Bourbaki

Proposition 9

Tag 0065, permanent. Cite it as [Bourbaki, Tag 0065].

Algebra, VIII, p. 89, § 5. Commutation

Let L be a field and E a subring of EndZ(L)_{\mathbf{Z}}(L) that contains the mapping γa:x↦ax\boldsymbol{\gamma }_a:x\mapsto ax for every a∈La\in L. Denote by K the set of elements aa of L such that u(xa)=u(x)au(xa) =u(x)a for every xx in L and every uu in E; it is a subfield of L.

Let V be a finite-dimensional linear subspace of the right K-vector space L, and let hh be a K-linear mapping from V to L. There exists an element of E that coincides with hh on V.

We view L as a left E-module. Since E contains the left multiplications γa\boldsymbol{\gamma }_a, every E-submodule of L is a left ideal of the field L; the E-module L is therefore simple. Every endomorphism of the additive group of L that commutes with the γa\boldsymbol{\gamma }_a is of the form δb:x↦xb\boldsymbol{\delta }_b:x\mapsto xb with bb in L. Consequently, b↦δbb\mapsto \boldsymbol{\delta }_b is an isomorphism from K to the opposite ring of EndE_E(L), which is a field.

The bicommutant E′′E'' of the E-module L therefore consists of the endomorphisms of the right K-vector space L. Let vv be an endomorphism of the K-vector space L whose restriction to V coincides with hh; it is an element of E′′E''. Let (xi)i∈I(x_i)_{i\in I} be a basis of V over K; by Theorem 3 (VIII, p. 88), there exists an element uu of E such that u(xi)=v(xi)=h(xi)u(x_i) =v(x_i) =h(x_i) for i∈Ii\in I. By linearity, it follows that u(x)=h(x)u(x) =h(x) for every xx in V.

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 edition, 2023, Springer Nature. Not translated.