Exercise 1, § 3

Algebra, X, p. 178

Let $a,b$ be two elements of A, such that the left annihilator of $a$ (resp. $b$ ) is the ideal $Ab$ (resp. $Aa$ ). Prove that the sequence

\cdots\longrightarrow A_s\xrightarrow{\delta_a}A_s\xrightarrow{\delta_b}A_s\xrightarrow{\delta_a}A_s\longrightarrow\cdots\longrightarrow A_s\xrightarrow{\delta_a}A_s\longrightarrow A/Aa\longrightarrow0,

where $\delta_a$ (resp. $\delta_b$ ) denotes right multiplication by $a$ (resp. $b$ ), defines a free resolution of the a-module $A/Aa$ . Prove that this construction applies in the following two cases:

$\alpha)$ Let $A_0$ be a ring; one takes $A=A_0(\varepsilon)$ (X, p. 27), $a=b=\varepsilon$ .

$\beta)$ Let $G$ be a finite cyclic group, $\sigma$ a generator of $G$ , $k$ a commutative ring. One takes

A=k^{(G)},\qquad a=1-e_\sigma,\qquad b=\sum_{g\in G}e_g.

Solution

No solution has been written yet.