# Algèbre: Chapitres 1 à 3 by N. Bourbaki By N. Bourbaki

This can be the softcover reprint of the English translation of 1974 (available from Springer considering 1989) of the 1st three chapters of Bourbaki's 'Algèbre'. It provides a radical exposition of the basics of common, linear and multilinear algebra. the 1st bankruptcy introduces the elemental items: teams, activities, jewelry, fields. the second one bankruptcy experiences the homes of modules and linear maps, in particular with admire to the tensor product and duality structures. The 3rd bankruptcy investigates algebras, particularly tensor algebras. Determinants, norms, lines and derivations also are studied.

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Additional info for Algèbre: Chapitres 1 à 3

Example text

F 0. However, we can make the positive statement given in the following proposition. 7. Cartan's Criterion for Semisimplicity 25 Proposition lAO. In the above notation, if C is nondegenerate, then dim u +dim u1. =dim v. PRooF. Define 1/1 : V ~ (1/f(v), u) U* by = C(v, u) for v e V, u e U. Then ker 1/1 = U 1.. To see that image 1/1 = U*, let U1 be a linear complement for U in V. Let u* be in U*, and define v• e v• by v• = { u• onU 0 onU1• Since C is nondegenerate, qJ is onto qJ(v) = v•. Then v•.

Since acts as nilpotent endomorphisms, the inductive hypothesis shows that V0 is not 0. If vis in V0 , then ~ HX0 v = [H, Xo]v + XoHv = 0 + 0 = 0 since~ is an ideal. Thus X0 {V0 ) ~ V0 • By assumption Xo is nilpotent, and thus 0 is its only eigenvalue. Hence Xo has an eigenvector v0 in V0 • Then X0 (v0 ) = 0 and ~(v0 ) = 0, so that g(v0 ) = 0. Consequently (b) holds for g, and the induction is complete. 38. If g is a Lie algebra such that each ad X for X e g is nilpotent, then g is a nilpotent Lie algebra.

Existence of an algebra structure with [X, X] = 0 is clear, and the problem is to prove the Jacobi identity. Thus let X, Y, z e g. If all three are in a or all three are in b, we are done. By skew symmetry we reduce to two cases: (i) X andY are in a and Z is in b. Then 1r([X, Y]) = 1r(X)1r(Y) - 1r(Y)1r(X). If we apply both sides to Z, we get [[X, Y], Z] = [X, [Y, Z]] - [Y, [X, Z]], which implies the Jacobi identity. (ii) X is in a and Y and Z are in b. Then 1r(X) is a derivation of b, and so 1r(X)[Y, Z] or = [1r(X)Y, Z] + [Y, 1r(X)Z] [X, [Y, Z]] = [[X, Y], Z] + [Y, [X, Z]], which implies the Jacobi identity.