## Left ideals generated by idempotents

Posted: June 10, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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Problem 1. (Richard Brauer) Let $I$ be a minimal left ideal of a ring $R$. Then either $I^2= \{0\}$ or $I=Re,$ for some non-zero idempotent $e \in I.$

Solution. Suppose that $I^2 \neq \{0\}.$ Then there exists some $x \in I$ such that $Ix \neq \{0\}.$ Thus $Ix=I,$ because $Ix \subseteq I$ is a non-zero left ideal of $R$ and $I$ is a minimal left ideal of $R.$ Hence there exists $0 \neq e \in I$ such that $ex=x$ and so

$(e^2-e)x=0. \ \ \ \ (1)$

On the other hand, $J= \{r \in I: \ rx = 0 \}$ is obviously a left ideal of $R$ which is contained in $I.$ Since $Ix \neq \{0\},$ we have $J \neq I$ and thus, by the minimality of $I,$ we must have $J=\{0\}$. Therefore $e^2-e=0$, by $(1).$ So $e \in I$ is a non-zero idempotent. Now $Re$ is a left ideal of $R$ which is contained in $I.$ Also $0 \neq e=e^2 \in Re$ and so $Re \neq \{0\}.$ Therefore $Re=I,$ by the minimality of $I. \ \Box$

Note that we did not need $R$ to have $1.$ Also, a similar result holds for minimal right ideals of $R$, i.e. if $I$ is a minimal right ideal of $R,$ then either $I^2 = \{0\}$ or $I=eR,$ for some non-zero idempotent $e \in I.$  If $R$ is a semisimple ring (with $1$), for example full matrix rings over division algebras, then every left (resp. right) ideal of $R$ is generated by some idempotent, as the next problem shows.

Problem 2. Let $R$ be a semisimple ring and $I$ any left (resp. right) ideal of $R$. Then there exists some idempotent $e \in I$ such that $I=Re$ (resp. $I=eR$).

Solution. We’ll only prove the claim for left ideals of $R.$ The proof for right ideals is similar. Since $R$ is semisimple, there exists a left ideal $J$ of $R$ such that $R=I \oplus J.$ So $1=e+f,$ for some $e \in I$ and $f \in J.$ Hence $e=e^2+ef$ and so $e^2=e$ and $ef=0$, because $e, e^2 \in I$ and $ef \in J$ and the sum is direct. So $e$ is an idempotent. It is clear that $Re \subseteq I$. Now if $x \in I$, then $x=xe+xf$ and therefore $x=xe.$ Thus $I \subseteq Re$ and we’re done.