= Solution
With the usual commutative, unital conventions, the chain is
$$
\boxed{\text{field}\ \Longrightarrow\ \text{principal ideal domain}
\ \Longrightarrow\ \text{unique factorization domain}\ \Longrightarrow\ \text{integral domain}.}
$$
A <field> is an <integral domain>, and every nonzero <ideal> contains an invertible element and is consequently the whole ring. Its only <ideals> are $(0)$ and $(1)$, so it is a <principal ideal domain>.
To prove the middle implication, first show that a <principal ideal domain> has the <ascending chain condition> on <ideals>. The union of an ascending chain is an <ideal>, hence is $(a)$; its generator belongs to one member of the chain, and all later members equal $(a)$. Now suppose a nonzero nonunit admitted no factorization into <irreducible elements>. It would factor into two nonunits, at least one of which again has no such factorization. Repeating would produce a strictly increasing chain of <principal ideals>: if $a=bc$ with $c$ a nonunit, $(a)\subsetneq(b)$, since equality and cancellation would make $c$ invertible. This contradicts the <ascending chain condition>. Existence of factorization follows. By the permitted fact that every <irreducible element> is a <prime element> in a <principal ideal domain>, any irreducible factor in one factorization divides a factor in the other; cancellation then proves uniqueness up to units and order. This is a <unique factorization domain>. The final implication is part of the definition of a <unique factorization domain>.
None of the adjacent implications reverses. The <principal ideal domain> $\mathbb Z$ is not a <field>, since $2$ is not invertible. The <polynomial ring> $\mathbb Q[x,y]$ is a <unique factorization domain>, by applying the polynomial extension theorem for <unique factorization domains> twice, but its <ideal> $(x,y)$ is not principal: a generator would divide both $x$ and $y$, so would be a unit, whereas every member of this ideal vanishes at the origin. Finally, $\mathbb Z[\sqrt{-5}]$ is an <integral domain> as a subring of $\mathbb C$, but is not a <unique factorization domain>. Its multiplicative <norm> is $N(a+b\sqrt{-5})=a^2+5b^2$. There is no element of norm $2$ or $3$. Hence $2,3,1+\sqrt{-5},1-\sqrt{-5}$, of norms $4,9,6,6$, are all irreducible: a nontrivial factorization would require one of those missing norms. The only units have norm one and are $\pm1$. The two factorizations
$$
6=2\cdot3=(1+\sqrt{-5})(1-\sqrt{-5})
$$
are therefore not related by associates and reordering.
Back to article page
