*Existence of a factorization.* Let \(D\) be a PID and \(a\) be a nonzero element in \(D\) that is not a unit. If \(a\) is irreducible, then we are done. If not, then there exists a factorization \(a = a_1 b_1\text{,}\) where neither \(a_1\) nor \(b_1\) is a unit. Hence, \(\langle a \rangle \subset \langle a_1 \rangle\text{.}\) By Lemma 18.11, we know that \(\langle a \rangle \neq \langle a_1 \rangle\text{;}\) otherwise, \(a\) and \(a_1\) would be associates and \(b_1\) would be a unit, which would contradict our assumption. Now suppose that \(a_1 = a_2 b_2\text{,}\) where neither \(a_2\) nor \(b_2\) is a unit. By the same argument as before, \(\langle a_1 \rangle \subset \langle a_2 \rangle\text{.}\) We can continue with this construction to obtain an ascending chain of ideals

\begin{equation*}
\langle a \rangle \subset \langle a_1 \rangle \subset \langle a_2 \rangle \subset \cdots.
\end{equation*}

By Lemma 18.14, there exists a positive integer \(N\) such that \(\langle a_n \rangle = \langle a_N \rangle\) for all \(n \geq N\text{.}\) Consequently, \(a_N\) must be irreducible. We have now shown that \(a\) is the product of two elements, one of which must be irreducible.

Now suppose that \(a = c_1 p_1\text{,}\) where \(p_1\) is irreducible. If \(c_1\) is not a unit, we can repeat the preceding argument to conclude that \(\langle a \rangle \subset \langle c_1 \rangle\text{.}\) Either \(c_1\) is irreducible or \(c_1 = c_2 p_2\text{,}\) where \(p_2\) is irreducible and \(c_2\) is not a unit. Continuing in this manner, we obtain another chain of ideals

\begin{equation*}
\langle a \rangle \subset \langle c_1 \rangle \subset \langle c_2 \rangle \subset \cdots.
\end{equation*}

This chain must satisfy the ascending chain condition; therefore,

\begin{equation*}
a = p_1 p_2 \cdots p_r
\end{equation*}

for irreducible elements \(p_1, \ldots, p_r\text{.}\)

*Uniqueness of the factorization.* To show uniqueness, let

\begin{equation*}
a = p_1 p_2 \cdots p_r = q_1 q_2 \cdots q_s,
\end{equation*}

where each \(p_i\) and each \(q_i\) is irreducible. Without loss of generality, we can assume that \(r \lt s\text{.}\) Since \(p_1\) divides \(q_1 q_2 \cdots q_s\text{,}\) by Corollary 18.13 it must divide some \(q_i\text{.}\) By rearranging the \(q_i\)'s, we can assume that \(p_1 \mid q_1\text{;}\) hence, \(q_1 = u_1 p_1\) for some unit \(u_1\) in \(D\text{.}\) Therefore,

\begin{equation*}
a = p_1 p_2 \cdots p_r = u_1 p_1 q_2 \cdots q_s
\end{equation*}

or

\begin{equation*}
p_2 \cdots p_r = u_1 q_2 \cdots q_s.
\end{equation*}

Continuing in this manner, we can arrange the \(q_i\)'s such that \(p_2 = q_2, p_3 = q_3, \ldots, p_r = q_r\text{,}\) to obtain

\begin{equation*}
u_1 u_2 \cdots u_r q_{r + 1} \cdots q_s = 1.
\end{equation*}

In this case \(q_{r + 1} \cdots q_s\) is a unit, which contradicts the fact that \(q_{r + 1}, \ldots, q_s\) are irreducibles. Therefore, \(r = s\) and the factorization of \(a\) is unique.