Bourbaki

Exercise 13, § 10

Algebra, II, 421

Bourbaki marks this one as harder than the rest.

Let E be a right vector space of dimension ⩾2\geqslant 2 and H0H_0 a hyperplane of E.

(a) Show that every automorphism u belonging to F(E) (Exercise 12) is the product of an automorphism of SL(E)\mathbf{SL}(E) and possibly a dilatation of the hyperplane H0H_0 (proceed by induction on the codimension of the kernel of 1E−u1_E - u, using Exercises 12(a) and 12(f)).

(b) Show that SL(E)\mathbf{SL}(E) contains the commutator group of F(E) and is identical with this group except when E is a space of dimension 2 over the field with 2 elements. (To show that SL(E)\mathbf{SL}(E) contains the commutator group of F(E), use (a) and Exercise 12(f). To see that SL(E)\mathbf{SL}(E) is contained in the commutator group of F(E), except in the case indicated, show that for every homomorphism of F(E) into a commutative group, the image of a transvection is the identity element, using Exercises 11(b) and 12(c).)

(c) Show that SL(E)\mathbf{SL}(E) is equal to its commutator group except when dim E = 2 and K has 2 or 3 elements. (If dim E ⩾3\geqslant 3, note that every transvection u can be written as vvw−1w−1vvw^{-1}w^{-1}, where v is a transvection and w∈SL(E)w \in \mathbf{SL}(E), using Exercise 12(d). If dim E = 2, note first that every automorphism u whose matrix with respect to a basis of E is of the form (λ00λ−1)\begin{pmatrix} \lambda & 0 \\ 0 & \lambda^{-1} \end{pmatrix} is a product of transvections, arguing as in (a); then consider the commutator uvu−1v−1uvu^{-1}v^{-1}, where v is the transvection whose matrix with respect to the same basis is (1101)\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}.)

Solution

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Transcribed from the printed volume. Not translated.