Bourbaki

Exercise 7, § 1

Lie Groups and Lie Algebras, I, 74

Bourbaki marks this one as harder than the rest.

Let A be a ring (possibly without unit element) such that the relation 2a=02a = 0 in A implies a=0a = 0. Let U be a subring of A and an ideal of the Lie Z\mathbf{Z}-algebra associated with A.

(a) If x,yx, y belong to U, then (xy−yx)A⊂U(xy - yx)A \subset U (write:

(xy−yx)s=x(ys)−(ys)x−y(xs−sx) (xy - yx)s = x(ys) - (ys)x - y(xs - sx)
and A(xy−yx)A⊂UA(xy - yx)A \subset U (write:
r(xy−yx)s=(xy−yx)sr+r[(xy−yx)s]−[(xy−yx)s]r). r(xy - yx)s = (xy - yx)sr + r[(xy - yx)s] - [(xy - yx)s]r).

(b) Suppose that U is commutative. Let x∈U,s∈Ax \in U, s \in A and y=xs−sxy = xs - sx. Show that (yt)3=0(yt)^3 = 0 for all t∈At \in A. (The element x commutes with xs−sxxs - sx and xs2−s2xxs^2 - s^2x; deduce that 2(xs−sx)2=02(xs - sx)^2 = 0, whence y2=0y^2 = 0. Similarly (yt−ty)2=0(yt - ty)^2 = 0 for all t∈At \in A.)

(c) Suppose henceforth that the only two-sided ideals of A are {0} and A and that for every non-zero element yy of A there exists t∈At \in A such that ytyt is not nilpotent. Show that, if U≠AU \neq A, U is contained in the centre of A. (Show, using (a), that U is commutative and then use (b).)

(d) Let V be an ideal of the Lie Z\mathbf{Z}-algebra associated with A. Then either V is contained in the centre of A or V⊃[A,A]V \supset [A, A]. (Apply (c) to the set U of t∈At \in A such that [t,A]⊂V[t, A] \subset V; use the identity [xy,z]=[x,yz]+[y,zx][xy, z] = [x, yz] + [y, zx].)

Solution

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Transcribed from the printed volume. Not translated.