Bourbaki

Proposition 6

Tag 00SU, permanent. Cite it as [Bourbaki, Tag 00SU].

Lie Groups and Lie Algebras, chapter VII, § 1. Primary decomposition of linear representations

Assume that kk is infinite and that condition (AC) is satisfied. Let k′k' be an extension of kk. Put V′=V⊗kk′,S′=S⊗kk′V'= V\otimes_kk',S'= S\otimes_kk'. Let r′:S′→r': S'\rightarrow End(V′') be the map obtained from rr by extension of scalars. Then

V0(S)⊗kk′=V′0(S)=V′0(S′) V^0(S)\otimes_kk'{= V'}^0{(S) = V'}^0(S')

The first equality follows from Prop. 1. To prove the second, we can assume that V=V0(S)V = V^0(S) and so V′=V′0(S)V'{= V'}^0(S). Let (s1,...,sm)(s_1, . . . , s_m) be a basis of S and (e1,...,en)(e_1, . . . , e_n) a basis of V. There exist polynomials Pij(X1,...,Xm)P_{ij}(X_1, . . . ,X_m) such that

r′(a1s1+⋯+amsm)nej=∑i=1nPij(a1,...,am)ei r'(a_1s_1+\cdots +a_ms_m)^ne_j=\sum_{i=1}^nP_{ij}(a_1, . . . , a_m)e_i

for 1≤j≤n1\leq j\leq n and a1,...,am∈k′a_1, . . . , a_m\in k'. By hypothesis, r′(s)n=0r'(s)^n= 0 for all s∈Ss\in S, in other words Pij(a1,...,am)=0P_{ij}(a_1, . . . , a_m) = 0 for 1≤i,j≤n1\leq i, j\leq n and a1,...,am∈ka_1, . . . , a_m\in k. Since kk is infinite, Pij=0P_{ij}= 0. Consequently, every element of r′(S′)r'(S') is nilpotent and V′=V′0(S′)V'{= V'}^0(S').

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Transcribed from the printed volume. Not translated.