In the artice [A short proof of the Wedderburn-Artin theorem - Tsiu-Kwen Lee], the author says:
A right ideal $ρ$ of ring $R$ is minimal if and only if $ρ= eR$ for some idempotent $e\in R$ with $eRe$ is a division ring.
(Throughout ring $R$ is simple and associative but not necessarily with unity.)
I have question with the "only if" part.
In this post, prove that the case $ρ^2=ρ$ is done, but in the case $ρ^2=0$ , how to find the idempotent $e\in R$？
Solution:
Consider $Rρ \neq {0}$(since$ρ \neq {0}$) a two side ideal of $R$ ,then $Rρ=R$(by the simplicity),we have $Rρ^2=Rρ=R$ implies $ρ^2 \neq 0$.
I answer inside the context of rings with identity because that is where most (all?) discussions of the Artin-Wedderburn theorem take place.
The claim
is incorrect, even for rings with identity.
It should read
Obviously when $\rho$ is nonzero and $\rho^2=\{0\}$ it is impossible to find an idempotent generator: it would have to satisfy $0\neq e=e^2=0$.
The context of the post you linked is "minimal right ideal of a simple ring". In a simple ring, the case $\rho^2=\{0\}$ doesn't happen except when $\rho=\{0\}$ already. Obviously this also is true more generally for semiprime rings, of which the semisimple rings of the Artin-Wedderburn theorem are examples.
It is also not true, in general that "$eRe$ a division ring implies $eR$ is simple."
As a counterexample, consider $R=\begin{bmatrix}F&F\\ 0&F\end{bmatrix}$ with the idempotent $e=\begin{bmatrix}1&0 \\ 0&0\end{bmatrix}$.
There simply has to be more constraints on $R$ to make these things true.
I suspect the paper you're reading from is only concerned with semiprime rings, and in that case the following existing answers cover both directions:
$\impliedby$
$\implies$