Exercise 25, § 2

With the notation of Exercise 24, suppose that $A$ is a left neat ring.

(a) Show that the canonical mapping $j_1 : A \to A_{\mathscr{F}}$ is injective, which allows us to identify $A$ with a subgroup of $A_{\mathscr{F}}$. Let $\mathscr{F}'$ be the set of left ideals $l'$ of $A_{\mathscr{F}}$ such that $l'_{\mathscr{F}} = A_{\mathscr{F}}$; show that $\mathscr{F}'$ is the set of left ideals $l'$ of $A_{\mathscr{F}}$ such that $l' \cap m' \neq 0$ for every left ideal $m' \# 0$ of $A_{\mathscr{F}}$. (Show first that, for every left ideal $m' \neq 0$ of $A_{\mathscr{F}}$, $A \cap m' \neq 0$; note on the other hand that the ideals $l' \in \mathscr{F}'$ are precisely the left ideals of $A_{\mathscr{F}}$ containing an ideal $m \in \mathscr{F}$.) Show that the ring $A_{\mathscr{F}}$ is left neat.

(b) Show that $A_{\mathscr{F}}$ is an \emph{injective} $A$-module, that is (\emph{Algebra}, Chapter II, §2, Exercise 11) that, for every left ideal $l$ of $A$ and all

there exists $a' \in A_{\mathscr{F}}$ such that $u(x) = xa'$ for all $x \in 1$. (Use Exercise 24 (b) and Exercise 19 (c).) Deduce that every endomorphism of the A-module $A_{\mathscr{F}}$ is of form $u : x \mapsto xa'$ where $a' \in A_{\mathscr{F}}$ (note that $\operatorname{Hom}_A(A, A_{\mathscr{F}}) = \operatorname{Hom}_{A_{\mathscr{F}}}(A_{\mathscr{F}}, A_{\mathscr{F}})$ by Exercise 19 (e)); also $\operatorname{Ker}(u) = (\operatorname{Ker}(u))_{\mathscr{F}}$.

(c) Show that, for every left ideal 1 of A, $1_{\mathscr{F}}$ is a direct factor of $A_{\mathscr{F}}$. (Let m be a left ideal of A such that $m \cap 1 = 0$ and $m + 1 \in \mathscr{F}$; extend the projection of $m + 1$ onto m, which is zero on 1, to an endomorphism $p$ of $A_{\mathscr{F}}$; show that $p^2 = p$ and that $\operatorname{Ker}(p) = 19$.)

(d) Show that $A_{\mathscr{F}}$ is an absolutely flat ring (Chapter I, § 2, Exercise 17). (Reduce it to the case where $A = A_{\mathscr{F}}$. If $u$ is the endomorphism $x \mapsto xa$ of the left A-module $A_s$, then $\operatorname{Ker}(u) = (\operatorname{Ker}(u))_{\mathscr{F}}$, and $\operatorname{Ker}(u)$ therefore admits a complementary ideal m in A.; note that $\operatorname{Im}(u)$ is isomorphic to m, hence injective (Algebra, Chapter II, § 2, Exercise 11) and therefore a direct factor of A, (loc. cit.). Finally, use Exercise 13 of Algebra, Chapter 11, § 1.)