Exercise 24, § 3

A ring (commutative or not) $A$, not reduced to 0, is called a principal ideal ring if it has no divisors of zero and every left or right ideal of $A$ is monogenic. Such a ring is left and right Noetherian.

\* Exercises 22 to 25 were communicated to us by P. Gabriel.

(a) Show that, for every element $c \neq 0$ of $A$, $A/Ac$ is an $A$-module of finite length. (Observe that, if a decreasing sequence of ideals $Aa_n$ of $A$ contains $Ac$, then $c = b_n a_n$ for all $n$ and consider the right ideals $b_n A$.)

(b) Show that every submodule of a (left or right) free $A$-module is free (same argument as in Algebra, Chapter VII, § 3, Theorem 1).

(c) In every $A$-module $M$, the set of elements of $M$ which are not free is a submodule $T$ of $M$ called the torsion submodule of $M$ (use Chapter II, §2, Exercise 14 (a)); $M$ is called a torsion module if $T = M$; $M$ is called torsion-free if $T = 0$.

(d) Show that every finitely generated torsion-free $A$-module is free (use Chapter II, § 2, Exercise 23 (b)).

(e) Show that every finitely generated $A$-module is the direct sum of a free module and a torsion module.

(f) Let $a$ be a two-sided ideal $\neq 0$ of $A$. Show that there exists an element $a \in a$ and an automorphism $\sigma$ of $A$ such that $a = Aa = aA$ and $ax = \sigma(x)a$ for all $x \in A$. (If $b$ is a generator of the left ideal $a$, there exists an endomorphism $\tau$ of $A$ such that $bx = \tau(x)b$ for all $x \in A$; show that, if $a = ub$ is a generator of the right ideal $a$, $u$ is invertible, using Algebra, Chapter VIIT, § 2, Exercise 8 (b).)