By Alexander J. Hahn, O.Timothy O'Meara, J. Dieudonne
It is a brilliant pride for a mathematician to witness the expansion and growth of a thought within which he has taken a few half in the course of its early years. whilst H. Weyl coined the phrases "classical groups", most effective in his brain have been their connections with invariant conception, which his recognized booklet helped to restore. even though his method in that e-book was once intentionally algebraic, his curiosity in those teams without delay derived from his pioneering learn of the distinctive case during which the scalars are actual or advanced numbers, the place for the 1st time he injected Topology into Lie idea. yet ever because the definition of Lie teams, the analogy among uncomplicated classical teams over finite fields and easy classical teams over IR or C have been saw, whether the idea that of "simplicity" was once no longer relatively an analogous in either circumstances. With the invention of the outstanding easy advanced Lie algebras by way of Killing and E. Cartan, it used to be traditional to seem for corresponding teams over finite fields, and already round 1900 this used to be performed by means of Dickson for the outstanding Lie algebras G and E • even though, a deep reason behind this 2 6 parallelism was once lacking, and it is just Chevalley who, in 1955 and 1961, chanced on that to every advanced uncomplicated Lie algebra corresponds, by way of a uniform technique, a gaggle scheme (fj over the hoop Z of integers, from which, for any box okay, will be derived a gaggle (fj(K).
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Example text
R ~ a) with (i,j) entry C ij = (A ij , Bij). It is clear that C satisfies PI C = A and pzC = B. Since the original square of rings is Cartesian, the matrix C is unique with this property. (R) ~ Matn(a). We will therefore denote the matrix C by (A, B). 20. (R) [, GLnlR/a) of linear groups is Cartesian. Proof. (R ~ a) is invertible. Since j(A -1) = (jA) - 1 = (jB) -1 = j(B - 1), there is a unique matrix C' = (A -l,B- 1 ) in Matn(R ~ a) such that PI C' = A-I and pzC' = B- 1 • Observe that PI (CC') = I and pz(CC') = I.
1t remains to show that the Yij(r) also satisfy (E1). This follows from (iii) and (ii) above: Yij(r + s) = [Yim(S + r), Ymj(l)] = [Yim(S)Yim(r), Ymj(l)] = [Yim(S), Yij(r)] [Yim(r), Ymj(l)] [Yim(S), Ymj(l)] = Yij(r)Yij(s). The existence of the homomorphism () is therefore established. That () is unique follows easily: Let ()': St(R) -+ G be another homomorphism such that tjJ()' =
R --+ R' be a ring homomorphism. Let M®RR' be the module obtained by change of rings to R'. For (J in End(M), (J®id R, is in End(M®RR'), and the assignment (J--+(J®id R, defines a ring homomorphism End(M) --+ End(M ® R R'). Assume that M is free with finite basis X = {Xl"'" x n}. Then X' = {Xl ® 1, ... , xn® 1} is a basis of M ®RR', and it is easy to check that the diagram End(M) Mat~ 1 -----+ End(M ® RR') 1Mat~' Matn(R) ~ Matn(R') commutes. Assume that p. is injective. Then so is End(M) --+ End(M ® RR').