Bourbaki

Exercise 3, § 1

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

Lie Groups and Lie Algebras, VII, 51

We make the assumptions in no. 2. Assume that kk is infinite and that condition (AC) is satisfied. Let k′k' be a perfect extension of kk. Let λ:S→k\lambda : S\rightarrow k be such that Vλ(S)≠0V^{\lambda}(S)\not= 0. 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 linear map obtained from rr by extension of scalars. There exists a unique map λ′:S′→k′\lambda ': S'\rightarrow k' such that Vλ(S)⊗kk′=V^{\lambda}(S)\otimes_kk'= V′λ′(S′){V'}^{\lambda'}(S'). (Reduce to the case where V=Vλ(S)V = V^{\lambda}(S). Let P be a polynomial function on S and qq a power of the characteristic exponent of kk dividing dim V, such that λq=P\lambda^q= P. Let P′P' be the polynomial function on S′S' which extends P. For each s′∈S′s'\in S', there exists a λ′(s′)∈k′\lambda '(s')\in k' such that λ′(s′)q=P′(s′)\lambda '(s')^q= P'(s'). Show that the characteristic polynomial of r′(s′)r'(s') is (X−λ′(s′))dimV.(X-\lambda '(s'))^{dim V}.)

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