Bourbaki

Exercise 3, § 5

Lie Groups and Lie Algebras, II, 194

Bourbaki marks this one as harder than the rest.

We preserve the notation of Exercise 2.

(a) Let J be a right ideal of K[F] contained in I and let u be an element of I. Show that

u∈J.I⇔Dx(u)∈Jfor all x∈X. u \in J . I \Leftrightarrow D_x(u) \in J \quad \text{for all } x \in X.

In particular, an element of I belongs to InI^n (n⩾1n \geqslant 1) if and only if all its partial derivatives belong to In−1I^{n-1}.

† For more details on the DxD_x, see R. Fox, Ann of Math., 57 (1953), pp. 547–560.

(b) Let R be a normal subgroup of F, let G = F/R and let J be the kernel of the canonical homomorphism γ:K[F]→K[G]\gamma : K[F] \to K[G]. Show that J is generated as a left (resp. right) ideal by the elements r−1r - 1, where r∈Rr \in R. Prove the exactness of the sequence

(*)

0⟶J/(J.I)→αI/(J.I)→βK[G]→δK⟶0, 0 \longrightarrow J/(J.I) \xrightarrow{\alpha} I/(J.I) \xrightarrow{\beta} K[G] \xrightarrow{\delta} K \longrightarrow 0,

where α\alpha is derived by taking quotients from the inclusion of J in I and β\beta is derived by taking quotients from the restriction of γ\gamma to I.

(c) For u∈I,x∈Xu \in I, x \in X, let D‾x(u)\overline{D}_x(u) denote the image of Dx(u)D_x(u) in K[G]K[G] under γ\gamma. Show, using (a), that the family (D‾x)x∈X(\overline{D}_x)_{x \in X} defines on taking quotients an isomorphism of I/(J.I)I/(J.I) onto K[G]∞K[G]^{\infty}.

(d) For r∈Rr \in R, let θ(r)\theta(r) be the image of r−1r - 1 in J/(J.I)J/(J.I). Show that

θ(rr′)=θ(r)+θ(r′)andθ(ryr−1)=y.θ(r)for r,r′ in R,y∈F. \theta(rr') = \theta(r) + \theta(r') \quad \text{and} \quad \theta(ryr^{-1}) = y.\theta(r) \quad \text{for } r, r' \text{ in } R, y \in F.

(Use the identities

rr′−1=(r−1)(r′−1)+(r−1)+(r′−1) rr' - 1 = (r - 1)(r' - 1) + (r - 1) + (r' - 1)
yry−1−1=y(r−1)(y−1−1)+y(r−1). yry^{-1} - 1 = y(r - 1)(y^{-1} - 1) + y(r - 1).
)

Deduce that θ\theta defines a homomorphism of R/(R,R)R/(R, R) into J/(J.I)J/(J.I) compatible with the action of G and show that the image of this homomorphism generates the K-module J/(J.I)J/(J.I); when K=ZK = \mathbf{Z}, show that an isomorphism is thus obtained of R/(R,R)R/(R, R) onto J/(J.I)J/(J.I) (define the inverse homomorphism directly).

(e) Let (rα)α∈A(r_\alpha)_{\alpha \in A} be a family of elements of R generating R as a normal subgroup of F. Show that the θ(rα)\theta(r_\alpha) generate the K[G]K[G]-module J/(J.I)J/(J.I).

(f) The matrix (D‾x(rα))x∈X,α∈A(\overline{D}_x(r_\alpha))_{x \in X, \alpha \in A} defines a homomorphism

ρ:K[G](A)→K[G]∞ \rho : K[G]^{(A)} \to K[G]^{\infty}

of left K[G]K[G]-modules. Show that the sequence

(**)

K[G](A)→ρK[G]∞→δK[G]→εK⟶0 K[G]^{(A)} \xrightarrow{\rho} K[G]^{\infty} \xrightarrow{\delta} K[G] \xrightarrow{\varepsilon} K \longrightarrow 0

is exact, where δ\delta is the homomorphism (ux)↦∑x∈Xux(γ(x)−1)(u_x) \mapsto \sum_{x \in X} u_x(\gamma(x) - 1).

(Transform the exact sequence (*) using (c), (d), (e) above.)†

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.