Bourbaki

Proposition 8

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

Lie Groups and Lie Algebras, chapter VIII, § 2. Root system of a split semi-simple Lie algebra

Let (Xα)α∈R(X_{\alpha})_{\alpha\in R} be a Chevalley system for (g,h)(\mathfrak{g},\mathfrak{h}). Let M be a Z\mathbf{Z}-submodule of h\mathfrak{h} containing the HαH_{\alpha} and contained in the group of weights of R∨R^{\vee}. Let gZ\mathfrak{g}_{\mathbf{Z}} be the Z\mathbf{Z}-submodule of g\mathfrak{g} generated by M and the XαX_{\alpha}. Then gZ\mathfrak{g}_{\mathbf{Z}} is a Z\mathbf{Z}-Lie subalgebra of g\mathfrak{g}, and the canonical map from gZ⊗Zk\mathfrak{g}_{\mathbf{Z}}\otimes_{\mathbf{Z}}k to g\mathfrak{g} is an isomorphism.

If α,β∈R\alpha , \beta \in R are such that α+β∈R\alpha +\beta \in R, then Nα,β∈ZN_{\alpha ,\beta}\in \mathbf{Z} (Prop. 7). On the other hand, if α∈R\alpha \in R and h∈Mh\in M, then α(h)∈Z\alpha (h)\in \mathbf{Z} (Chap. VI, §1, no. 9). This proves that gZ\mathfrak{g}_{\mathbf{Z}} is a Z\mathbf{Z}-Lie subalgebra of g\mathfrak{g}. On the other hand, M is a free abelian group of rank dimh(\mathfrak{h}(Algebra, Chap. VII, §3, Th. 1), so gZ\mathfrak{g}_{\mathbf{Z}} is a free abelian group of rank dimg\mathfrak{g}; this implies the last assertion.

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.