**Problem 1**. (Richard Brauer) Let be a minimal left ideal of a ring . Then either or for some non-zero idempotent

**Solution.** Suppose that Then there exists some such that Thus because is a non-zero left ideal of and is a minimal left ideal of Hence there exists such that and so

On the other hand, is obviously a left ideal of which is contained in Since we have and thus, by the minimality of we must have . Therefore , by So is a non-zero idempotent. Now is a left ideal of which is contained in Also and so Therefore by the minimality of

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

**Problem 2**. Let be a semisimple ring and any left (resp. right) ideal of . Then there exists some idempotent such that (resp. ).

**Solution**. We’ll only prove the claim for left ideals of The proof for right ideals is similar. Since is semisimple, there exists a left ideal of such that So for some and Hence and so and , because and and the sum is direct. So is an idempotent. It is clear that . Now if , then and therefore Thus and we’re done.