Exercise 6, § 1

A commutative ring A is called a principal ideal ring if every ideal of A is principal. a) Show that a quotient ring of a principal ideal ring and a finite product of principal ideal rings are principal ideal rings. b) Show that every principal ideal ring A is a direct product of a finite family $(A_i)$, $1 \leq i \leq m$ of principal ideal rings in each of which the unit element is the only nonzero idempotent (use Ex. 5, b)). c) Suppose that A is a principal ideal ring in which the unit element is the only nonzero idempotent, and which contains zero divisors other than 0. Show that there exists an indivisible nilpotent element p in A. (Using Ex. 5, a), show that there exists an indivisible element p ≠ 0 in A which is a zero divisor ; apply Ex. 5 d), then prove that the principal ideal Ab of y ∈ A such that yp^m ∈ Ap^{m-1} is the whole of A ; finally apply Ex. 5, c)). d) With the hypotheses and notation as in c), suppose that $p^m = 0$ and $p^{m'} \neq 0$. Show that the annihilator of $p^{m'}$ is Ap (otherwise it would be a principal ideal Ac with p ∈ Ac and c ∉ Ap ; using the fact that p is indivisible, deduce that c would be invertible) ; deduce that, for $1 \leq k \leq m$, the annihilator of $p^{m-k}$ is $Ap^k$. Conclude that the ideal Ap is maximal (argue as for the first assertion of d), using the fact that 1 − xp is invertible in A for all x ∈ A). e) Deduce from d) that for all x ∈ A there exists an integer k ≥ 0 and an invertible element u such that x = up^k, and that the only ideals of A are the $Ap^k$ ($0 \leq k \leq m$) (observe that every element of A not belonging to Ap is invertible in A). f) Let A be a principal ideal ring, let $(x_i)$ be a family of elements of A, and let Ad be the ideal generated by the family $(x_i)$ ; show that there exist elements $x'_i$ of A such that $x_i = dx'_i$ for all i, and $\sum A x'_i = A$ (reduce to the case where 1 is the only nonzero idempotent of A). g) Give an example of a principal ideal ring and an indivisible element p of A such that the ideal Ap is not maximal (use Ex. 5, b)).