Exercise 26, § 2
Commutative Algebra, II, 135
Bourbaki marks this one as harder than the rest.
(a) Show that every Zorn ring A (Algebra, Chapter VIII, § 6, Exercise 13) with no nil ideal is left neat; in particular, every absolutely flat ring and every primitive ring whose socle is not reduced to 0 is a left neat ring.
(b) Let A be a left primitive ring whose socle is not reduced to 0; show that, if A is represented as a dense subring of a ring of endomorphisms of a vector space (Algebra, Chapter VIII, § 5, Exercise 10), the ring corresponding to A (Exercise 25) is isomorphic to .
(c) Show that every quasi-simple ring A (Algebra, Chapter VIII, § 5, Exercise 5) is left and right neat and that the corresponding ring is quasi-simple. Deduce that, for every ring B not reduced to 0, there exists a homomorphism to a ring R which is quasi-simple, absolutely flat and such that is an injective R-module.
(d) Let A be a left Noetherian ring with no nilpotent ideals. Show that A is left neat and that the corresponding ring is a semi-simple ring ("Goldie's Theorem"; use Exercise 24 (a) and, on the other hand, use Algebra, Chapter VIII, § 2, Exercise 7, noting that in the absolutely flat ring there cannot exist any family of idempotents which are pairwise orthogonal, unless A contained an infinite direct sum of left ideals ). Show that, if is not a right divisor of 0 in A, it is invertible in and conversely; the set S of these elements is a multiplicative subset of A. Show that is the set of left ideals of A which meet S (to see that the ideals of A meeting S belong to , note that the elements of S are not left divisors of 0 in A; to see that every contains an element of S, represent each simple component of as an endomorphism ring of an n-dimensional vector space E and define by induction n elements such that for and , the being of rank 1; finally show that the kernel of is zero). Deduce that is a ring of left fractions of A with denominators in S (Exercise 22).
Solution
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