Bourbaki

Exercise 2, § 8

Algebra, I, 171

Let AA be a ring. (a) The mapping which associates with a∈Aa \in A the left homothety x↦axx \mapsto ax is an isomorphism of AA onto the ring of endomorphisms of the additive group of AA which commute with right homotheties. (b) Let MM be the set of sequences (a,b,c,d)(a, b, c, d) of four elements of AA which may be represented in the form of a table or matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}. With every element (a\ b\\c\ d) of M we associate the mapping (x,y)↦(ax+by,cx+dy)(x, y) \mapsto (ax + by, cx + dy) of A×AA \times A into itself. Show that a bijection is thus obtained of M onto the ring B of endomorphisms of the commutative group A×AA \times A which commute with the operations (x,y)↦(xr,yr)(x, y) \mapsto (xr, yr) for all r∈Ar \in A. Henceforth M is identified with the ring B. Calculate the product of two matrices.

(c) Show that the subset T of M consisting of the matrices of the form (ab0c)\begin{pmatrix} a & b \\ 0 & c \end{pmatrix} is the subring of M which leaves stable the subgroup A×0A \times 0 of A×AA \times A.

The mappings (ab0c)↦a\begin{pmatrix} a & b \\ 0 & c \end{pmatrix} \mapsto a and (ab0c)↦c\begin{pmatrix} a & b \\ 0 & c \end{pmatrix} \mapsto c are homomorphisms of T into A; let aa and bb be their kernels. Show that a+b=Ta + b = T, that ab={0}ab = \{0\} and that ba=b∩aba = b \cap a is ≠{0}\neq \{0\} if A≠{0}A \neq \{0\} (cf. no. 9, Proposition 6).

(d) If A=Z/2ZA = \mathbf{Z}/2\mathbf{Z}, T is not commutative, but the multiplicative group of invertible elements of T is commutative.

Solution

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Transcribed from the printed volume. Not translated.