Bourbaki

Exercise 15, § 10

Algebra, II, 422

Bourbaki marks this one as harder than the rest.

Let AA be a ring; for every integer n⩾1n \geqslant 1, AnA^n is canonically identified with the submodule of A(N)A^{(\mathbf{N})} consisting of the elements whose coordinates of index ⩾n\geqslant n are 0 and An′A'_n denotes the complementary submodule consisting of the elements whose coordinates of index <n< n are 0. Every endomorphism u∈GLn(A)u \in \mathbf{GL}_n(A) is identified with the automorphism of A(N)A^{(\mathbf{N})} whose restriction to AnA^n is uu and whose restriction to An′A'_n is the identity, so that GLn(A)\mathbf{GL}_n(A) is identified with a subgroup of GL(A(N))\mathbf{GL}(A^{(\mathbf{N})}); let FF denote the subgroup of GL(A(N))\mathbf{GL}(A^{(\mathbf{N})}) the union of the GLn(A)\mathbf{GL}_n(A), in other words the subgroup leaving invariant the elements of the canonical basis of A(N)A^{(\mathbf{N})} except for a finite number of them (cf. Exercise 12).

Let TnT_n be the subgroup of GLn(A)\mathbf{GL}_n(A) generated by the Bij(λ)B_{ij}(\lambda) for 0≤i<n,0≤j<n,i≠j,λ∈A0 \leq i < n, 0 \leq j < n, i \neq j, \lambda \in A (no. 13) and let TT be the subgroup of FF the union of the TnT_n.

(a) Show that for n≥3,Tnn \geq 3, T_n is equal to its commutator group. (b) Let a,ba, b be two invertible elements of AA. Show that

(ab001)=P(a00b)Qand(a00b)=P′(b00a)Q′ \begin{pmatrix} ab & 0 \\ 0 & 1 \end{pmatrix} = P \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} Q \quad \text{and} \quad \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} = P' \begin{pmatrix} b & 0 \\ 0 & a \end{pmatrix} Q'
where P,Q,P′,Q′P, Q, P', Q' belong to T2T_2. Deduce that the matrix (aba−1b−1001)\begin{pmatrix} aba^{-1}b^{-1} & 0 \\ 0 & 1 \end{pmatrix} belongs to T2T_2.

(c) Deduce from (b) that the commutator group of GLn(A)\mathbf{GL}_n(A) is contained in T2nT_{2n} (replace AA by Mn(A)\mathbf{M}_n(A) in (b)). Conclude (using (a)) that the commutator group of FF is equal to TT.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.