How do you prove Schur's lemma?

Answer 1

Schur's theorem:Let (J,+) be an abelian group with more than one element, and let K be a primitive ring with endomorphisms, E. Then the centralizer, C, of K, in the ring ξ(E), which is defined as the set of all endomorphisms of (J,+), is a subdivision of ξ(E). Proof:First off, it needs to be stated that C is a non-zero set because it contains the identity function, I, which obviously fits the definition of a centralizer:

CK(J) = {x Є K: jx = xj for all j Є J}.

Also, using the established theorem that "if J is a subset of ring K, then C(J) is a subring of K, and if an invertible element a of K belongs to C(J), then k-1 Є C(J)," we need only show that if g Є C* (C* being the set of non-zero elements of C), then g is an automorphism of E. Assuming non-triviality, g � 0, and there exists

b Є E such that g(b) � 0. For each h Є E there exists m Є K such that m(g(b)) = y since K is primitive. Thus: g(m(b)) = m(g(b)) = y, showing that g is surjective.

Finally to show g is also injective and thus an automorphism, we take a non-zero element w belonging to the kernel of g. For each z Є E some endomorphism would exist u Є K such that u(w) = z as K is primitive. Therefore:

g(z) = g(u(w)) = u(g(w)) = u(0) = 0, the zero endomorphism which is a contradiction. Hence g is injective and an automorphism of E.

Q.E.D.