= Solution
Take the <integral domain> $R=\mathbb Z[\sqrt{-5}]$, embedded in $\mathbb C$. It exhibits <failure of unique factorization in the ring of integers generated by square root of minus five>:
$$
\boxed{6=2\cdot3=(1+\sqrt{-5})(1-\sqrt{-5}).}
$$
Here is why these are genuinely distinct factorizations into <irreducible elements>. The multiplicative <algebraic norm> $N(a+b\sqrt{-5})=a^2+5b^2$ is a nonnegative integer and equals one only for $\pm1$, which are therefore the only <units>. Neither two nor three is a value of $N$: if $b\ne0$ the value is at least five, and if $b=0$ it is a square. A factorization of $2$ into nonunits would require norms two and two; one of $3$ would require norms three and three; one of $1\pm\sqrt{-5}$ would require norms two and three. All are impossible. Thus all four factors are irreducible. The factors of norms four and nine cannot be <associate elements> of those of norm six, and $1+\sqrt{-5}$ and $1-\sqrt{-5}$ are not related by a sign. Therefore $R$ is \b[an <integral domain> which is not a <unique factorization domain>].
Back to article page
