ED implies PID implies UFD

Theorem: Every Euclidean domain is a principal ideal domain.

Proof: For any ideal $I$, take a nonzero element of minimal norm $b$. Then $I$ must be generated by $b$, because for any $a \in I$ we have $a = b q + r $ for some $q,r$ with $N(r) \lt N(b)$, and we must have $r = 0$ otherwise $r$ would be a nonzero element of smaller norm than $b$, which is a contradiction.

Fact: If $R$ is a UFD then $R[x]$ is also a UFD.

Theorem: Every principal ideal domain is a unique factorization domain.

Proof: We show it is impossible to find an infinite sequence $a_1, a_2,...$ such that $a_i$ is divisible by $a_{i+1}$ but is not an associate. Once done we can iteratively factor an element as we are guaranteed this process terminates.

Suppose such a sequence exists. Then the $a_i$ generate the sequence of distinct principal ideals $(a_1) \subset (a_2) \subset ...$. The union of these ideals is some principal ideal $(a)$. So $a \in (a_n)$ for some $n$, which implies $(a_i) = (a_n)$ for all $i \ge n$, a contradiction.

Uniqueness: each irreducible $p$ generates a maximal ideal $(p)$ because if $(p) \subset (a) \subset R$ then $p =a b$ for some $b \in R$ implying that $a$ or $b$ is a unit, thus $(a) = (p)$ or $(a) = R$. Thus $R/(p)$ is a field. Next suppose a member of $R$ has two factorizations

\[ p_1 ... p_r = q_1 ... q_s \]

Consider the ideals $(p_i), (q_i)$. Relabel so that $p_1$ generates a minimal ideal amongst these (in other words, $(p_1)$ does not strictly contain another one of the ideals). Now we show $(p_1) = (q_i)$ for some $i$. Suppose not. Then $(p_1)$ does not contain any $q_i$, thus $q_i$ is nonzero modulo $(p_1)$ for all $i$, which is a contradiction because the left-hand side of the above equation is zero modulo $(p_1)$.

Relabel so that $(p_1) = (q_1)$. Then $p_1 = u q_1$ for some unit $u$. Cancelling gives $u p_2 ... p_r = q_2 ... q_s$. The element $u p_2$ is also irreducible, so by induction we have that factorization is unique.∎

The converse of the above theorem is not always true. Consider the ring $\mathbb{Z}[x]$. The ideal $(2,x)$ is not principal: suppose $(2,x) = (a)$ for some $a$. Since this ideal contains the even integers, $a$ must be some integer (multiplication never reduces the degree of an element), and in fact it must be (an associate of) 2. But $(2)$ does not contain polynomials with odd coefficients, so $(2,x) \ne (2)$.