Bourbaki

Exercise 9, § 5

Algebra, III, 630

Bourbaki marks this one as harder than the rest.

Let A be a commutative ring, F the free associative algebra over A of the set of integers I={1,n}I = \{1, n\}, where n≥2n \geq 2, and XiX_i (1≤i≤n1 \leq i \leq n) the corresponding indeterminates, so that a canonical basis of F over A consists of the products Xi1Xi2…XirX_{i_1} X_{i_2} \ldots X_{i_r}, where (i1,…,ir)(i_1, \ldots, i_r) is an arbitrary finite sequence of elements of II.

(a) Let C2C_2 denote the set of commutators [Xi,Xj]=XiXj−XjXi[X_i, X_j] = X_i X_j - X_j X_i for 1≤i<j≤n1 \leq i < j \leq n. Assuming that Cm−1C_{m-1} is defined, CmC_m denotes the set of elements [Xi,P]=XiP−PXi[X_i, P] = X_i P - P X_i, where 1≤i≤n1 \leq i \leq n and P runs through Cm−1C_{m-1}. Let U be the graded subalgebra of F generated by 1 and the union of the CmC_m for m≥2m \geq 2. Show that, for 1≤i≤n1 \leq i \leq n, [Xi,P]∈U[X_i, P] \in U for all P∈UP \in U (confine it to the case where P is a product of elements of ⋃mCm\bigcup_m C_m and argue by induction on the number of factors).

(b) Suppose from now on that A is a field. For every integer m≥0m \geq 0, let UmU_m be the vector A-space of homogeneous elements of degree m in U (so that U0=A,U1={0}U_0 = A, U_1 = \{0\}); choose a basis Rm,1,…,Rm,dmR_{m,1}, \ldots, R_{m,d_m} in each UmU_m, consisting of products of elements of ⋃k≤mCk\bigcup_{k \leq m} C_k. For every ordered pair of integers (m,k)(m, k) such that m≥2,1≤k≤dmm \geq 2, 1 \leq k \leq d_m, let Em,kE_{m, k} be the vector sub-A-space of U generated by the Rp,jR_{p, j} such that p>mp > m or p=mp = m and j≥kj \geq k; show that Em,kE_{m, k} is a two-sided ideal of U and that, if Lm,kL_{m, k} is the left ideal of F generated by Em,kE_{m, k}, Lm,kL_{m, k} is a two-sided ideal of F.

(c) For every multiindex α=(α1,…,αn)∈Nn\alpha = (\alpha_1, \ldots, \alpha_n) \in \mathbf{N}^n, we write

Xα=X1α1X2α2…Xnαn. X^\alpha = X_1^{\alpha_1} X_2^{\alpha_2} \ldots X_n^{\alpha_n}.
Show that the elements XαRm,kX^\alpha R_{m, k} (m>0,1≤k≤dmm > 0, 1 \leq k \leq d_m) form a basis of F over K. (To prove that these elements form a generating system of the vector space F, prove that every element Xi1…XirX_{i_1} \ldots X_{i_r} is a linear combination of them by induction on the degree r. To see that the XαRm,kX^\alpha R_{m, k} are linearly independent, argue by reductio ad absurdum by considering the maximum degree with respect to one of the XiX_i of the monomials XαX^\alpha which would appear in a linear relation between the XαRm,kX^\alpha R_{m, k} and proceed by induction on this maximum degree; for this, reduce it to the case where K is infinite (considering if necessary F⊗KK′F \otimes_K K' for an extension field K′K' of K) and observe that when Xi+λX_i + \lambda is substituted in an Rm,kR_{m, k} for XiX_i, for some λ∈K′\lambda \in K', the element Rm,kR_{m, k} is unaltered).

(d) Show that the two-sided ideal J(F)\mathfrak{J}(F) of F generated by the commutators [u,v]=uv−vu[u, v] = uv - vu of elements of F is the sum of the UmU_m for m≥2m \geq 2; F/J(F)F/\mathfrak{J}(F) is isomorphic to A[X1,…,Xn]A[X_1, \ldots, X_n] and hence is an integral domain.

(e) Show that the algebra U is not a free associative algebra by proving that the quotient U/J(U)U/\mathfrak{J}(U), where J(U)\mathfrak{J}(U) is generated by the commutators of elements of U, admits nilpotent elements ≠0\neq 0. (Consider the image z′z' in U/J(U)U/\mathfrak{J}(U) of the element z=[Xi,Xj]z = [X_i, X_j] of U; z′2≠0{z'}^2 \neq 0 but z′4=0{z'}^4 = 0.)

Solution

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Transcribed from the printed volume. Not translated.