Left ideals generated by idempotents

Posted: June 10, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , , ,

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.

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