose nonzero entries are in the same positions as $A$, but $B's$ nonzero entries are simply $1$'s, so that, since $B$ is in the ideal $I$, we can get any matrix with the nonzero entries in the same positions as $A$ but being any elements of $R$. Hence, since there are finitely many possible choices for the positions of the nonzero entries of a matrix $3$ x $3$, it seems to follow that $S$ has finitely many minimal left ideals... contradicting what I'm supposed to prove.
Any help would be appreciated.

It appears that you are misinterpreting the use of the symbol $R$ in the definition of $S$.

Each instance of the symbol $R$ in that definition is intended as an independent, arbitrary choice of an element of $R$.

With that understanding, we can proceed as follows . . .

For each $r\in R$, let $I_r$ be the left ideal of $S$ generated by the matrix
$$
T_r
=
\pmatrix{
0&1&r\\
0&0&0\\
0&0&0\\
}
$$
Then for any $A\in S$ we have
$$
AT_r=
\pmatrix{
0&a&ar\\
0&0&0\\
0&0&0\\
}
$$
where $a=A_{1,1}$.

It follows that

*

*Each of the left ideals $I_r$ is minimal.$\\[4pt]$

*No two of the left ideals $I_r$ are equal.

