If a right ideal ρ of ring R is minimal then ρ= eR for some idempotent e in R

abstract-algebranoncommutative-algebraring-theory

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 $ρ^2 \neq 0$.

Best Answer

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

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.

is incorrect, even for rings with identity.

It should read

If $\rho$ is a minimal right ideal of a ring $R$, then either $\rho^2=\{0\}$ or $\rho$ is generated by an idempotent (and it additionally follows $eRe$ is a division ring where $eR=\rho$.)

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:

$\impliedby$

$\implies$

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