Reid, Undergraduate Commutative Algebra, Exercise 0.23 Let $f \in A$; if $f$ is reducible then the principal ideal $(f)$ is contained in a bigger principal ideal $(f_1)$. Consider the following conditions on a ring $A$.
(a) $A$ is a UFD;
(b) every increasing chain $(f_1) \subset (f_2) \subset \cdots \subset (f_n) \subset\cdots$ of principal ideal is eventually stationary, that is $(f_k) = (f_{k+1}) = \cdots$ for some $k$;
(c) any element can be written as a product of irreducible elements.
Prove that (a) $\Rightarrow$ (b), (a) $\Rightarrow$ (c), and that (b) $\Rightarrow$ (c).
 A: (a) $\Rightarrow$ (b): Suppose $A$ is a UFD and suppose, by way of contradiction, that the ascending chain condition on principal ideals does not hold. Let $(f_1) \subsetneq (f_2) \subsetneq \cdots \subsetneq (f_n) \subsetneq \cdots$ be a chain of of principal ideals in $A$ which is not stationary. Suppose, without loss of generality, that $f_1$ is a nonzero nonunit, then by definition of UFD, it can be written as a finite product of irreduicbles, say $f_1 = p_1^{k_1} \cdots p_n^{k_n}$ where $p_i \nsim p_j$ if $i \neq j$ (i.e., they are not associates of each other). 
Notice that $f_{k + 1} \mid f_k$ for each $k$ with $f_k = f_{k + 1} \cdot t$ where $t$ is not a unit. This means $f_{k + 1}$ must have strictly fewer primes than $f_k$. Since $f_1$ has $k_1 + \cdots + k_n$ primes, it follows that the chain can be at most $k_1 + \cdots + k_n$ ideals long, which means it is stationary, a contradiction to our assumption that it wasn't. Conclude that the ascending chain condition is satisfied on principal ideals in a UFD.

(a) $\Rightarrow$ (c): Suppose $A$ is a UFD, then by definition every element can be written as a product of irreducibles (uniquely up to units).

(b) $\Rightarrow$ (c): Suppose every chain $(f_1) \subseteq (f_2) \subseteq \cdots \subseteq (f_n) \subseteq \cdots$ of principal ideals is eventually stationary. 
We want to show that $A$ is a factorization domain, i.e., every nonzero nonunit can be written as a finite product of irreducible elements. 
Suppose, by way of contradiction, that it is not, then there exists $a \in A$ such that $a$ cannot be written as a finite product of irreducible elements. Let $$\Sigma = \{(a) : a \text{ cannot be written as a finite product of irreducible elements}\}.$$ Notice that $\Sigma \neq \emptyset$ by our hypothesis above. Let $\{(a_n)\}$ be an increasing chain in $\Sigma$. By hypothesis, it has an upper bound. By Zorn's lemma, $\Sigma$ has a maximal element, say $(b)$ where $b$ cannot be written as a finite product of irreducibles. In particular, $b$ is not irreducible, so we can write $b = cd$ where neither $c, d$ are units. In particular, we have $b \subsetneq c$ and $b \subsetneq d$ which means $(c), (d) \notin \Sigma$, by maximality of $(b)$, so $c, d$ can both be written as a finite product of irreducibles which means $b$ can be written as a finite product of irreducibles, a contradiction.
Conclude that every element can be written as a finite product of irreducibles.
