Number Fields - ED implies PID implies UFD

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 $N (r) < N (b)$ , 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 $i \geq n$ , 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 $ℤ [x]$ . 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 $(2, x) \neq (2)$ .