Bourbaki

Exercise 8, § 9

Algebra, II, 412

Let E be a left affine space over a field K and T its translation space.

(a) For a mapping f of E into a left vector space L over K to be affine, it is necessary and sufficient that

f(t+x)−f(x)=f(t+y)−f(y) f(t + x) - f(x) = f(t + y) - f(y)
f(λt+x)−f(x)=λf(t+x)−f(x)) f(\lambda t + x) - f(x) = \lambda f(t + x) - f(x))

for all λ∈K,t∈T,x,y\lambda \in K, t \in T, x, y in E. Let x↦[x]x \mapsto [x] be the canonical injection of E into the vector space Ks(E)K_s^{(E)} of formal linear combinations of elements of E and let N be the subspace of Ks(E)K_s^{(E)} generated by the elements

[t+x]−[x]−[t+y]+[y] [t + x] - [x] - [t + y] + [y]
[λt+x]−[x]−λ[t+x]+λ[x] [\lambda t + x] - [x] - \lambda[t + x] + \lambda[x]

for λ∈K,t∈T,x,y\lambda \in K, t \in T, x, y in EE; finally, let VV be the quotient space of Ks(E)K_s^{(E)} by NN and ϕ\phi the mapping of EE into VV which associates with every x∈Ex \in E the class of [x][x] modulo NN. Then ϕ\phi is an affine mapping and for every affine mapping ff of EE into a left vector space LL over KK, there exists one and only one linear mapping gg of VV into LL such that f=g∘ϕf = g \circ \phi.

(b) Let ϕ0:T→V\phi_0 : T \to V be the linear mapping associated with ϕ\phi, so that ϕ(t+x)−ϕ(x)=ϕ0(t)\phi(t + x) - \phi(x) = \phi_0(t). Show that ϕ\phi (and therefore also ϕ0\phi_0) is injective (consider the mapping x↦x−ax \mapsto x - a of EE into TT for some a∈Ea \in E); for every family (λi)i∈I(\lambda_i)_{i \in I} of elements of KK with finite support and every family (xi)i∈I(x_i)_{i \in I} of elements of EE,

∑iλiϕ(xi)=ϕ0(∑iλixi)if ∑iλi=0 \sum_i \lambda_i \phi(x_i) = \phi_0 \left( \sum_i \lambda_i x_i \right) \quad \text{if } \sum_i \lambda_i = 0
∑iλiϕ(xi)=μϕ0(∑iμ−1λixi)if ∑iλi=μ≠0. \sum_i \lambda_i \phi(x_i) = \mu \phi_0 \left( \sum_i \mu^{-1} \lambda_i x_i \right) \quad \text{if } \sum_i \lambda_i = \mu \neq 0.

Deduce that ϕ0(T)\phi_0(T) is a hyperplane passing through 0 in VV and ϕ(E)\phi(E) a hyperplane parallel to ϕ0(T)\phi_0(T).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.